Gravitation

Class 12 Physics, unit 15, book pages 7–25

Gravitation is the force of attraction between any two masses. This unit uses Newton's law of gravitation to describe the gravitational field of the Earth and other bodies: how strong it is at the surface and at any height, and why it is nearly constant near the ground. It then applies these ideas to satellites (orbital velocity, period and geostationary orbits) and introduces gravitational potential and potential energy, the tools needed when gravity cannot be treated as constant.

Learning outcomes (9)

Define and calculate gravitational field strength, including more challenging problems than in Grade 9, using $g = \frac{GM}{r ^{2}}$ .
Analyse gravitational fields by means of field lines, knowing that for a point outside a uniform sphere the mass of the sphere may be considered to be a point mass at its centre.
Apply Newton's law of gravitation, $F = G \frac{m _{1} m _{2}}{r ^{2}}$ for the force between two point masses, to solve problems.
Analyse circular orbits in gravitational fields by relating the gravitational force to the centripetal acceleration it causes.
Analyse the motion of geostationary satellites, knowing that a geostationary orbit remains at the same point above the Earth's surface, with an orbital period of 24 hours, orbiting from west to east, directly above the equator.
Derive the equation $g = \frac{GM}{r ^{2}}$ for the gravitational field strength due to a point mass, from Newton's law of gravitation and the definition of gravitational field.
Analyse why $g$ is approximately constant for small changes in height near the Earth's surface.
Define and calculate gravitational potential at a point as the work done per unit mass in bringing a small test mass from infinity to the point, using $ϕ = - \frac{GM}{r}$ for the potential in the field of a point mass.
Justify how the concept of gravitational potential leads to the gravitational potential energy of two point masses, and use $E_{p} = - G \frac{M m}{r}$ in problems.

15.1Newton's law of universal gravitationOften examined

Gravity is the force that keeps us on the ground, holds the Moon in its path around the Earth and keeps the planets moving around the Sun. It also shapes the universe on a large scale. Our galaxy, the Milky Way, is a huge disk of stars, dust and gas held together by gravity. The Sun lies near its edge, about $26000$ light-years from the centre, and the nearest large galaxy, Andromeda, is about $2.5$ million light-years away.

Newton realised that the force that pulls a falling apple to the ground is the same force that keeps the Moon in its orbit. He described it with one law that works for every pair of objects in the universe.

Force of gravity (gravitational force): The force of attraction that any two objects exert on each other because of their masses.
Newton's law of universal gravitation: Every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Derivation Newton's law in equation form

Aim. Write the law of gravitation as an equation for two bodies of masses $m_{1}$ and $m_{2}$ whose centres are a distance $r$ apart.

Assumptions.

The bodies are point masses or uniform spheres, so $r$ is measured from centre to centre.
No other masses act on them.

Figures 15.1 and 15.3: two bodies attract each other with forces of equal size and opposite direction. The distance

r

is measured between the centres.

The force is directly proportional to the product of the masses: doubling either mass doubles the force.
$F_{g} \propto m_{1} m_{2} (i)$
The force is inversely proportional to the square of the distance between the centres.
$F_{g} \propto \frac{1}{r ^{2}} (ii)$
Combine (i) and (ii) into a single proportionality.
$F_{g} \propto \frac{m _{1} m _{2}}{r ^{2}}$
Replace the proportionality sign by an equals sign and a constant of proportionality $G$ , the gravitational constant.
$F_{g} = G \frac{m _{1} m _{2}}{r ^{2}}$

Result

F_{g} = G \frac{m _{1} m _{2}}{r ^{2}}, G = 6.67 \times 1 0^{- 11} N m^{2} k g^{- 2}

The force grows with the masses and falls off quickly with distance. $G$ is so small that the force is noticeable only when at least one of the masses is huge, like a planet or a star.

The gravitational constant G

$G$ has the same value everywhere in the universe and for all kinds of matter, so it is also called the universal gravitational constant. Put $m_{1} = m_{2} = 1 kg$ and $r = 1 m$ in the law and you get $F_{g} = G$ . So $G$ is numerically equal to the force of attraction between two $1 kg$ masses placed $1 m$ apart: only $6.67 \times 1 0^{- 11} N$ .

G = \frac{F _{g} r ^{2}}{m _{1} m _{2}} \Rightarrow unit of G = \frac{N m ^{2}}{k g ^{2}} = N m^{2} k g^{- 2} = m^{3} k g^{- 1} s^{- 2}

The last form uses $1 N = 1 kg m s^{- 2}$ .

Features of the gravitational force

Always attractive. Masses pull each other; gravity never pushes. (Electric and magnetic forces can attract or repel.)
Independent of the medium. The force is the same whether air, water or vacuum lies between the masses.
Acts along the line joining the centres of the two bodies.
Inverse-square law. When the distance doubles the force falls to $\frac{1}{4}$ ; when it triples the force falls to $\frac{1}{9}$ (Figure 15.2).
Very weak between everyday objects, because $G$ is tiny. It becomes important only when a very large mass is involved.
Long range. It never becomes exactly zero, however far apart the bodies are, and it needs no contact.
Mutual. The forces on the two bodies are equal in size and opposite in direction: an action and reaction pair.

Gravitational force against distance

r

2 r

3 r

F

\frac{1}{4} F

\frac{1}{9} F

d

F_{g}

F

\frac{1}{4} F

\frac{1}{9} F

F_{g} \propto \frac{1}{r ^{2}}

Figure 15.2:

d

is the distance between the centres. The force is

F

d = r

, falls to

\frac{1}{4} F

2 r

and to

\frac{1}{9} F

3 r

The forces form an action and reaction pair

Let $F_{21}$ be the force on $m_{2}$ caused by $m_{1}$ , and $F_{12}$ the force on $m_{1}$ caused by $m_{2}$ (Figure 15.3). The law gives the size of each force with the same expression $G m_{1} m_{2} / r^{2}$ , so the two forces are equal in magnitude. $F_{21}$ points from $m_{2}$ towards $m_{1}$ and $F_{12}$ points from $m_{1}$ towards $m_{2}$ , so they are opposite in direction.

F_{12} = - F_{21}, F_{12} = F_{21} = G \frac{m _{1} m _{2}}{r ^{2}}

Two forces that are equal in size, opposite in direction, act on different bodies and come from the same interaction form an action and reaction pair. So Newton's law of gravitation is consistent with Newton's third law of motion.

Example 15.1

Two bodies A and B are placed at a distance of

1 m

from each other. What will be the force of attraction between them if their masses are

45 kg

and

50 kg

respectively?

Given

$r = 1 m$
$m_{1} = 45 kg$
$m_{2} = 50 kg$
$G = 6.67 \times 1 0^{- 11} N m^{2} k g^{- 2}$

To find

$F_{g}$

Use Newton's law of gravitation.
$F_{g} = G \frac{m _{1} m _{2}}{r ^{2}}$
Substitute the values in SI units.
$F_{g} = \frac{6.67 \times 1 0 ^{- 11} \times 45 \times 50}{( 1 ) ^{2}}$
Evaluate.
$F_{g} = 1.5 \times 1 0^{- 7} N$

The force is so small that it goes unnoticed unless very sensitive instruments are used. Friction on the bodies is far larger, so they do not move towards each other.

F_{g} = 1.5 \times 1 0^{- 7} N

Assignment 15.1 (1), solved

The mass of Earth is

6 \times 1 0^{24} kg

and that of the moon is

7.4 \times 1 0^{22} kg

, with a distance of

3.84 \times 1 0^{5} km

between them. Calculate the force exerted by the Earth on the moon.

Given

$M_{E} = 6 \times 1 0^{24} kg$
$m = 7.4 \times 1 0^{22} kg$ mass of the Moon
$r = 3.84 \times 1 0^{5} km = 3.84 \times 1 0^{8} m$

To find

$F_{g}$

Convert the distance to metres first, then use the law.
$F_{g} = G \frac{M _{E} m}{r ^{2}} = \frac{6.67 \times 1 0 ^{- 11} \times 6 \times 1 0 ^{24} \times 7.4 \times 1 0 ^{22}}{( 3.84 \times 1 0 ^{8} ) ^{2}}$
Evaluate the numerator and the denominator.
$F_{g} = \frac{2.96 \times 1 0 ^{37}}{1.47 \times 1 0 ^{17}} = 2.0 \times 1 0^{20} N$

By Newton's third law the Moon pulls the Earth with a force of the same size. This force provides the centripetal force that keeps the Moon in its orbit.

F_{g} \approx 2.0 \times 1 0^{20} N

Assignment 15.1 (2), solved

Why do we say that the law of gravitation is a universal law?

The law applies to every pair of objects everywhere in the universe, with the same constant $G$ . The same force that makes an apple fall to the ground keeps the Moon moving around the Earth, keeps the planets moving around the Sun, holds the stars of a galaxy together and makes galaxies attract each other.

It does not depend on the kind of matter, on the medium between the bodies, or on where the bodies are. Because one law with one constant describes gravity for all bodies, on Earth and in space, it is called a universal law.

15.2Gravitational field and field strengthOften examined

A mass does not need to touch another mass to pull it. We say that every mass sets up a gravitational field in the space around it, and any other mass placed in this field feels a force. The Earth's field reaches the Moon; the Sun's field reaches every planet.

Gravitational field: The region (space) around a massive object, such as the Earth or the Sun, in which its gravitational force acts on other masses.
Gravitational field strength ( $g$ ): The gravitational force acting on a unit mass placed at a point in the field, $g = F_{g} / m$ . It is a vector that points in the direction of the gravitational force. SI unit: $N k g^{- 1}$ , which is the same as $m s^{- 2}$ .
Test mass: A small mass placed at a point to measure the field there. It is small so that it does not disturb the field it measures.
Point mass: A body treated as if all its mass were concentrated at a single point in space.

Field lines

A gravitational field is shown by field lines. The arrow on a line shows the direction of the force on a mass placed there. The spacing of the lines shows the strength: closer lines mean a stronger field, lines farther apart mean a weaker field. For a point mass the lines point radially inward from all directions, forming a symmetric pattern (Figure 15.5, in 'Gravitational field lines').

Field strength equals the acceleration due to gravity

The field acts between a test mass and the source mass. By Newton's second law, $a = F / m$ . If gravity is the only force on the test mass $m$ , its acceleration is

a_{g} = \frac{F _{g}}{m} = g

So the gravitational field strength at a point is equal to the free-fall acceleration there, and the two units are the same:

1 N k g^{- 1} = 1 \frac{kg m s ^{- 2}}{kg} = 1 m s^{- 2}

Derivation Field strength due to a point mass

Aim. Derive $g = \frac{GM}{r ^{2}}$ for the field strength at a distance $r$ from a point mass $M$ .

Assumptions.

$M$ is a point mass, or a uniform sphere with $r$ measured from its centre.
The test mass $m$ is small, so it does not disturb $M$ or its field.

A test mass

m

at P, a distance

r

from the point mass

M

. The force on it, and so the field

g

, points towards

M

Place a small test mass $m$ at point P, a distance $r$ from $M$ . Newton's law of gravitation gives the force on it.
$F_{g} = \frac{GM m}{r ^{2}}$
By definition, field strength is the force per unit mass.
$g = \frac{F _{g}}{m}$
Substitute the force. The test mass $m$ cancels.
$g = \frac{GM m / r ^{2}}{m} = \frac{GM}{r ^{2}}$
Direction: the force is attractive, so $g$ at P points from P towards $M$ . With $\overset{r}{^}$ a unit vector pointing away from $M$ , the vector form is
$g = - \frac{GM}{r ^{2}} \overset{r}{^}$

Result

g = \frac{GM}{r ^{2}}

The field strength depends only on the mass $M$ that causes the field and the distance $r$ from it, not on the test mass. Doubling $M$ doubles $g$ ; doubling $r$ reduces $g$ to one quarter.

Do not confuse $g$ with $G$ .
	$g$ (field strength)	$G$ (gravitational constant)
Meaning	force per unit mass (free-fall acceleration) at a point	constant in Newton's law of gravitation
Value	about $9.8 m s^{- 2}$ at the Earth's surface; changes from place to place	$6.67 \times 1 0^{- 11}$ everywhere in the universe
Unit	$N k g^{- 1}$ or $m s^{- 2}$	$N m^{2} k g^{- 2}$
Nature	vector	scalar constant
Depends on	mass causing the field and distance from its centre	nothing: it is universal

15.2Value of g on the surface of the EarthOften examined

At the Earth's surface every kilogram is pulled with a force of about 9.8\,\mathrm{N}. Newton's law shows where this number comes from: it depends only on the Earth's mass and radius.

Derivation Value of g on the surface of the Earth

Aim. Derive g=\dfrac{GM_E}{R_E^{2}} and find its value.

Assumptions.

The Earth is a uniform sphere of mass M_E and radius R_E, so it attracts as if its mass were at its centre.
The stone is tiny compared with the Earth, so the distance between their centres is R_E.

A stone of mass m_0 on the surface is a distance R_E from the Earth's centre O. The Earth's pull on it is its weight, F_g=W=m_0g.

Newton's law of gravitation gives the force between the stone and the Earth, whose centres are R_E apart.
F_g=G\,\dfrac{m_0M_E}{R_E^{2}}\qquad(15.1)
This force is the weight of the stone. By Newton's second law, weight equals mass times the acceleration due to gravity.
F_g=W=m_0g\qquad(15.2)
Both equations describe the same force, so equate them.
m_0g=G\,\dfrac{m_0M_E}{R_E^{2}}
Cancel m_0 from both sides.
g=G\,\dfrac{M_E}{R_E^{2}}\qquad(15.3)
Put in G=6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}, M_E=6\times10^{24}\,\mathrm{kg} and R_E=6.4\times10^{6}\,\mathrm{m}.
g=\dfrac{6.67\times10^{-11}\times6\times10^{24}}{(6.4\times10^{6})^{2}}=9.77\,\mathrm{m\,s^{-2}}\approx9.8\,\mathrm{m\,s^{-2}}

Resultg=\dfrac{GM_E}{R_E^{2}}\approx9.8\,\mathrm{m\,s^{-2}}

The mass m_0 of the body has cancelled, so (without air resistance) light and heavy bodies fall with the same acceleration. g depends only on the mass and radius of the Earth.

g on any planet

The same working applies to any planet or moon of mass m and radius r. The book writes the general form as

g=G\,\dfrac{m}{r^{2}}\qquad(15.4)

So the field at a planet's surface is decided by two factors: its mass (g\propto m) and its radius (g\propto1/r^{2}). A planet with more mass, or the same mass packed into a smaller radius, has a stronger surface field. Your weight W=mg therefore changes from planet to planet, while your mass stays the same.

Mass of the Earth

Rearranging Eq. (15.3) gives the mass of the Earth from quantities measured at its surface, once G is known. The middle form, GM_E=gR_E^{2} (Eq. 15.6), is used often because it replaces GM_E by quantities you usually know.

GM_E=gR_E^{2}\quad\Rightarrow\quad M_E=\dfrac{gR_E^{2}}{G}=\dfrac{9.8\times(6.4\times10^{6})^{2}}{6.67\times10^{-11}}=6.0\times10^{24}\,\mathrm{kg}

Example 15.3

The largest planet in our solar system is Jupiter. It has a mass of 1.898\times10^{27}\,\mathrm{kg} and radius 7.15\times10^{7}\,\mathrm{m}. Determine the free-fall acceleration at the surface of Jupiter. Also calculate the weight of a 60\,\mathrm{kg} astronaut on it.

Given

m_J = 1.898\times10^{27}\,\mathrm{kg}
r_J = 7.15\times10^{7}\,\mathrm{m}
m = 60\,\mathrm{kg} mass of the astronaut

To find

Field strength at the surface of Jupiter, from Eq. (15.4).
g_J=G\,\dfrac{m_J}{r_J^{2}}=6.67\times10^{-11}\times\dfrac{1.898\times10^{27}}{(7.15\times10^{7})^{2}}=24.8\,\mathrm{m\,s^{-2}}
Compare with the Earth.
\dfrac{g_J}{g_E}=\dfrac{24.8}{9.8}\approx2.53
Weight is mass times the local field strength. The astronaut's mass is the same everywhere.
W_A=mg_J=60\times24.8=1.49\times10^{3}\,\mathrm{N}
On Earth the same astronaut weighs
W=mg=60\times9.8=588\,\mathrm{N}

g_J\approx24.8\,\mathrm{m\,s^{-2}}\ (\text{about }2.5\,g),\qquad W_A\approx1.49\times10^{3}\,\mathrm{N}

Assignment 15.3, solved

The gravitational field strength on the surface of the moon is 1.6\,\mathrm{N\,kg^{-1}}. The mass of the moon is 7.3\times10^{22}\,\mathrm{kg}. What is its radius?

Given

g_m = 1.6\,\mathrm{N\,kg^{-1}}
M = 7.3\times10^{22}\,\mathrm{kg}

To find

Start from the surface field and make R the subject.
g_m=\dfrac{GM}{R^{2}}\ \Rightarrow\ R=\sqrt{\dfrac{GM}{g_m}}
Substitute and evaluate.
R=\sqrt{\dfrac{6.67\times10^{-11}\times7.3\times10^{22}}{1.6}}=\sqrt{3.04\times10^{12}}=1.74\times10^{6}\,\mathrm{m}

R\approx1.74\times10^{6}\,\mathrm{m}\approx1740\,\mathrm{km}

15.2Variation of g with altitudeOften examined

Moving away from the Earth's surface increases the distance from the Earth's centre, so g becomes smaller and our weight decreases. The value of g at a place depends on its distance from the centre of the Earth (Figure 15.4).

Derivation g at height h above the surface

Aim. Express the field strength g_h at height h in terms of the surface value g.

Assumptions.

The Earth is a uniform sphere of mass M_E and radius R_E, so it acts as a point mass at its centre O.
The body at height h is small.

Figure 15.4: a body at P, height h above the surface, is a distance R_E+h from the centre O.

At height h the distance from the centre is R_E+h, so replace R_E by R_E+h in Eq. (15.3).
g_h=G\,\dfrac{M_E}{(R_E+h)^{2}}\qquad(15.5)
At the surface g=GM_E/R_E^{2}. Rearrange it.
GM_E=gR_E^{2}\qquad(15.6)
Put Eq. (15.6) in Eq. (15.5). This removes G and M_E, leaving only distances.
g_h=\dfrac{gR_E^{2}}{(R_E+h)^{2}}=g\left(\dfrac{R_E}{R_E+h}\right)^{2}\qquad(15.7)
This is the exact result. Rewrite it with h/R_E to see how small heights behave.
g_h=g\left(1+\dfrac{h}{R_E}\right)^{-2}
Close to the surface h\ll R_E. The binomial approximation (1+x)^{n}\approx1+nx, with x=h/R_E and n=-2, gives a simple straight-line rule.
g_h\approx g\left(1-\dfrac{2h}{R_E}\right)
Use the exact form (15.7) whenever h is not small compared with R_E, for example for satellites.

Resultg_h=g\left(\dfrac{R_E}{R_E+h}\right)^{2}\;\approx\;g\left(1-\dfrac{2h}{R_E}\right)\ \text{ for } h\ll R_E

As we go farther from the centre (or the surface) of the Earth, g decreases. At h=R_E it is exactly g/4. Near the surface it drops by only about 0.03\% per kilometre.

Table 15.1: variation in g with altitude. The third column checks the book's values using g=9.8\,\mathrm{m\,s^{-2}} and R_E=6.4\times10^{6}\,\mathrm{m}.
Altitude (km)	g_h in the book (\mathrm{m\,s^{-2}})	g_h from Eq. (15.7)	Example
0	9.8	9.80	average Earth radius
8.8	9.8	9.77	Mount Everest
36.6	9.7	9.69	highest manned balloon
400	8.7	8.68	space shuttle orbit
35 700	0.2	0.23	communication satellites

The change in g becomes noticeable only at large heights, as Table 15.1 shows.

Field strength against height

The straight-line approximation matches the exact curve only for small heights.

Why g is nearly constant near the surface

The approximate form shows that the fractional decrease in g at height h is 2h/R_E. The Earth's radius is very large, so for everyday heights this is tiny.

\dfrac{g-g_h}{g}\approx\dfrac{2h}{R_E}

For h=1\,\mathrm{km}: \dfrac{2h}{R_E}=\dfrac{2\times1000}{6.4\times10^{6}}=3.1\times10^{-4}, only about 0.03\%.
At the top of Mount Everest (h\approx8.8\,\mathrm{km}): \dfrac{2h}{R_E}\approx0.0028, so g falls by less than 0.3\%, from 9.80 to about 9.77\,\mathrm{m\,s^{-2}}.

So for small changes in height (buildings, mountains, aircraft) we may treat g as constant. For heights that are a sizeable fraction of R_E (hundreds of kilometres or more) we must use the exact Eq. (15.7).

Example 15.2

Shooting stars are meteors that burn up in Earth's atmosphere. Meteors become visible between about 75 to 120 km above Earth. What is the gravitational field strength at 120 km above earth surface?

Given

g = 9.8\,\mathrm{m\,s^{-2}}
R_E = 6.4\times10^{6}\,\mathrm{m}
h = 120\,\mathrm{km}=0.12\times10^{6}\,\mathrm{m}

To find

Use the exact relation (15.7).
g_h=\dfrac{gR_E^{2}}{(R_E+h)^{2}}
Substitute.
g_h=\dfrac{9.8\times(6.4\times10^{6})^{2}}{(6.4\times10^{6}+0.12\times10^{6})^{2}}=9.8\times\left(\dfrac{6.40}{6.52}\right)^{2}
Evaluate.
g_h=9.8\times0.964=9.4\,\mathrm{m\,s^{-2}}

At 120\,\mathrm{km} the field is only slightly less than at the surface.

g_h\approx9.4\,\mathrm{m\,s^{-2}}

Assignment 15.2, solved

What will be the value of gravitational field strength g at 35\,700\,\mathrm{km}, where geostationary satellites orbit around the Earth?

35\,700\,\mathrm{km} is the height of the orbit above the surface, so the distance from the centre is R_E+h=6.4\times10^{6}+3.57\times10^{7}=4.21\times10^{7}\,\mathrm{m}.

Use Eq. (15.7).
g_h=g\left(\dfrac{R_E}{R_E+h}\right)^{2}=9.8\times\left(\dfrac{6.4\times10^{6}}{4.21\times10^{7}}\right)^{2}
Evaluate.
g_h=9.8\times(0.152)^{2}=9.8\times0.0231=0.23\,\mathrm{m\,s^{-2}}

This agrees with the value 0.2\,\mathrm{m\,s^{-2}} in Table 15.1. The field is not zero there: this weak field is exactly what keeps the satellite in its orbit.

g_h\approx0.23\,\mathrm{m\,s^{-2}}

15.2Gravitational field lines

To understand the field around the Earth we model the planet as a perfectly uniform sphere: a sphere with the same density throughout. This makes calculations for stars, planets and moons much easier, without considering differences in their density.

Figure 15.5: field lines of a uniform sphere point radially inward. They crowd together near the mass (stronger field) and spread out farther away (weaker field). Over a small region near the surface they are almost parallel.

A. Outside the sphere

Outside a uniform sphere the field is exactly the same as if all its mass were concentrated at its centre, like a point mass with the same total mass. This is why every distance in this unit is measured from the centre of the Earth.

g=\dfrac{GM}{r^{2}}\qquad(r\ge R)

The field lines spread out as they go away from the Earth, so the field weakens rapidly. This agrees with the inverse-square nature of Newton's law of gravitation.

B. Inside the sphere

At a point inside the sphere only the mass within the radius of that point affects the field; the pulls of the mass outside it cancel. Going from the centre towards the surface, more and more mass lies inside, so the field strength increases steadily. It is zero at the centre and greatest at the surface.

Derivation Field inside a uniform sphere

Aim. Show that inside a uniform sphere g increases in proportion to the distance r from the centre.

Assumptions.

The sphere has mass M, radius R and uniform density.
Only the mass within radius r produces a field at a point inside.

Volume is proportional to (radius)^3 and the density is uniform, so the mass inside radius r is the fraction r^{3}/R^{3} of the whole mass.
M_r=M\,\dfrac{r^{3}}{R^{3}}
This inner mass acts as a point mass at the centre.
g_r=\dfrac{GM_r}{r^{2}}=\dfrac{GM}{r^{2}}\cdot\dfrac{r^{3}}{R^{3}}=\dfrac{GM}{R^{3}}\,r
Use g=GM/R^{2} for the surface value.
g_r=g\,\dfrac{r}{R}

Resultg_r=g\,\dfrac{r}{R}\qquad(r\le R)

Inside the sphere g rises in a straight line from zero at the centre to its largest value at the surface. Outside, it falls as 1/r^{2}.

Field strength inside and outside a uniform sphere

g is greatest at the surface. It falls linearly to zero towards the centre and as 1/r^{2} outside.

15.3Satellites and orbits

A satellite is any object that orbits a planet or another body because of the force of gravity, keeping a stable path around it. The Moon is a satellite of the Earth, and the Earth itself is a satellite of the Sun.

Satellite: Any object that orbits another object due to the force of gravity, maintaining a stable path around it.
Natural satellite: A natural body orbiting a planet, dwarf planet or minor planet, where the larger body's gravity dominates the system. Six of the major planets have natural satellites, often called moons.
Artificial satellite: An object intentionally placed into orbit around the Earth or another celestial body.
Orbit: The closed path that a satellite follows around the body it circles.

Artificial satellites

The first artificial satellite was launched in 1957. Since then thousands have been sent into orbit. Once in orbit a satellite needs no engine to keep going: the Earth's gravity holds it in its path, and at such heights there is almost no air resistance to slow it down. Engineers build different types for different missions:

Communication satellites carry television, telephone and internet signals.
Weather satellites photograph clouds and measure temperatures for forecasts.
Navigation satellites, such as GPS, give positions on the Earth.
Scientific research: space telescopes such as Hubble, and space stations.
Military operations: communication and observation.

Types of orbit

Figure 15.6: different orbits for artificial satellites. 1: in the plane of the equator; 2: inclined orbit; 3: polar orbit; 4: geostationary orbit.

Equatorial orbit: lies in the plane of the equator.
Inclined orbit: tilted at an angle to the equator.
Polar orbit: passes over (or near) both poles. As the Earth turns beneath it, the satellite scans the whole surface, so polar orbits suit mapping and weather observation.
Geostationary orbit: a circular equatorial orbit about 36\,000\,\mathrm{km} above the surface, in which a satellite takes 24 hours to go round (section 15.4).

Launching a satellite

A satellite is put into orbit by first lifting it to a high altitude and then accelerating it to a sufficiently high speed parallel to the Earth's surface (a tangential speed) with the help of rockets. What happens next depends on that speed (Figure 15.7):

Figure 15.7: artificial satellites launched from the same height at different speeds.

If the speed is too low, the satellite falls back to the Earth.
If the speed is adjusted to exactly the orbital velocity, it moves in a circular orbit and keeps going round.
If the speed is higher, it moves in an elliptical orbit.
If the speed reaches the escape speed, it escapes from the Earth's gravity and never returns.

Satellites are usually placed in circular (or nearly circular) orbits, because such orbits require the least energy.

15.3Orbital velocityOften examined

A satellite moving in a circle keeps changing direction, so it needs a centripetal force towards the centre of the circle. For a satellite this force is the gravitational pull of the Earth. The speed at which gravity supplies exactly the centripetal force needed is the orbital velocity.

Orbital velocity (v_o): The minimum velocity required for a satellite to orbit the Earth or another celestial body at a specific altitude. In a circular orbit the satellite moves with this constant tangential speed.

Derivation Orbital velocity

Aim. Find the speed v_o needed for a circular orbit of radius r around the Earth.

Assumptions.

The orbit is a circle of radius r measured from the Earth's centre.
Gravity is the only force on the satellite (no air resistance).
The satellite is tiny compared with the Earth.

Figure 15.8: a satellite launched at orbital speed. Gravity towards O provides the centripetal force; the velocity v_o is along the tangent.

A body moving with speed v_o in a circle of radius r has a centripetal acceleration towards the centre.
a_c=\dfrac{v_o^{2}}{r}
This acceleration is supplied by gravity, so it equals the field strength g at the orbit.
a_c=g\quad\Rightarrow\quad g=\dfrac{v_o^{2}}{r}
Make v_o the subject.
v_o^{2}=gr\quad\Rightarrow\quad v_o=\sqrt{gr}
At distance r from the centre the field is g=GM_E/r^{2}. Substitute it.
v_o^{2}=\dfrac{GM_E}{r^{2}}\cdot r=\dfrac{GM_E}{r}\quad\Rightarrow\quad v_o=\sqrt{\dfrac{GM_E}{r}}\qquad(15.8)
The satellite is small, so r is the Earth's radius plus the height h of the satellite above the surface.
v_o=\sqrt{\dfrac{GM_E}{R_E+h}}\qquad(15.9)

Resultv_o=\sqrt{\dfrac{GM_E}{r}}=\sqrt{\dfrac{GM_E}{R_E+h}}

The orbital velocity depends on the mass of the central body and on the distance from its centre to the satellite. It does not depend on the satellite's mass. A satellite moves faster in a low orbit and slower in a high orbit.

Features of orbital velocity

v_o\propto1/\sqrt r: the higher the orbit, the lower the orbital speed. Four times the radius gives half the speed.
v_o\propto\sqrt{M}: a more massive planet needs a faster satellite at the same radius.
v_o does not depend on the satellite's mass: a small satellite and the huge ISS at the same height move at the same speed.

Satellite just above the surface

For an orbit just above the surface (ignoring air resistance), r=R_E and g=9.8\,\mathrm{m\,s^{-2}}:

v_o=\sqrt{gR_E}=\sqrt{9.8\times6.4\times10^{6}}=7.9\times10^{3}\,\mathrm{m\,s^{-1}}\approx7.9\,\mathrm{km\,s^{-1}}

This is the greatest speed for any circular orbit around the Earth; every higher orbit is slower.

Orbital speed (\mathrm{km\,s^{-1}}) against orbit radius

Orbital speed falls as the orbit gets larger (v_o\propto1/\sqrt r). The ISS, only 400\,\mathrm{km} up (r=1.06R_E), moves at 7.7\,\mathrm{km\,s^{-1}}.

Time period of a satellite

Derivation Period of a satellite

Aim. Find the time T for one complete circular orbit of radius r.

In one orbit the satellite covers the circumference 2\pi r at constant speed v_o.
T=\dfrac{2\pi r}{v_o}
Substitute v_o=\sqrt{GM/r}.
T=2\pi r\sqrt{\dfrac{r}{GM}}=2\pi\sqrt{\dfrac{r^{3}}{GM}}
Square both sides.
T^{2}=\dfrac{4\pi^{2}}{GM}\,r^{3}\quad\Rightarrow\quad T^{2}\propto r^{3}

ResultT=\dfrac{2\pi r}{v_o}=2\pi\sqrt{\dfrac{r^{3}}{GM}}

Higher satellites take longer to go round: they move more slowly and along a longer path. T^{2}\propto r^{3} is Kepler's third law of planetary motion. Solved for r, the same equation gives the radius of the geostationary orbit (section 15.4).

Kinetic energy of a satellite

Multiply v_o^{2}=GM/r by \tfrac12m:

K.E.=\tfrac12mv_o^{2}=\dfrac{GMm}{2r}

So when a satellite is moved to a higher orbit (larger r), its speed and kinetic energy decrease. Its gravitational potential energy U=-GMm/r (section 15.5) increases, becoming less negative, and so does its total energy E=K.E.+U=-GMm/2r. Rockets must supply energy to lift a satellite to a higher orbit.

Example 15.4

The International Space Station (ISS) is the biggest structure ever placed in space (even visible in night sky with unaided eye) and serves as a space laboratory. It orbits at an average altitude of about 400\,\mathrm{km} above the Earth's surface in Low Earth Orbit (LEO). What is its orbital velocity?

Given

h = 400\,\mathrm{km}=4.0\times10^{5}\,\mathrm{m}
M_E = 6\times10^{24}\,\mathrm{kg}
R_E = 6.4\times10^{6}\,\mathrm{m}

To find

Use the orbital velocity with r=R_E+h.
v_o=\sqrt{\dfrac{GM_E}{R_E+h}}
Substitute.
v_o=\sqrt{\dfrac{6.67\times10^{-11}\times6\times10^{24}}{6.4\times10^{6}+0.4\times10^{6}}}=\sqrt{\dfrac{4.00\times10^{14}}{6.8\times10^{6}}}
Evaluate.
v_o=\sqrt{5.89\times10^{7}}=7.67\times10^{3}\,\mathrm{m\,s^{-1}}=7.67\,\mathrm{km\,s^{-1}}
Check the book's remark about orbits per day with the period.
T=\dfrac{2\pi r}{v_o}=\dfrac{2\pi\times6.8\times10^{6}}{7.67\times10^{3}}=5.57\times10^{3}\,\mathrm{s}\approx93\,\mathrm{min},\qquad \dfrac{86\,400}{5570}\approx15.5

So in 24 hours the ISS goes round about 16 times, and its crew see about 16 sunrises and sunsets a day. The book rounds the speed to 7.66\,\mathrm{km\,s^{-1}}.

v_o\approx7.67\,\mathrm{km\,s^{-1}}

Example 15.5

The mass of the Sun is 1.99\times10^{30}\,\mathrm{kg}, and the radius of the Earth's orbit around the Sun is 1.5\times10^{11}\,\mathrm{m}. What is the orbital velocity of the Earth?

Given

M_S = 1.99\times10^{30}\,\mathrm{kg}
r = 1.5\times10^{11}\,\mathrm{m}

To find

Here the Sun is the central body, so use its mass.
v_o=\sqrt{\dfrac{GM_S}{r}}=\sqrt{\dfrac{6.67\times10^{-11}\times1.99\times10^{30}}{1.5\times10^{11}}}
Evaluate.
v_o=\sqrt{8.85\times10^{8}}=2.97\times10^{4}\,\mathrm{m\,s^{-1}}\approx30\,\mathrm{km\,s^{-1}}

This is a very high speed: all of us, sitting still on the Earth's surface, are actually moving around the Sun at about 30\,\mathrm{km\,s^{-1}}.

v_o\approx30\,\mathrm{km\,s^{-1}}

Assignment 15.4, solved

Calculate the orbital speed of a satellite orbiting the Earth at an altitude equal to Earth's radius.

At h=R_E the orbit radius is r=2R_E=1.28\times10^{7}\,\mathrm{m}.
v_o=\sqrt{\dfrac{GM_E}{2R_E}}=\sqrt{\dfrac{6.67\times10^{-11}\times6\times10^{24}}{1.28\times10^{7}}}
Evaluate.
v_o=\sqrt{3.13\times10^{7}}=5.59\times10^{3}\,\mathrm{m\,s^{-1}}
Check with the ratio method: doubling r divides v_o by \sqrt2.
v_o=\dfrac{7.91}{\sqrt2}=5.59\,\mathrm{km\,s^{-1}}

v_o\approx5.6\,\mathrm{km\,s^{-1}}

15.4Geostationary satellitesOften examined

A higher satellite takes longer to go round. At one particular distance from the Earth's centre a satellite takes exactly 24 hours to circle the Earth, the same time the Earth takes to turn once on its axis. If it also moves in the direction of the Earth's rotation, directly above the equator, it stays above the same point on the Earth. From the ground it seems to hang motionless in the sky.

Geostationary satellite: A satellite that orbits the Earth directly above the equator, from west to east, with an orbital period equal to the Earth's rotation period (24 hours), so that it appears stationary above a fixed point on the Earth's surface. It is also called a geosynchronous equatorial orbit (GEO) satellite.
Geostationary orbit: The circular orbit in the plane of the equator, of radius about 4.23\times10^{7}\,\mathrm{m} from the Earth's centre, in which a satellite has a period of 24 hours.

Conditions for a geostationary orbit

The orbit is circular and lies in the plane of the equator: the satellite is directly above the equator.
The satellite moves from west to east, the direction in which the Earth rotates.
Its period is 24 hours (T=86\,400\,\mathrm{s}), equal to the Earth's rotation period.
As a result it has one fixed orbital radius (about 42\,300\,\mathrm{km} from the centre, or 35\,900\,\mathrm{km} above the surface) and one fixed speed (about 3.1\,\mathrm{km\,s^{-1}}).

Figure 15.9: seen from above the North Pole, the Earth and a geostationary satellite turn through the same angle in the same time, so the satellite S stays directly above the same point P on the equator.

Derivation Orbital radius of a geostationary satellite

Aim. Find the radius r of a circular orbit whose period is T (24 hours for a geostationary satellite).

Assumptions.

The orbit is a circle of radius r measured from the Earth's centre.
The satellite (mass m) moves under gravity alone.

Gravity provides the centripetal force, so the orbital velocity is
v_o=\sqrt{\dfrac{GM_E}{r}}\qquad(\mathrm{i})
In one period T the satellite goes once round the circle, a distance equal to the circumference 2\pi r.
v_o=\dfrac{S}{t}=\dfrac{2\pi r}{T}\qquad(\mathrm{ii})
Both expressions give the same speed, so equate (i) and (ii).
\dfrac{2\pi r}{T}=\sqrt{\dfrac{GM_E}{r}}
Square both sides to remove the root.
\dfrac{4\pi^{2}r^{2}}{T^{2}}=\dfrac{GM_E}{r}
Multiply both sides by rT^{2}/4\pi^{2} to collect the powers of r.
r^{3}=\dfrac{GM_ET^{2}}{4\pi^{2}}
Take the cube root of both sides.
r=\left(\dfrac{GM_ET^{2}}{4\pi^{2}}\right)^{1/3}\qquad(15.10)

Resultr=\left(\dfrac{GM_ET^{2}}{4\pi^{2}}\right)^{1/3}

The radius depends only on the period and the Earth's mass. Every geostationary satellite has T=24\,\mathrm{h}, so they all share one orbit radius, whatever their mass.

Orbital radius, height and speed of a geostationary satellite

Calculate the orbital radius r, the height h above the Earth's surface and the orbital speed v_o of a geostationary satellite. Use G=6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}, M_E=6\times10^{24}\,\mathrm{kg}, R_E=6.4\times10^{6}\,\mathrm{m}, T=24\,\mathrm{h} and \pi=3.14.

Convert the period to seconds.
T=24\times3600=86\,400\,\mathrm{s}
Radius from Eq. (15.10). This is measured from the centre of the Earth.
r=\left(\dfrac{6.67\times10^{-11}\times6\times10^{24}\times(86\,400)^{2}}{4\times(3.14)^{2}}\right)^{1/3}=\left(7.58\times10^{22}\right)^{1/3}=4.23\times10^{7}\,\mathrm{m}
Height above the surface (this is Assignment 15.5).
h=r-R_E=4.23\times10^{7}-0.64\times10^{7}=3.59\times10^{7}\,\mathrm{m}\approx35\,900\,\mathrm{km}
Orbital speed from v_o=\sqrt{GM_E/r}.
v_o=\sqrt{\dfrac{6.67\times10^{-11}\times6\times10^{24}}{4.23\times10^{7}}}=\sqrt{9.46\times10^{6}}=3.08\times10^{3}\,\mathrm{m\,s^{-1}}
Check with v_o=2\pi r/T.
v_o=\dfrac{2\pi\times4.23\times10^{7}}{86\,400}=3.08\times10^{3}\,\mathrm{m\,s^{-1}}

r=4.23\times10^{7}\,\mathrm{m},\quad h=3.59\times10^{7}\,\mathrm{m},\quad v_o\approx3.1\times10^{3}\,\mathrm{m\,s^{-1}}

Why three satellites give global coverage

A geostationary satellite is so high that it can see a large part of the Earth, about 42\% of the surface. Three satellites spaced 120^\circ apart above the equator can together see the whole equatorial belt and nearly the whole Earth; only small regions near the poles are missed. So the minimum number of geostationary satellites for global coverage is 3.

Three geostationary satellites 120^\circ apart cover the whole equatorial belt (not to scale).

Uses of geostationary satellites

Communication: television broadcasting, telephone and internet links. A dish antenna on the ground points in one fixed direction and never needs to track the satellite.
Weather monitoring: continuous pictures of the same large region.
Navigation and positioning services that need a satellite fixed in the sky.

Example 15.6

Venus has a mass of 4.867\times10^{24}\,\mathrm{kg} and has a period of 243 days. What would be the radius of a synchronous satellite for this planet?

Given

M_V = 4.867\times10^{24}\,\mathrm{kg}
T = 243\ \text{days}=243\times86\,400\,\mathrm{s}=2.10\times10^{7}\,\mathrm{s}

To find

A synchronous satellite has the same period as the planet's rotation. Use Eq. (15.10) with the mass of Venus.
r=\left(\dfrac{GM_VT^{2}}{4\pi^{2}}\right)^{1/3}
Substitute.
r=\left(\dfrac{6.67\times10^{-11}\times4.867\times10^{24}\times(2.10\times10^{7})^{2}}{4\times(3.14)^{2}}\right)^{1/3}
Evaluate.
r=1.54\times10^{9}\,\mathrm{m}=1.54\times10^{6}\,\mathrm{km}

Venus rotates very slowly, so the required orbit is enormous. At about 1.5 million km the satellite would be outside Venus's Hill sphere (which is small because Venus has less mass than the Earth), and the Sun's gravity would make its orbit unstable. So although the formula gives an answer, a synchronous satellite around Venus is not practically possible.

r\approx1.54\times10^{9}\,\mathrm{m}\approx1.5\times10^{6}\,\mathrm{km}

15.5Gravitational potential energyOften examined

Near the Earth's surface we use \Delta P.E.=mgh. This formula assumes that g is constant, which is true only for small heights. Farther away gravity weakens, so mgh becomes wrong (Assignment 15.6 in the next section shows by how much). For large distances we need an expression based on Newton's law of gravitation.

Potential energy is always measured from a chosen zero (reference) point. For mgh the zero is the Earth's surface. For large distances the natural zero is a point at infinity, where the Earth's pull has fallen to zero. With this choice the potential energy at every finite distance is negative: gravity does positive work on a body as it comes in from infinity, so the body's potential energy falls below zero.

Absolute gravitational potential energy: The gravitational potential energy U of a body at a point, measured from the zero at infinity: the work done in bringing the body from infinity to that point slowly, without changing its kinetic energy. It is negative at every finite distance. Its size at the Earth's surface, GM_Em/R_E, is the work needed to move the body from the surface to infinity, where g is negligible.

Derivation Gravitational potential energy U=-\dfrac{GM_Em}{r}

Aim. Find the gravitational potential energy of a mass m at the Earth's surface, and then at any distance r from the centre, taking U=0 at infinity.

Assumptions.

The Earth is a uniform sphere of mass M_E and radius R_E; it acts as a point mass at its centre O.
The mass m is moved slowly, so its kinetic energy does not change and the work done on it by the agent goes into potential energy.
U=0 at infinity.

Figure 15.10: the path from the surface (point 0, distance R_E from the centre O) to a very distant point n is divided into small steps \Delta r. In each step the force is taken as constant. F_g is the pull of gravity on m; F is the equal and opposite force of the agent.

By the work and energy principle, when m is moved slowly the work done on it by the external agent equals the change in its potential energy.
\Delta P.E.=W_{net}\qquad(1)
The force changes with distance, so W=Fd cannot be used for the whole path at once. Divide the path between the surface (point 0) and a very distant point n into n small steps of length \Delta r, so small that the force is almost constant in each. The total work is the sum of the works in the steps.
W_{net}=W_1+W_2+W_3+\dots+W_{n-1}+W_n\qquad(2)
Take the step between point 1 (distance r_1) and point 0 (distance R_E). To bring m in slowly, the agent pulls outward with a force F_{av} equal to the pull of gravity, while m moves inward through \Delta r. Force and displacement are opposite: \theta=180^\circ and \cos180^\circ=-1.
W_1=F_{av}\,\Delta r\cos180^\circ=-F_{av}\,\Delta r\qquad(3)
The average force in this step follows from Newton's law, using the average distance r_{av} of the step.
F_{av}=\dfrac{GM_Em}{r_{av}^{2}}\qquad(4)
The average distance is neither R_E nor r_1 but their mean. From the figure \Delta r=r_1-R_E, so r_1=R_E+\Delta r.
r_{av}=\dfrac{R_E+r_1}{2}=\dfrac{2R_E+\Delta r}{2}=R_E+\dfrac{\Delta r}{2}\qquad(5\text{ to }7)
Square the average distance.
r_{av}^{2}=\left(R_E+\dfrac{\Delta r}{2}\right)^{2}=R_E^{2}+R_E\,\Delta r+\dfrac{(\Delta r)^{2}}{4}
\Delta r is very small, so the term (\Delta r)^{2}/4 is negligible compared with the others. Then put \Delta r=r_1-R_E.
r_{av}^{2}\approx R_E^{2}+R_E\,\Delta r=R_E^{2}+R_E(r_1-R_E)=R_Er_1\qquad(8,\ 9)
Put Eq. (9) in Eq. (4).
F_{av}=\dfrac{GM_Em}{R_Er_1}\qquad(10)
Put Eq. (10) and \Delta r=r_1-R_E (Eq. 11) in Eq. (3), then split the fraction.
W_1=-\dfrac{GM_Em}{R_Er_1}\,(r_1-R_E)=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_1}\right)\qquad(12)
The same working applies to every other step.
\begin{aligned}W_2&=-GM_Em\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\\ W_3&=-GM_Em\left(\dfrac{1}{r_2}-\dfrac{1}{r_3}\right)\\ &\ \ \vdots\\ W_{n-1}&=-GM_Em\left(\dfrac{1}{r_{n-2}}-\dfrac{1}{r_{n-1}}\right)\\ W_n&=-GM_Em\left(\dfrac{1}{r_{n-1}}-\dfrac{1}{r_n}\right)\end{aligned}\qquad(13\text{ to }16)
Add all the steps as in Eq. (2). Every middle term appears once with + and once with -, so they cancel in pairs; only the first and last terms remain.
\begin{aligned}W_{net}&=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_1}+\dfrac{1}{r_1}-\dfrac{1}{r_2}+\dots+\dfrac{1}{r_{n-1}}-\dfrac{1}{r_n}\right)\\ &=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_n}\right)\end{aligned}\qquad(17)
By Eq. (1), this is the change in potential energy when m is brought from point n to the surface.
\Delta P.E.=U_{\text{surface}}-U_n=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_n}\right)\qquad(18)
Now take point n at infinity: r_n=\infty, so 1/r_n=0, and U_n=0 (our zero). The work done in bringing m from infinity to the surface is then
U=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{\infty}\right)=-\dfrac{GM_Em}{R_E}
The same steps, started from any point P at distance r from the centre instead of the surface, give the general result.
U=-\dfrac{GM_Em}{r}\qquad(15.9)

ResultU=-\dfrac{GM_Em}{r}

U is zero at infinity and negative at every finite distance, and it increases (becomes less negative) as r increases. The same form, U=-Gm_1m_2/r, gives the potential energy of any two point masses; it belongs to the pair, not to one body alone.

Gravitational potential energy (in units of GM_Em/R_E) against distance

Figure 15.11: U is most negative at the surface and rises towards zero as the distance r between the centres increases.

Meaning of negative potential energy

U=-GMm/r is negative because we chose U=0 at infinity and gravity is attractive: the Earth does positive work on a body as it falls in from infinity, so the body ends with less energy than it had at infinity. Negative U means the body is bound to the Earth. To free it completely (take it to infinity) we must give it at least GMm/r of energy. A body whose total energy (K.E.+U) is negative cannot escape: it falls back or moves in an orbit, like the Moon and artificial satellites.

Change in potential energy between two points

Moving m from distance r_1 out to r_2 changes its potential energy by

\Delta U=U_2-U_1=\left(-\dfrac{GMm}{r_2}\right)-\left(-\dfrac{GMm}{r_1}\right)=GMm\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)

This is positive when r_2>r_1: moving away from the Earth increases the potential energy.

Derivation \Delta U reduces to mgh near the surface

Aim. Show that for a small height h above the surface the change in potential energy becomes mgh.

Put r_1=R_E and r_2=R_E+h in \Delta U and combine the fractions.
\Delta U=GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{R_E+h}\right)=\dfrac{GM_Em\,h}{R_E(R_E+h)}
For h\ll R_E, R_E+h\approx R_E.
\Delta U\approx\dfrac{GM_Em\,h}{R_E^{2}}
Use GM_E/R_E^{2}=g.
\Delta U\approx mgh

Result\Delta U\approx mgh\qquad(h\ll R_E)

mgh is the small-height special case of the general formula. For large heights mgh overestimates the change, because it assumes g keeps its surface value all the way up.

15.5Gravitational potentialOften examined

The potential energy U depends on the mass m placed at a point. Dividing by m gives a quantity that describes the point itself, whatever mass is put there: the gravitational potential.

Gravitational potential (V or \phi): The gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point. It is the gravitational potential energy per unit mass, V=U/m. It is a scalar with SI unit \mathrm{J\,kg^{-1}} (the same as \mathrm{N\,m\,kg^{-1}}).

Derivation Gravitational potential V=-\dfrac{GM_E}{r}

Aim. Find the gravitational potential at a point P at distance r from the centre of the Earth.

By definition, potential is potential energy per unit mass.
V=\dfrac{U}{m}\qquad(15.10)
The potential energy of a small test mass m at distance r, from the step derivation in 'Gravitational potential energy', is
U=-\dfrac{GM_Em}{r}\qquad(15.9)
Put Eq. (15.9) in Eq. (15.10). The test mass m cancels.
V=\dfrac{-GM_Em/r}{m}=-\dfrac{GM_E}{r}\qquad(15.11)

ResultV=-\dfrac{GM_E}{r}

The potential depends only on the mass M_E that creates the field and the distance r from its centre. It is zero at infinity, negative everywhere else, and rises towards zero as r increases. For any point mass M, \phi=-GM/r.

Potential at the surface of the Earth

Eq. (15.11) does not depend on the mass of the object, so it gives the potential at the Earth's surface directly:

V=-\dfrac{GM_E}{R_E}=-\dfrac{6.67\times10^{-11}\times6\times10^{24}}{6.4\times10^{6}}=-6.25\times10^{7}\,\mathrm{J\,kg^{-1}}\approx-62.5\,\mathrm{MJ\,kg^{-1}}

This means that 62.5\,\mathrm{MJ} of work is needed to take each kilogram from the Earth's surface to infinity. The equations were derived for the Earth, but they are equally valid for any other gravitating body.

Gravitational potential of the Earth (\mathrm{MJ\,kg^{-1}})

The potential is -62.5\,\mathrm{MJ\,kg^{-1}} at the surface and rises towards zero with distance.

From potential to the potential energy of two point masses

The concept of potential leads directly to the potential energy of two point masses. A point mass M creates the potential V=-GM/r at distance r. Since V is potential energy per unit mass, a mass m placed at that point has

U=mV=m\left(-\dfrac{GM}{r}\right)=-\dfrac{GMm}{r}

Starting instead from the potential of m at the position of M gives U=M(-Gm/r), the same value. So U is a shared property of the two masses and their separation.

Field strength, potential and potential energy compared.
	Field strength g	Potential V (\phi)	Potential energy U
Definition	force per unit mass	work done per unit mass in bringing a test mass from infinity	work done in bringing the mass m from infinity
For a point mass M	\dfrac{GM}{r^{2}}	-\dfrac{GM}{r}	-\dfrac{GMm}{r}
Depends on the test mass	no	no	yes
Nature	vector	scalar	scalar
SI unit	\mathrm{N\,kg^{-1}}	\mathrm{J\,kg^{-1}}	\mathrm{J}
With distance	falls as 1/r^{2}	rises towards 0 as -1/r	rises towards 0 as -1/r

Example 15.7

What is the gravitational potential energy and gravitational potential with respect to the Sun at the position of the Earth? The mass of the Sun is 1.99\times10^{30}\,\mathrm{kg} and the mass of the Earth is 6\times10^{24}\,\mathrm{kg}. The mean Earth-to-Sun distance is 1.5\times10^{11}\,\mathrm{m}.

Given

M_S = 1.99\times10^{30}\,\mathrm{kg}
M_E = 6\times10^{24}\,\mathrm{kg}
r = 1.5\times10^{11}\,\mathrm{m}

To find

Potential energy of the Sun and Earth pair.
U=-\dfrac{GM_SM_E}{r}=-\dfrac{6.67\times10^{-11}\times1.99\times10^{30}\times6\times10^{24}}{1.5\times10^{11}}=-5.31\times10^{33}\,\mathrm{J}
Potential of the Sun at the Earth's position (it does not involve the Earth's mass).
V=-\dfrac{GM_S}{r}=-\dfrac{6.67\times10^{-11}\times1.99\times10^{30}}{1.5\times10^{11}}=-8.85\times10^{8}\,\mathrm{J\,kg^{-1}}=-885\,\mathrm{MJ\,kg^{-1}}
Check: U=M_EV.
U=6\times10^{24}\times(-8.85\times10^{8})=-5.31\times10^{33}\,\mathrm{J}

U\approx-5.31\times10^{33}\,\mathrm{J},\qquad V\approx-8.85\times10^{8}\,\mathrm{J\,kg^{-1}}=-885\,\mathrm{MJ\,kg^{-1}}

Assignment 15.6, solved

Calculate the value of gravitational potential at 1000\,\mathrm{km}, 50\,000\,\mathrm{km} and 100\,000\,\mathrm{km} from the surface of Earth. Compare the values obtained with potential energy formula E_P=mgh.

The distances are heights above the surface, so r=R_E+h. Use V=-GM_E/r with GM_E=6.67\times10^{-11}\times6\times10^{24}=4.00\times10^{14}\,\mathrm{N\,m^{2}\,kg^{-1}}. To compare with mgh, work per kilogram: the exact energy needed to lift 1 kg from the surface is V-V_{\text{surface}} (with V_{\text{surface}}=-62.5\,\mathrm{MJ\,kg^{-1}}), while mgh per kilogram is gh.

h	r=R_E+h	V=-GM_E/r	exact lift energy V-V_{\text{surface}}	gh (from mgh)
1000\,\mathrm{km}	7.40\times10^{6}\,\mathrm{m}	-54.1\,\mathrm{MJ\,kg^{-1}}	8.45\,\mathrm{MJ\,kg^{-1}}	9.8\,\mathrm{MJ\,kg^{-1}} (16% too big)
50\,000\,\mathrm{km}	5.64\times10^{7}\,\mathrm{m}	-7.10\,\mathrm{MJ\,kg^{-1}}	55.4\,\mathrm{MJ\,kg^{-1}}	490\,\mathrm{MJ\,kg^{-1}} (9 times too big)
100\,000\,\mathrm{km}	1.064\times10^{8}\,\mathrm{m}	-3.76\,\mathrm{MJ\,kg^{-1}}	58.8\,\mathrm{MJ\,kg^{-1}}	980\,\mathrm{MJ\,kg^{-1}} (17 times too big)

mgh measures energy from the surface (zero at the surface) and assumes g keeps its surface value. The potential V is measured from infinity and allows for the weakening of g. The exact lift energy is close to gh only at the lowest height; at large heights mgh grossly overestimates it. The exact value can never exceed 62.5\,\mathrm{MJ\,kg^{-1}}, the energy needed to reach infinity.

V_{1000\,\mathrm{km}}=-54.1,\quad V_{50\,000\,\mathrm{km}}=-7.10,\quad V_{100\,000\,\mathrm{km}}=-3.76\ \mathrm{MJ\,kg^{-1}}

Formula sheet

Newton's law of gravitation

F_g=G\,\dfrac{m_1m_2}{r^{2}}

m_1,m_2: masses; r: distance between their centres; G: gravitational constant

Unit: \mathrm{N}

Gravitational constant

G=6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}

same value everywhere in the universe

Unit: \mathrm{N\,m^{2}\,kg^{-2}}=\mathrm{m^{3}\,kg^{-1}\,s^{-2}}

Action and reaction

\vec F_{12}=-\vec F_{21}

\vec F_{12}: force on m_1 due to m_2; \vec F_{21}: force on m_2 due to m_1

Unit: \mathrm{N}

Gravitational field strength (definition)

g=\dfrac{F_g}{m}

F_g: gravitational force on a small test mass m

Unit: \mathrm{N\,kg^{-1}}=\mathrm{m\,s^{-2}}

Field strength of a point mass

g=\dfrac{GM}{r^{2}}

M: mass causing the field; r: distance from its centre

Unit: \mathrm{N\,kg^{-1}}

g at the Earth's surface

g=\dfrac{GM_E}{R_E^{2}}\approx9.8\,\mathrm{m\,s^{-2}}

M_E=6\times10^{24}\,\mathrm{kg}: mass of the Earth; R_E=6.4\times10^{6}\,\mathrm{m}: radius of the Earth

Unit: \mathrm{m\,s^{-2}}

g on any planet

g=G\,\dfrac{m}{r^{2}}

m: mass of the planet; r: its radius

Unit: \mathrm{m\,s^{-2}}

Mass of the Earth

GM_E=gR_E^{2}\ \Rightarrow\ M_E=\dfrac{gR_E^{2}}{G}

g: surface field strength; R_E: radius of the Earth

Unit: \mathrm{kg}

g at height h

g_h=\dfrac{gR_E^{2}}{(R_E+h)^{2}}=g\left(\dfrac{R_E}{R_E+h}\right)^{2}

g_h: field strength at height h above the surface; g: surface value

Unit: \mathrm{m\,s^{-2}}

g at small heights

g_h\approx g\left(1-\dfrac{2h}{R_E}\right)\quad(h\ll R_E)

fractional decrease 2h/R_E

Unit: \mathrm{m\,s^{-2}}

g inside a uniform sphere

g_r=g\,\dfrac{r}{R}\quad(r\le R)

r: distance from the centre; R: radius of the sphere

Unit: \mathrm{m\,s^{-2}}

Orbital velocity

v_o=\sqrt{gr}=\sqrt{\dfrac{GM_E}{r}}=\sqrt{\dfrac{GM_E}{R_E+h}}

r=R_E+h: orbit radius from the Earth's centre; g: field strength at the orbit

Unit: \mathrm{m\,s^{-1}}

Orbital velocity just above the surface

v_o=\sqrt{gR_E}\approx7.9\,\mathrm{km\,s^{-1}}

g=9.8\,\mathrm{m\,s^{-2}}, R_E=6.4\times10^{6}\,\mathrm{m}

Unit: \mathrm{m\,s^{-1}}

Period of a satellite

T=\dfrac{2\pi r}{v_o}=2\pi\sqrt{\dfrac{r^{3}}{GM}}

T: time for one orbit; r: orbit radius; M: mass of the central body

Unit: \mathrm{s}

Kinetic energy of a satellite

K.E.=\tfrac12mv_o^{2}=\dfrac{GMm}{2r}

m: mass of the satellite

Unit: \mathrm{J}

Radius of a geostationary (synchronous) orbit

r=\left(\dfrac{GM_ET^{2}}{4\pi^{2}}\right)^{1/3}

T=24\,\mathrm{h}=86\,400\,\mathrm{s} for the Earth; gives r=4.23\times10^{7}\,\mathrm{m}, h=3.59\times10^{7}\,\mathrm{m}, v_o\approx3.07\,\mathrm{km\,s^{-1}}

Unit: \mathrm{m}

Gravitational potential energy

U=-\dfrac{GM_Em}{r}

U=0 at infinity; r: distance from the Earth's centre (r=R_E at the surface)

Unit: \mathrm{J}

Change in potential energy

\Delta U=GMm\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\approx mgh\ \ (h\ll R_E)

moving from r_1 to r_2

Unit: \mathrm{J}

Gravitational potential (definition)

V=\dfrac{U}{m}

U: potential energy of a small test mass m at the point

Unit: \mathrm{J\,kg^{-1}}

Gravitational potential of the Earth (point mass)

V=\phi=-\dfrac{GM_E}{r}

r: distance from the centre; V=-62.5\,\mathrm{MJ\,kg^{-1}} at the surface

Unit: \mathrm{J\,kg^{-1}}

Potential energy from potential

U=mV=-\dfrac{GMm}{r}

V: potential due to M at the position of m

Unit: \mathrm{J}

Key terms

Force of gravity: The force of attraction that any two objects exert on each other because of their masses.
Newton's law of universal gravitation: Every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres: F_g=Gm_1m_2/r^{2}.
Gravitational constant (G): The constant in Newton's law of gravitation, 6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}; numerically the force between two 1\,\mathrm{kg} masses 1\,\mathrm{m} apart.
Inverse-square law: A law in which a quantity is proportional to 1/r^{2}: doubling the distance reduces it to one quarter.
Gravitational field: The region around a massive object in which its gravitational force acts on other masses.
Gravitational field strength (g): The gravitational force per unit mass at a point, g=F_g/m; a vector towards the attracting mass, unit \mathrm{N\,kg^{-1}} or \mathrm{m\,s^{-2}}. It equals the free-fall acceleration.
Field lines: Lines that show the direction of the gravitational field (arrows) and its strength (closer lines, stronger field).
Test mass: A small mass used to measure a field without disturbing it.
Point mass: A body treated as if all its mass were concentrated at one point.
Uniform sphere: A sphere with the same density throughout; outside it, its field is that of a point mass at its centre.
Altitude: Height h above the Earth's surface.
Satellite: Any object that orbits another object due to the force of gravity, maintaining a stable path around it.
Natural satellite: A natural body (such as a moon) orbiting a planet, dwarf planet or minor planet.
Artificial satellite: An object intentionally placed into orbit around the Earth or another celestial body.
Space station: A spacecraft that can support a crew, designed to stay in space for a long time, to which other spacecraft can dock (for example the ISS).
Polar orbit: An orbit that passes over or near both poles of the Earth.
Low Earth Orbit (LEO): An orbit a few hundred kilometres above the surface, such as that of the ISS at about 400\,\mathrm{km}.
Orbital velocity: The minimum velocity required for a satellite to orbit a body at a specific altitude; in a circular orbit it is the constant tangential speed v_o=\sqrt{GM/r}.
Escape speed: The launch speed at which a body escapes from the Earth's gravity and never returns.
Time period of a satellite: The time taken for one complete orbit, T=2\pi r/v_o.
Geostationary satellite: A satellite that orbits directly above the equator, from west to east, with a period of 24 hours, so it appears stationary above a fixed point on the Earth; also called a geosynchronous equatorial orbit (GEO) satellite.
Geostationary orbit: The equatorial circular orbit of radius about 4.23\times10^{7}\,\mathrm{m} (height about 3.59\times10^{7}\,\mathrm{m}) with a period of 24 hours.
Hill sphere: The region around a celestial body in which its gravitational pull is stronger than that of a larger nearby body.
Absolute gravitational potential energy: The potential energy of a body at a point with respect to the zero at infinity: the work done in bringing it from infinity to that point, U=-GMm/r.
Gravitational potential: The gravitational potential energy per unit mass at a point: the work done per unit mass in bringing a small test mass from infinity to that point, V=\phi=-GM/r; unit \mathrm{J\,kg^{-1}}.

Multiple choice questionsp. 23

Two identical balls of masses 1\,\mathrm{kg} each having distance of 1\,\mathrm{m} between their centres then gravitational force between them is:

A667\times10^{-9}\,\mathrm{N}
B6.67\times10^{-11}\,\mathrm{N}Answer
C667\times10^{11}\,\mathrm{N}
D6.67\times10^{-13}\,\mathrm{N}

Why

F=G\dfrac{m_1m_2}{r^{2}}=6.67\times10^{-11}\times\dfrac{1\times1}{1^{2}}=6.67\times10^{-11}\,\mathrm{N}. This is just the value of G: G is numerically the force between two 1\,\mathrm{kg} masses 1\,\mathrm{m} apart. Option A equals 6.67\times10^{-7}\,\mathrm{N}, which is 10^{4} times too large.

Gravitational force between two objects is F. If masses of bodies are doubled and distance between their centres is reduced to half, then gravitational force is:

AF
B4F
CF/4
D16FAnswer

Why

F'=G\dfrac{(2m_1)(2m_2)}{(r/2)^{2}}=\dfrac{4}{1/4}\,G\dfrac{m_1m_2}{r^{2}}=16F. Doubling both masses gives a factor of 4 and halving the distance gives another factor of 4. Choosing B forgets the effect of the distance.

32026 Annual I, Set B, Q.2 (iv) OR

The value of g at height of 1500\,\mathrm{km} above the surface of Earth in \mathrm{m\,s^{-2}} is:

A0
B6.4Answer
C9.8
D12.2

Why

g_h=g\left(\dfrac{R_E}{R_E+h}\right)^{2}=9.8\left(\dfrac{6400}{7900}\right)^{2}=6.4\,\mathrm{m\,s^{-2}}. g must decrease with height, so C and D are impossible, and it becomes zero only at infinity, so A is wrong.

If we consider Earth as perfect sphere and ignore the presence of air resistance. What will be the orbital velocity to launch a satellite in circular orbit just above the surface at r=R_E=6.4\times10^{6}\,\mathrm{m}:

A0\,\mathrm{km\,s^{-1}}
B1.19\,\mathrm{km\,s^{-1}}
C3.07\,\mathrm{km\,s^{-1}}
D7.9\,\mathrm{km\,s^{-1}}Answer

Why

v_o=\sqrt{gR_E}=\sqrt{9.8\times6.4\times10^{6}}=7.9\times10^{3}\,\mathrm{m\,s^{-1}}=7.9\,\mathrm{km\,s^{-1}}. Option C (3.07\,\mathrm{km\,s^{-1}}) is the speed of a geostationary satellite, which is much farther out.

When a satellite is put into a higher circular orbit, its kinetic energy:

Aincreases
BdecreasesAnswer
Cis zero
Dremains the same

Why

K.E.=\tfrac12mv_o^{2}=\dfrac{GMm}{2r}. A larger orbit radius r means a smaller orbital speed and so a smaller kinetic energy. (Its potential energy and its total energy increase, which is why rockets must supply energy to raise the orbit.)

The orbital velocity of geostationary satellite is

A0\,\mathrm{km\,s^{-1}}
B1.19\,\mathrm{km\,s^{-1}}
C3.07\,\mathrm{km\,s^{-1}}Answer
D7.9\,\mathrm{km\,s^{-1}}

Why

v_o=\dfrac{2\pi r}{T}=\dfrac{2\pi\times4.23\times10^{7}}{86\,400}\approx3.07\times10^{3}\,\mathrm{m\,s^{-1}}. Option A is tempting because the satellite looks stationary from the ground, but it only keeps pace with the rotating Earth; it still moves at about 3\,\mathrm{km\,s^{-1}} around the Earth's centre.

The minimum number of geostationary satellites for the complete global coverage is

A1
B2
C3Answer
D4

Why

One geostationary satellite sees about 42\% of the Earth's surface, a band about 163^\circ wide along the equator. Two such bands (326^\circ) leave gaps; three satellites 120^\circ apart overlap and cover the whole equatorial belt, all except small regions near the poles.

When an object moves away from a massive body, its gravitational potential energy:

AincreasesAnswer
Bdecreases
Cremains constant
Dbecomes zero

Why

U=-GMm/r is negative and becomes less negative as r increases, so it increases towards zero. Work has to be done against gravity to move the object away. It becomes zero only at infinity, so D is wrong.

The value of gravitational potential at an altitude of 35\,700\,\mathrm{km} above the Earth's surface, where communication satellites orbit the Earth is approximately:

A-94.7\,\mathrm{MJ\,kg^{-1}}Answer
B-62.3\,\mathrm{MJ\,kg^{-1}}
C+947.11\,\mathrm{J\,kg^{-1}}
D-947.11\,\mathrm{J\,kg^{-1}}

Why

r=R_E+h=6.4\times10^{6}+3.57\times10^{7}=4.21\times10^{7}\,\mathrm{m}, so V=-\dfrac{GM_E}{r}=-\dfrac{6.67\times10^{-11}\times6\times10^{24}}{4.21\times10^{7}}=-9.5\times10^{6}\,\mathrm{J\,kg^{-1}}\approx-9.5\,\mathrm{MJ\,kg^{-1}} (the book's M_E=5.98\times10^{24}\,\mathrm{kg} gives -9.47\,\mathrm{MJ\,kg^{-1}}).

None of the printed options equals this. Option A has the right digits and unit but a misplaced decimal point: it should read -9.47\,\mathrm{MJ\,kg^{-1}}. As printed, -94.7\,\mathrm{MJ\,kg^{-1}} is impossible, because it is more negative than the surface value -62.5\,\mathrm{MJ\,kg^{-1}}. B is the surface value, C has the wrong sign (the potential is always negative) and D is 10^{4} times too small.

Note. Misprint in the book: no option is correct as printed. The intended answer is A, which should read -9.47\,\mathrm{MJ\,kg^{-1}}.

Short questionsp. 24

13 marks

Why can gravitation not account for the formation of molecules?

Solution

Molecules are held together by electric (electromagnetic) forces between the charged particles of atoms: electrons are attracted by nuclei, and like charges repel. Gravitation is far too weak to do this, because the masses of atoms are tiny (about 10^{-27} to 10^{-25}\,\mathrm{kg}) and G is very small. For two protons the electric force is about 10^{36} times the gravitational force:

\dfrac{F_e}{F_g}=\dfrac{ke^{2}}{Gm_p^{2}}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^{2}}{6.67\times10^{-11}\times(1.67\times10^{-27})^{2}}\approx1.2\times10^{36}

Gravity is also always attractive, so it cannot provide the balance of attraction and repulsion that fixes bond lengths and shapes. Chemical bonding is therefore electrical, not gravitational.

23 marks

Why don't two books on your desk attract each other gravitationally, despite Newton's law of gravitation?

Solution

They do attract each other, but the force is extremely small because G is tiny and the masses of books are small. For two 1\,\mathrm{kg} books whose centres are 0.5\,\mathrm{m} apart:

F=G\dfrac{m_1m_2}{r^{2}}=\dfrac{6.67\times10^{-11}\times1\times1}{(0.5)^{2}}=2.7\times10^{-10}\,\mathrm{N}

The static friction between a book and the desk can be a few newtons, about 10^{10} times larger. So the gravitational pull cannot overcome friction and the books do not move towards each other. Gravity becomes noticeable only when at least one body is very massive, like the Earth.

33 marks

Why does an apple fall towards the Earth due to gravity, while the Earth doesn't move towards the apple?

Solution

By Newton's third law the apple pulls the Earth with a force equal and opposite to the Earth's pull on the apple; for a 0.1\,\mathrm{kg} apple each force is about 0.98\,\mathrm{N}. The effect of a force depends on the mass it acts on, a=F/m:

a_{\text{apple}}=\dfrac{0.98}{0.1}=9.8\,\mathrm{m\,s^{-2}},\qquad a_{\text{Earth}}=\dfrac{0.98}{6\times10^{24}}\approx1.6\times10^{-25}\,\mathrm{m\,s^{-2}}

So the Earth does move towards the apple, but its acceleration is far too small to notice because its mass is enormous. The apple, with its tiny mass, is accelerated strongly and visibly falls.

43 marks

Can gravitational field strength be negative? Explain.

Solution

The magnitude of the gravitational field strength, g=GM/r^{2}, can never be negative, because G, M and r^{2} are all positive. It can become zero (far away at infinity, or at the centre of the Earth), but not negative.

However, \vec g is a vector, and a minus sign can appear in its component. Gravity is always attractive, so \vec g always points towards the mass. If we take the outward (upward) direction as positive, then

\vec g=-\dfrac{GM}{r^{2}}\,\hat r\qquad\text{or}\qquad g=-9.8\,\mathrm{m\,s^{-2}}\ \text{(taking upward as positive)}

The negative sign only shows the direction (towards the centre); it does not mean a negative strength.

52026 Annual I, Set B, Q.2 (xii)3 marks

What factors determine the strength of the gravitational field around a planet?

Solution

The field strength around a planet of mass M is g=\dfrac{GM}{r^{2}}, where r is the distance from the planet's centre. So it is determined by:

The mass of the planet: g\propto M. The more massive the planet, the stronger its field.
The distance from the planet's centre: g\propto1/r^{2}. At the surface r=R, so a planet with a smaller radius (for the same mass) has a stronger surface field. Above the surface g_h=GM/(R+h)^{2}, so g also falls with height.

It does not depend on the mass, shape or velocity of the body placed in the field. Example: Jupiter's mass is 316 times the Earth's and its radius about 11.2 times, so its surface field is 316/11.2^{2}\approx2.5 times the Earth's, about 24.8\,\mathrm{m\,s^{-2}}.

63 marks

If two planets have the same mass but different radii. How would their gravitational field strengths compare?

Solution

At the surface of a planet, g=\dfrac{GM}{R^{2}}. With the same mass M, the surface field strength is inversely proportional to the square of the radius:

\dfrac{g_1}{g_2}=\dfrac{R_2^{2}}{R_1^{2}}

So the smaller planet has the stronger surface field. For example, if planet 2 has twice the radius of planet 1, its surface field is only one quarter as strong: g_2=g_1/4. However, at the same distance r from their centres (outside both planets) the two fields are equal, because the field outside a sphere depends only on its mass and on r.

72026 Annual I, Set B, Q.2 (i)2026 Annual I, Set D, Q.2 (i)3 marks

Why satellites in higher orbits have lower orbital velocities?

Solution

For a circular orbit, gravity provides the centripetal force:

\dfrac{GMm}{r^{2}}=\dfrac{mv_o^{2}}{r}\quad\Rightarrow\quad v_o=\sqrt{\dfrac{GM}{r}}\quad\Rightarrow\quad v_o\propto\dfrac{1}{\sqrt r}

In a higher orbit r is larger and the Earth's pull is weaker (it falls as 1/r^{2}). A weaker pull can hold a satellite on a circle only if it moves more slowly; a fast satellite would fly outward. Example: at 400\,\mathrm{km} (ISS) v_o=7.7\,\mathrm{km\,s^{-1}}, while in the geostationary orbit (r=4.23\times10^{7}\,\mathrm{m}) v_o=3.1\,\mathrm{km\,s^{-1}}. Four times the radius gives half the speed.

83 marks

How does the mass of the Earth affect the orbital velocity required for a satellite to stay in orbit?

Solution

The orbital velocity is

v_o=\sqrt{\dfrac{GM_E}{r}}\quad\Rightarrow\quad v_o\propto\sqrt{M_E}\ \text{ (for the same orbit radius)}

A more massive Earth would pull the satellite more strongly. To stay on the same circle the satellite would then need a larger centripetal force mv_o^{2}/r, that is, a higher speed; otherwise it would spiral inward. For example, if the Earth's mass were 4 times larger, the orbital velocity at the same radius would double. A less massive planet needs a lower orbital speed: around Mars, 300\,\mathrm{km} up, only about 3.4\,\mathrm{km\,s^{-1}} is needed. The satellite's own mass does not matter, because it cancels.

93 marks

Is it possible for an object's gravitational potential energy to become negative? If so, what does this mean for the object's motion?

Solution

Yes. We take the gravitational potential energy to be zero at infinity. Then at any finite distance r from the Earth's centre

U=-\dfrac{GM_Em}{r}<0

It is negative because gravity does positive work as the object comes in from infinity, lowering its energy below zero. For the motion, negative U means the object is bound to the Earth: it cannot escape to infinity unless it is given at least GM_Em/r of extra energy. If its total energy K.E.+U is negative, it either falls back or keeps moving in an orbit, like the Moon and satellites. As it moves closer, U becomes more negative and its kinetic energy increases.

103 marks

How the gravitational potential energy of two point masses is related to concept of gravitational potential?

Solution

The gravitational potential at a point is the potential energy per unit mass, V=U/m. So the potential energy of a mass m at a point is its mass times the potential there, U=mV.

A point mass M creates the potential V=-\dfrac{GM}{r} at distance r. Placing a second point mass m at that point gives the potential energy of the pair:

U=mV=m\left(-\dfrac{GM}{r}\right)=-\dfrac{GMm}{r}

The potential describes the point itself (it depends only on M and r), while the potential energy also depends on the mass placed there. Using the potential of m at the position of M gives the same U, so U belongs to the pair of masses.

Comprehensive questionsp. 24

What is force of gravity? State and explain Newton's law of gravitation. Also show that Newton's law of gravitation is consistent with Newton's 3rd law of motion.

Solution

Force of gravity

The force of gravity (gravitational force) is the force of attraction that any two objects exert on each other because of their masses. It keeps us on the ground, holds the Moon in orbit around the Earth and the planets around the Sun.

Newton's law of universal gravitation

Statement: every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Derivation Newton's law in equation form

Aim. Write the law of gravitation as an equation for two bodies of masses m_1 and m_2 whose centres are a distance r apart.

Assumptions.

The bodies are point masses or uniform spheres, so r is measured from centre to centre.
No other masses act on them.

Two bodies of masses m_1 and m_2, centres r apart, attract each other with forces \vec F_{12} and \vec F_{21}.

The force is directly proportional to the product of the masses: doubling either mass doubles the force.
F_g\propto m_1m_2\qquad(\mathrm{i})
The force is inversely proportional to the square of the distance between the centres.
F_g\propto\dfrac{1}{r^{2}}\qquad(\mathrm{ii})
Combine (i) and (ii) into a single proportionality.
F_g\propto\dfrac{m_1m_2}{r^{2}}
Replace the proportionality sign by an equals sign and a constant of proportionality G, the gravitational constant.
F_g=G\,\dfrac{m_1m_2}{r^{2}}

ResultF_g=G\,\dfrac{m_1m_2}{r^{2}},\qquad G=6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}

The force grows with the masses and falls off quickly with distance. G is so small that the force is noticeable only when at least one of the masses is huge, like a planet or a star.

G is the universal gravitational constant, G=6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}. It is numerically equal to the force between two 1\,\mathrm{kg} masses placed 1\,\mathrm{m} apart.

Explanation

The force is always attractive and acts along the line joining the centres of the two bodies.
It does not depend on the medium between the bodies.
It obeys the inverse-square law: if the distance is doubled the force becomes \tfrac14; if it is tripled the force becomes \tfrac19.
It is universal: the same law and the same G apply to an apple and the Earth, the Moon and the Earth, the planets and the Sun.
It is very weak for ordinary objects because G is tiny (two 45\,\mathrm{kg} and 50\,\mathrm{kg} bodies 1\,\mathrm{m} apart attract with only 1.5\times10^{-7}\,\mathrm{N}), but very large when a planet or star is involved.

Consistency with Newton's third law

Let \vec F_{21} be the force on m_2 due to m_1 and \vec F_{12} the force on m_1 due to m_2. The law gives the same magnitude for both, because the expression Gm_1m_2/r^{2} is symmetric in the two masses:

\left|\vec F_{12}\right|=\left|\vec F_{21}\right|=G\,\dfrac{m_1m_2}{r^{2}}

\vec F_{12} points from m_1 towards m_2 and \vec F_{21} points from m_2 towards m_1, so they are opposite in direction:

\vec F_{12}=-\vec F_{21}

The two forces are equal in magnitude, opposite in direction, act on different bodies and arise from the same interaction. This is exactly an action and reaction pair, so Newton's law of gravitation is consistent with Newton's third law of motion. For example, the Earth pulls a falling apple down and the apple pulls the Earth up with the same force; the Earth's acceleration is negligible only because its mass is huge.

What are gravitational field and gravitational field strength? Explain. Derive the formula for gravitational acceleration g on the surface of Earth and discuss the variation of g with altitude.

Solution

Gravitational field

The gravitational field is the region (space) around a massive body, such as the Earth, in which its gravitational force acts on other masses. It is represented by field lines. The arrows show the direction of the force and the closeness of the lines shows the strength. Around a point mass (or a uniform sphere) the lines point radially inward, so the field is strongest near the body and weakens with distance.

Gravitational field strength

The gravitational field strength \vec g at a point is the gravitational force acting on a unit mass placed at that point:

\vec g=\dfrac{\vec F_g}{m}

It is a vector pointing towards the attracting mass. Its SI unit is \mathrm{N\,kg^{-1}}, which equals \mathrm{m\,s^{-2}}. By Newton's second law a freely falling mass has acceleration a=F_g/m=g, so the field strength equals the acceleration due to gravity. For a point mass M at distance r, g=GM/r^{2}.

Derivation of g on the surface of the Earth

Derivation Value of g on the surface of the Earth

Aim. Derive g=\dfrac{GM_E}{R_E^{2}} and find its value.

Assumptions.

The Earth is a uniform sphere of mass M_E and radius R_E, so it attracts as if its mass were at its centre.
The stone is tiny compared with the Earth, so the distance between their centres is R_E.

A stone of mass m_0 on the surface, a distance R_E from the centre O.

Newton's law of gravitation gives the force between the stone and the Earth, whose centres are R_E apart.
F_g=G\,\dfrac{m_0M_E}{R_E^{2}}\qquad(15.1)
This force is the weight of the stone. By Newton's second law, weight equals mass times the acceleration due to gravity.
F_g=W=m_0g\qquad(15.2)
Both equations describe the same force, so equate them.
m_0g=G\,\dfrac{m_0M_E}{R_E^{2}}
Cancel m_0 from both sides.
g=G\,\dfrac{M_E}{R_E^{2}}\qquad(15.3)
Put in G=6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}, M_E=6\times10^{24}\,\mathrm{kg} and R_E=6.4\times10^{6}\,\mathrm{m}.
g=\dfrac{6.67\times10^{-11}\times6\times10^{24}}{(6.4\times10^{6})^{2}}=9.77\,\mathrm{m\,s^{-2}}\approx9.8\,\mathrm{m\,s^{-2}}

Resultg=\dfrac{GM_E}{R_E^{2}}\approx9.8\,\mathrm{m\,s^{-2}}

The mass m_0 of the body has cancelled, so (without air resistance) light and heavy bodies fall with the same acceleration. g depends only on the mass and radius of the Earth.

Variation of g with altitude

Derivation g at height h above the surface

Aim. Express the field strength g_h at height h in terms of the surface value g.

Assumptions.

The Earth is a uniform sphere of mass M_E and radius R_E, so it acts as a point mass at its centre O.
The body at height h is small.

A body at height h is a distance R_E+h from the centre.

At height h the distance from the centre is R_E+h, so replace R_E by R_E+h in Eq. (15.3).
g_h=G\,\dfrac{M_E}{(R_E+h)^{2}}\qquad(15.5)
At the surface g=GM_E/R_E^{2}. Rearrange it.
GM_E=gR_E^{2}\qquad(15.6)
Put Eq. (15.6) in Eq. (15.5). This removes G and M_E, leaving only distances.
g_h=\dfrac{gR_E^{2}}{(R_E+h)^{2}}=g\left(\dfrac{R_E}{R_E+h}\right)^{2}\qquad(15.7)
This is the exact result. Rewrite it with h/R_E to see how small heights behave.
g_h=g\left(1+\dfrac{h}{R_E}\right)^{-2}
Close to the surface h\ll R_E. The binomial approximation (1+x)^{n}\approx1+nx, with x=h/R_E and n=-2, gives a simple straight-line rule.
g_h\approx g\left(1-\dfrac{2h}{R_E}\right)
Use the exact form (15.7) whenever h is not small compared with R_E, for example for satellites.

Resultg_h=g\left(\dfrac{R_E}{R_E+h}\right)^{2}\;\approx\;g\left(1-\dfrac{2h}{R_E}\right)\ \text{ for } h\ll R_E

As we go farther from the centre (or the surface) of the Earth, g decreases. At h=R_E it is exactly g/4. Near the surface it drops by only about 0.03\% per kilometre.

Discussion. g decreases as the altitude increases, because the distance from the centre increases. The decrease is very small near the surface (g=9.8\,\mathrm{m\,s^{-2}} on Mount Everest to 2 significant figures) but large far away: 8.7\,\mathrm{m\,s^{-2}} at 400\,\mathrm{km}, g/4 at h=R_E, and only about 0.2\,\mathrm{m\,s^{-2}} at 35\,700\,\mathrm{km} where communication satellites orbit.

How do engineers calculate the required velocity for a satellite to be placed into a specific orbit? Can you describe the steps involved in determining the orbital parameters necessary for a successful satellite launch and deployment?

Solution

Required velocity: the orbital velocity

In a circular orbit the only force on the satellite is gravity, and it must provide exactly the centripetal force needed. Engineers use this condition to find the required speed.

Derivation Orbital velocity

Aim. Find the speed v_o needed for a circular orbit of radius r around the Earth.

Assumptions.

The orbit is a circle of radius r measured from the Earth's centre.
Gravity is the only force on the satellite (no air resistance).
The satellite is tiny compared with the Earth.

Gravity towards the centre O provides the centripetal force for a satellite moving with speed v_o in an orbit of radius r=R_E+h.

A body moving with speed v_o in a circle of radius r has a centripetal acceleration towards the centre.
a_c=\dfrac{v_o^{2}}{r}
This acceleration is supplied by gravity, so it equals the field strength g at the orbit.
a_c=g\quad\Rightarrow\quad g=\dfrac{v_o^{2}}{r}
Make v_o the subject.
v_o^{2}=gr\quad\Rightarrow\quad v_o=\sqrt{gr}
At distance r from the centre the field is g=GM_E/r^{2}. Substitute it.
v_o^{2}=\dfrac{GM_E}{r^{2}}\cdot r=\dfrac{GM_E}{r}\quad\Rightarrow\quad v_o=\sqrt{\dfrac{GM_E}{r}}\qquad(15.8)
The satellite is small, so r is the Earth's radius plus the height h of the satellite above the surface.
v_o=\sqrt{\dfrac{GM_E}{R_E+h}}\qquad(15.9)

Resultv_o=\sqrt{\dfrac{GM_E}{r}}=\sqrt{\dfrac{GM_E}{R_E+h}}

Steps in finding the orbital parameters

Choose the orbit for the mission. Space stations and space telescopes use low orbits (a few hundred km); mapping and weather satellites use polar orbits; communication satellites use the geostationary orbit. This fixes the height h and the plane of the orbit.
Find the orbit radius from the Earth's centre: r=R_E+h.
Calculate the orbital velocity, v_o=\sqrt{GM_E/r}.
Calculate the period, T=2\pi r/v_o=2\pi\sqrt{r^{3}/GM_E}. If the mission needs a particular period (24 hours for a geostationary satellite), work backwards: r=\left(GM_ET^{2}/4\pi^{2}\right)^{1/3}.
Launch. Rockets lift the satellite to height h, then turn and accelerate it parallel to the surface until its speed is exactly v_o. Too little speed and it falls back; too much and it goes into an elliptical orbit or escapes. Launching towards the east, near the equator, adds the Earth's own rotation speed (about 0.46\,\mathrm{km\,s^{-1}} at the equator) and saves fuel.
Deploy and adjust. The satellite is released and small thrusters correct its speed and direction until the orbit is circular at the planned height (for a geostationary satellite: above the equator, moving west to east).

The launch speed decides the path: too slow (1), orbital speed (2), faster (3), escape speed (4).

Example

The Hubble Space Telescope orbits 613\,\mathrm{km} above the surface: r=6.4\times10^{6}+0.613\times10^{6}=7.01\times10^{6}\,\mathrm{m}, so v_o=\sqrt{\dfrac{6.67\times10^{-11}\times6\times10^{24}}{7.01\times10^{6}}}=7.55\times10^{3}\,\mathrm{m\,s^{-1}} and T=\dfrac{2\pi r}{v_o}=5.83\times10^{3}\,\mathrm{s}\approx97\,\mathrm{min} (Numerical 4).

What are geostationary satellites? Calculate the orbital radius r, orbital speed v_o and height h above surface of Earth for the geostationary orbit.

Solution

Geostationary satellites

A geostationary satellite is a satellite that orbits the Earth directly above the equator, from west to east, with a period of 24 hours. Because it turns at the same rate as the Earth, it remains stationary above a fixed point on the Earth's surface. Conditions:

circular orbit in the plane of the equator;
motion from west to east (the direction of the Earth's rotation);
period T=24\,\mathrm{h}=86\,400\,\mathrm{s}.

The satellite S and the point P below it turn through the same angle in the same time, so S stays above P.

Orbital radius

Derivation Orbital radius of a geostationary satellite

Aim. Find the radius r of a circular orbit whose period is T (24 hours for a geostationary satellite).

Assumptions.

The orbit is a circle of radius r measured from the Earth's centre.
The satellite (mass m) moves under gravity alone.

Gravity provides the centripetal force, so the orbital velocity is
v_o=\sqrt{\dfrac{GM_E}{r}}\qquad(\mathrm{i})
In one period T the satellite goes once round the circle, a distance equal to the circumference 2\pi r.
v_o=\dfrac{S}{t}=\dfrac{2\pi r}{T}\qquad(\mathrm{ii})
Both expressions give the same speed, so equate (i) and (ii).
\dfrac{2\pi r}{T}=\sqrt{\dfrac{GM_E}{r}}
Square both sides to remove the root.
\dfrac{4\pi^{2}r^{2}}{T^{2}}=\dfrac{GM_E}{r}
Multiply both sides by rT^{2}/4\pi^{2} to collect the powers of r.
r^{3}=\dfrac{GM_ET^{2}}{4\pi^{2}}
Take the cube root of both sides.
r=\left(\dfrac{GM_ET^{2}}{4\pi^{2}}\right)^{1/3}\qquad(15.10)

Resultr=\left(\dfrac{GM_ET^{2}}{4\pi^{2}}\right)^{1/3}

The radius depends only on the period and the Earth's mass. Every geostationary satellite has T=24\,\mathrm{h}, so they all share one orbit radius, whatever their mass.

Calculation

Data: G=6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}, M_E=6\times10^{24}\,\mathrm{kg}, R_E=6.4\times10^{6}\,\mathrm{m}, T=86\,400\,\mathrm{s}. Orbital radius:
r=\left(\dfrac{6.67\times10^{-11}\times6\times10^{24}\times(86\,400)^{2}}{4\times(3.14)^{2}}\right)^{1/3}=4.23\times10^{7}\,\mathrm{m}
Orbital speed (the book's 3.07 uses M_E=5.98\times10^{24}\,\mathrm{kg}):
v_o=\sqrt{\dfrac{GM_E}{r}}=\sqrt{\dfrac{6.67\times10^{-11}\times6\times10^{24}}{4.23\times10^{7}}}=3.08\times10^{3}\,\mathrm{m\,s^{-1}}\approx3.1\,\mathrm{km\,s^{-1}}
Check the speed with the circumference and the period.
v_o=\dfrac{2\pi r}{T}=\dfrac{2\pi\times4.23\times10^{7}}{86\,400}=3.08\times10^{3}\,\mathrm{m\,s^{-1}}
Height above the surface:
h=r-R_E=4.23\times10^{7}-0.64\times10^{7}=3.59\times10^{7}\,\mathrm{m}\approx36\,000\,\mathrm{km}

Uses: communication (television, telephone and internet, with fixed dish antennas) and weather monitoring of the same region. Three geostationary satellites 120^\circ apart can cover almost the whole Earth.

r=4.23\times10^{7}\,\mathrm{m},\qquad v_o\approx3.07\times10^{3}\,\mathrm{m\,s^{-1}},\qquad h=3.59\times10^{7}\,\mathrm{m}

52026 Annual I, Set B, Q.32026 Annual I, Set D, Q.3

If M_E is the mass of Earth and r is the distance of unit mass from the centre of Earth and G is the universal gravitational constant. Prove that gravitational potential V can be written as: V=-\dfrac{GM_E}{r}

Solution

Gravitational potential

The gravitational potential V at a point is the work done per unit mass in bringing a small test mass from infinity to that point. It is the gravitational potential energy per unit mass, V=U/m, a scalar measured in \mathrm{J\,kg^{-1}}. The zero of potential is at infinity.

Derivation

Gravity changes with distance, so the work cannot be found with W=Fd in one go. We split the path into small steps, find the work in each, and add.

Derivation Derivation of V=-\dfrac{GM_E}{r}

Aim. Find the work done per unit mass in bringing a small test mass m from infinity to a point at distance r from the centre of the Earth.

Assumptions.

The Earth is a uniform sphere of mass M_E and radius R_E; it acts as a point mass at its centre O.
The mass m is moved slowly, so its kinetic energy does not change and the work done on it by the agent goes into potential energy.
U=0 at infinity.

The path from the surface (point 0, distance R_E from the centre O) to a very distant point n is divided into small steps \Delta r; in each step the force is taken as constant.

By the work and energy principle, when m is moved slowly the work done on it by the external agent equals the change in its potential energy.
\Delta P.E.=W_{net}\qquad(1)
The force changes with distance, so W=Fd cannot be used for the whole path at once. Divide the path between the surface (point 0) and a very distant point n into n small steps of length \Delta r, so small that the force is almost constant in each. The total work is the sum of the works in the steps.
W_{net}=W_1+W_2+W_3+\dots+W_{n-1}+W_n\qquad(2)
Take the step between point 1 (distance r_1) and point 0 (distance R_E). To bring m in slowly, the agent pulls outward with a force F_{av} equal to the pull of gravity, while m moves inward through \Delta r. Force and displacement are opposite: \theta=180^\circ and \cos180^\circ=-1.
W_1=F_{av}\,\Delta r\cos180^\circ=-F_{av}\,\Delta r\qquad(3)
The average force in this step follows from Newton's law, using the average distance r_{av} of the step.
F_{av}=\dfrac{GM_Em}{r_{av}^{2}}\qquad(4)
The average distance is neither R_E nor r_1 but their mean. From the figure \Delta r=r_1-R_E, so r_1=R_E+\Delta r.
r_{av}=\dfrac{R_E+r_1}{2}=\dfrac{2R_E+\Delta r}{2}=R_E+\dfrac{\Delta r}{2}\qquad(5\text{ to }7)
Square the average distance.
r_{av}^{2}=\left(R_E+\dfrac{\Delta r}{2}\right)^{2}=R_E^{2}+R_E\,\Delta r+\dfrac{(\Delta r)^{2}}{4}
\Delta r is very small, so the term (\Delta r)^{2}/4 is negligible compared with the others. Then put \Delta r=r_1-R_E.
r_{av}^{2}\approx R_E^{2}+R_E\,\Delta r=R_E^{2}+R_E(r_1-R_E)=R_Er_1\qquad(8,\ 9)
Put Eq. (9) in Eq. (4).
F_{av}=\dfrac{GM_Em}{R_Er_1}\qquad(10)
Put Eq. (10) and \Delta r=r_1-R_E (Eq. 11) in Eq. (3), then split the fraction.
W_1=-\dfrac{GM_Em}{R_Er_1}\,(r_1-R_E)=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_1}\right)\qquad(12)
The same working applies to every other step.
\begin{aligned}W_2&=-GM_Em\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\\ W_3&=-GM_Em\left(\dfrac{1}{r_2}-\dfrac{1}{r_3}\right)\\ &\ \ \vdots\\ W_{n-1}&=-GM_Em\left(\dfrac{1}{r_{n-2}}-\dfrac{1}{r_{n-1}}\right)\\ W_n&=-GM_Em\left(\dfrac{1}{r_{n-1}}-\dfrac{1}{r_n}\right)\end{aligned}\qquad(13\text{ to }16)
Add all the steps as in Eq. (2). Every middle term appears once with + and once with -, so they cancel in pairs; only the first and last terms remain.
\begin{aligned}W_{net}&=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_1}+\dfrac{1}{r_1}-\dfrac{1}{r_2}+\dots+\dfrac{1}{r_{n-1}}-\dfrac{1}{r_n}\right)\\ &=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_n}\right)\end{aligned}\qquad(17)
By Eq. (1), this is the change in potential energy when m is brought from point n to the surface.
\Delta P.E.=U_{\text{surface}}-U_n=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_n}\right)\qquad(18)
Now take point n at infinity: r_n=\infty, so 1/r_n=0, and U_n=0 (our zero). The work done in bringing m from infinity to the surface is then
U=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{\infty}\right)=-\dfrac{GM_Em}{R_E}
The same steps, started from any point P at distance r from the centre instead of the surface, give the general result.
U=-\dfrac{GM_Em}{r}\qquad(15.9)
By definition the potential is the work done per unit mass, so divide by m. For a unit mass (m=1\,\mathrm{kg}) the work itself is the potential.
V=\dfrac{U}{m}=\dfrac{-GM_Em/r}{m}=-\dfrac{GM_E}{r}

ResultV=-\dfrac{GM_E}{r}

The potential is zero at infinity and negative at every finite distance; it rises towards zero as r increases. It depends only on the Earth's mass and the distance from its centre, not on the test mass. For any point mass M: V=-GM/r.

Sketch

Sketch: V (\mathrm{MJ\,kg^{-1}}) against distance from the centre

V is most negative (-62.5\,\mathrm{MJ\,kg^{-1}}) at the surface and rises towards zero as r increases.

Result. V=-GM_E/r is negative at every finite distance, zero at infinity, and increases (becomes less negative) with distance. At the Earth's surface V=-GM_E/R_E=-62.5\,\mathrm{MJ\,kg^{-1}}.

Numerical problemsp. 25

Take G=6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}, M_E=6\times10^{24}\,\mathrm{kg}, R_E=6.4\times10^{6}\,\mathrm{m} and g=9.8\,\mathrm{m\,s^{-2}} unless a question gives other values. Always measure distances from the centre of the attracting body.

Find the mass of the Earth by considering a scenario where a small object is placed on the Earth's surface such that distance between their centres is equal to the Earth's radius, i.e., 6.4\times10^{6}\,\mathrm{m}.

Solution

Given

r=R_E = 6.4\times10^{6}\,\mathrm{m}
g = 9.8\,\mathrm{m\,s^{-2}}
G = 6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}

To find

The weight of the object is the gravitational force of the Earth on it.
mg=G\,\dfrac{M_Em}{R_E^{2}}
Cancel m and make M_E the subject.
M_E=\dfrac{gR_E^{2}}{G}
Substitute.
M_E=\dfrac{9.8\times(6.4\times10^{6})^{2}}{6.67\times10^{-11}}=\dfrac{9.8\times4.096\times10^{13}}{6.67\times10^{-11}}
Evaluate.
M_E=\dfrac{4.01\times10^{14}}{6.67\times10^{-11}}=6.0\times10^{24}\,\mathrm{kg}

M_E\approx6.0\times10^{24}\,\mathrm{kg}

Book answer. 6.4\times10^{24}\,\mathrm{kg}

Note. The book's answer 6.4\times10^{24}\,\mathrm{kg} is a misprint (it repeats the digits of the radius). With g=9.8\,\mathrm{m\,s^{-2}} the mass is 6.0\times10^{24}\,\mathrm{kg}, the value the book itself uses for M_E everywhere else.

Alpha Centauri is a binary star system with two stars: Alpha Centauri A and Alpha Centauri B. The mass of Alpha Centauri A is 2.19\times10^{30}\,\mathrm{kg} and Alpha Centauri B is 1.80\times10^{30}\,\mathrm{kg}, they attract each other with a force of 2.24\times10^{25}\,\mathrm{N}. What is the distance between the two stars?

Solution

Given

m_A = 2.19\times10^{30}\,\mathrm{kg}
m_B = 1.80\times10^{30}\,\mathrm{kg}
F = 2.24\times10^{25}\,\mathrm{N}

To find

Make r the subject of Newton's law of gravitation.
F=G\,\dfrac{m_Am_B}{r^{2}}\ \Rightarrow\ r=\sqrt{\dfrac{Gm_Am_B}{F}}
Substitute.
r=\sqrt{\dfrac{6.67\times10^{-11}\times2.19\times10^{30}\times1.80\times10^{30}}{2.24\times10^{25}}}=\sqrt{\dfrac{2.63\times10^{50}}{2.24\times10^{25}}}
Evaluate.
r=\sqrt{1.17\times10^{25}}=3.43\times10^{12}\,\mathrm{m}

This is about 23 times the Earth to Sun distance (1.5\times10^{11}\,\mathrm{m}).

r\approx3.43\times10^{12}\,\mathrm{m}

Book answer. 3.43\times10^{12}\,\mathrm{m}

What will be value of g on an exoplanet (a planet outside the Solar System) whose mass is five times the mass of Earth and its radius is twice the radius of Earth?

Solution

Given

M_p = 5M_E
R_p = 2R_E
g_E = 9.8\,\mathrm{m\,s^{-2}} on the Earth

To find

Write the surface field for the exoplanet and for the Earth.
g_p=\dfrac{GM_p}{R_p^{2}},\qquad g_E=\dfrac{GM_E}{R_E^{2}}
Divide; G cancels and only the ratios remain.
\dfrac{g_p}{g_E}=\dfrac{M_p}{M_E}\times\left(\dfrac{R_E}{R_p}\right)^{2}=5\times\left(\dfrac12\right)^{2}=\dfrac54
Evaluate.
g_p=\dfrac54\times9.8=12.25\,\mathrm{m\,s^{-2}}

g_p=12.25\,\mathrm{m\,s^{-2}}

Book answer. 12.25\,\mathrm{m\,s^{-2}}

The Hubble Space Telescope is in a circular orbit 613\,\mathrm{km} above Earth's surface. The average radius of the Earth is 6.4\times10^{6}\,\mathrm{m} and the mass of Earth is 6\times10^{24}\,\mathrm{kg}. (a) What is the speed of the telescope in its orbit? (b) What is the period of the telescope's orbit?

Solution

Given

h = 613\,\mathrm{km}=6.13\times10^{5}\,\mathrm{m}
R_E = 6.4\times10^{6}\,\mathrm{m}
M_E = 6\times10^{24}\,\mathrm{kg}
G = 6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}

To find

(a) v_o
(b) T

Orbit radius: r=R_E+h=6.4\times10^{6}+0.613\times10^{6}=7.013\times10^{6}\,\mathrm{m}.

(a) Speed of the telescope in its orbit.

Use the orbital velocity.
v_o=\sqrt{\dfrac{GM_E}{r}}=\sqrt{\dfrac{6.67\times10^{-11}\times6\times10^{24}}{7.013\times10^{6}}}
Evaluate.
v_o=\sqrt{5.71\times10^{7}}=7.55\times10^{3}\,\mathrm{m\,s^{-1}}

v_o\approx7.55\times10^{3}\,\mathrm{m\,s^{-1}}

(b) Period of the telescope's orbit.

One orbit is one circumference travelled at speed v_o.
T=\dfrac{2\pi r}{v_o}=\dfrac{2\times3.14\times7.013\times10^{6}}{7.55\times10^{3}}
Evaluate.
T=5.83\times10^{3}\,\mathrm{s}\approx97\,\mathrm{min}

T\approx5.83\times10^{3}\,\mathrm{s}

Book answer. (a) 7550\,\mathrm{m\,s^{-1}}, (b) 5807\,\mathrm{s}

Note. Part (b): with the data given in the question, T=5833\,\mathrm{s}. The book's 5807\,\mathrm{s} matches slightly different Earth data (for example R_E=6.37\times10^{6}\,\mathrm{m} and M_E=5.98\times10^{24}\,\mathrm{kg}); the difference is under 0.5\%.

To launch a satellite around planet Mars at an altitude of 300\,\mathrm{km} above its surface. Mars has a mass of 6.42\times10^{23}\,\mathrm{kg} and has a radius of 3.39\times10^{6}\,\mathrm{m}, calculate its orbital velocity.

Solution

Given

M = 6.42\times10^{23}\,\mathrm{kg} mass of Mars
R = 3.39\times10^{6}\,\mathrm{m} radius of Mars
h = 300\,\mathrm{km}=0.30\times10^{6}\,\mathrm{m}

To find

The orbit radius is measured from the centre of Mars.
r=R+h=3.39\times10^{6}+0.30\times10^{6}=3.69\times10^{6}\,\mathrm{m}
Use the orbital velocity with the mass of Mars.
v_o=\sqrt{\dfrac{GM}{r}}=\sqrt{\dfrac{6.67\times10^{-11}\times6.42\times10^{23}}{3.69\times10^{6}}}
Evaluate.
v_o=\sqrt{1.16\times10^{7}}=3.41\times10^{3}\,\mathrm{m\,s^{-1}}\approx3.4\,\mathrm{km\,s^{-1}}

v_o\approx3.4\,\mathrm{km\,s^{-1}}

Book answer. 3.4\,\mathrm{km\,s^{-1}}

An asteroid orbits the sun 8.35\times10^{11}\,\mathrm{m} from it. (a) How fast must the asteroid travel to maintain its circular orbit around the sun? (b) How long will it take the asteroid to orbit the sun?

Solution

Given

r = 8.35\times10^{11}\,\mathrm{m}
M_S = 1.99\times10^{30}\,\mathrm{kg} mass of the Sun, from Example 15.5

To find

(a) v_o
(b) T

(a) Orbital speed of the asteroid.

The Sun is the central body.
v_o=\sqrt{\dfrac{GM_S}{r}}=\sqrt{\dfrac{6.67\times10^{-11}\times1.99\times10^{30}}{8.35\times10^{11}}}
Evaluate.
v_o=\sqrt{1.59\times10^{8}}=1.26\times10^{4}\,\mathrm{m\,s^{-1}}

v_o\approx1.26\times10^{4}\,\mathrm{m\,s^{-1}}

(b) Time for one orbit.

Period = circumference / speed.
T=\dfrac{2\pi r}{v_o}=\dfrac{2\pi\times8.35\times10^{11}}{1.26\times10^{4}}=4.16\times10^{8}\,\mathrm{s}
Convert to years (1\ \text{year}=365\times86\,400\,\mathrm{s}=3.15\times10^{7}\,\mathrm{s}).
T=\dfrac{4.16\times10^{8}}{3.15\times10^{7}}=13.2\ \text{years}

T\approx4.16\times10^{8}\,\mathrm{s}\approx13.2\ \text{years}

Book answer. 1.26\times10^{4}\,\mathrm{m\,s^{-1}}, 13.2 years

Note. The question does not give the mass of the Sun; the value 1.99\times10^{30}\,\mathrm{kg} from Example 15.5 reproduces the book's answers.

Mars has a mass of 6.42\times10^{23}\,\mathrm{kg} and has a period of 88\,642\,\mathrm{s}. What would be the radius of a stationary satellite for this planet?

Solution

Given

M = 6.42\times10^{23}\,\mathrm{kg}
T = 88\,642\,\mathrm{s} rotation period of Mars

To find

A stationary (synchronous) satellite has a period equal to the planet's rotation period. Use Eq. (15.10) with the mass of Mars.
r=\left(\dfrac{GMT^{2}}{4\pi^{2}}\right)^{1/3}
Substitute.
r=\left(\dfrac{6.67\times10^{-11}\times6.42\times10^{23}\times(88\,642)^{2}}{4\pi^{2}}\right)^{1/3}=\left(8.52\times10^{21}\right)^{1/3}
Evaluate.
r=2.04\times10^{7}\,\mathrm{m}\approx20\,400\,\mathrm{km}

This radius is measured from the centre of Mars; the height above its surface would be about 20\,400-3390\approx17\,000\,\mathrm{km}.

r\approx2.04\times10^{7}\,\mathrm{m}\approx20\,400\,\mathrm{km}

Book answer. 20\,428\,\mathrm{km}

Calculate the potential energy of the Moon having mass 7.35\times10^{22}\,\mathrm{kg}, relative to Earth with mass 6\times10^{24}\,\mathrm{kg} if the distance between their centres is 3.94\times10^{5}\,\mathrm{km}.

Solution

Given

m = 7.35\times10^{22}\,\mathrm{kg}
M_E = 6\times10^{24}\,\mathrm{kg}
r = 3.94\times10^{5}\,\mathrm{km}=3.94\times10^{8}\,\mathrm{m}

To find

Use the gravitational potential energy of two point masses (zero at infinity).
U=-\dfrac{GM_Em}{r}
Substitute, with r in metres.
U=-\dfrac{6.67\times10^{-11}\times6\times10^{24}\times7.35\times10^{22}}{3.94\times10^{8}}
Evaluate.
U=-\dfrac{2.94\times10^{37}}{3.94\times10^{8}}=-7.47\times10^{28}\,\mathrm{J}

The negative sign shows that the Moon is bound to the Earth.

U\approx-7.5\times10^{28}\,\mathrm{J}

Book answer. -7.4\times10^{28}\,\mathrm{J}

Note. The usual Earth to Moon distance is 3.84\times10^{5}\,\mathrm{km}, which would give -7.66\times10^{28}\,\mathrm{J}; the book's answer uses the printed 3.94\times10^{5}\,\mathrm{km}.

A 50\,\mathrm{kg} weather satellites move in circular orbits about the Earth at an altitude of 1000\,\mathrm{km}. A similar 50\,\mathrm{kg} communication satellite is at an altitude of 37\,000\,\mathrm{km} in circular orbits about the Earth. Calculate the difference in the gravitational potential energies of the two satellites in their respective orbits?

Solution

Given

m = 50\,\mathrm{kg}
h_1 = 1000\,\mathrm{km}=1.0\times10^{6}\,\mathrm{m}
h_2 = 37\,000\,\mathrm{km}=3.7\times10^{7}\,\mathrm{m}
R_E = 6.4\times10^{6}\,\mathrm{m}
M_E = 6\times10^{24}\,\mathrm{kg}

To find

\Delta U=U_2-U_1

Orbit radii from the Earth's centre.
r_1=R_E+h_1=7.4\times10^{6}\,\mathrm{m},\qquad r_2=R_E+h_2=4.34\times10^{7}\,\mathrm{m}
Potential energy of each satellite, U=-GM_Em/r, with GM_E=4.00\times10^{14}\,\mathrm{N\,m^{2}\,kg^{-1}}.
\begin{aligned}U_1&=-\dfrac{4.00\times10^{14}\times50}{7.4\times10^{6}}=-2.70\times10^{9}\,\mathrm{J}\\ U_2&=-\dfrac{4.00\times10^{14}\times50}{4.34\times10^{7}}=-4.61\times10^{8}\,\mathrm{J}\end{aligned}
Difference.
\Delta U=U_2-U_1=-4.61\times10^{8}-(-2.70\times10^{9})=2.24\times10^{9}\,\mathrm{J}\approx2.2\,\mathrm{GJ}

The higher (communication) satellite has the greater potential energy: its U is less negative.

\Delta U\approx2.2\times10^{9}\,\mathrm{J}=2.2\,\mathrm{GJ}

Book answer. 2.2\,\mathrm{GJ}

Note. The Earth's mass and radius are not given; the book's values M_E=6\times10^{24}\,\mathrm{kg} and R_E=6.4\times10^{6}\,\mathrm{m} are used.

What is the change in gravitational potential energy of a 64.5\,\mathrm{kg} astronaut, lifted from Earth's surface into a circular orbit of altitude 4.40\times10^{2}\,\mathrm{km}?

Solution

Given

m = 64.5\,\mathrm{kg}
h = 4.40\times10^{2}\,\mathrm{km}=4.40\times10^{5}\,\mathrm{m}
R_E = 6.4\times10^{6}\,\mathrm{m}
M_E = 6\times10^{24}\,\mathrm{kg}

To find

\Delta U

The height is not small compared with R_E, so use the exact change in potential energy, not mgh.
\Delta U=GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{R_E+h}\right)
Substitute; R_E+h=6.84\times10^{6}\,\mathrm{m}.
\Delta U=6.67\times10^{-11}\times6\times10^{24}\times64.5\left(\dfrac{1}{6.4\times10^{6}}-\dfrac{1}{6.84\times10^{6}}\right)
Evaluate the bracket, then multiply.
\Delta U=2.58\times10^{16}\times1.005\times10^{-8}=2.6\times10^{8}\,\mathrm{J}

\Delta U\approx2.6\times10^{8}\,\mathrm{J}

Book answer. 2.6\times10^{8}\,\mathrm{J}

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 11 mark

If the distance between the centers of two identical bodies is doubled, the gravitational force F between them will become:

AF
B2F
C\dfrac{F}{2}
D\dfrac{F}{4}Answer

Why

F\propto1/r^{2}. With the distance doubled, F'=G\dfrac{m^{2}}{(2r)^{2}}=\dfrac14\,G\dfrac{m^{2}}{r^{2}}=\dfrac{F}{4}. Option C (F/2) would be true only if the force were inversely proportional to the distance, not to its square.

2026 Annual I, Set B, MCQ 21 mark

If the gravitational field strength at the earth's surface is g. What is its value at a height, h=R_e from surface of earth?

A\dfrac{g}{2}
B\dfrac{g}{4}Answer
C2g
D\dfrac{g}{9}

Why

At h=R_e the distance from the centre is 2R_e, so g_h=g\left(\dfrac{R_e}{R_e+h}\right)^{2}=g\left(\dfrac{R_e}{2R_e}\right)^{2}=\dfrac{g}{4}. Option D (g/9) is the value at h=2R_e, where the distance from the centre is 3R_e.

2026 Annual I, Set B, Q.2 (i)1+2 = 3 marks

When a satellite is put into a higher orbit what will be the effect on its orbital velocity? Explain mathematically.

Solution

Effect (1 mark). The orbital velocity decreases: a satellite in a higher orbit moves more slowly.

Mathematical explanation (2 marks). For a circular orbit of radius r (from the Earth's centre), gravity provides the centripetal force:

\dfrac{GM_Em}{r^{2}}=\dfrac{mv_o^{2}}{r}\quad\Rightarrow\quad v_o=\sqrt{\dfrac{GM_E}{r}}\quad\Rightarrow\quad v_o\propto\dfrac{1}{\sqrt r}

A higher orbit has a larger r, so v_o is smaller. If r becomes 4 times larger, v_o becomes half. Example: the ISS at 400\,\mathrm{km} moves at 7.7\,\mathrm{km\,s^{-1}}, but a geostationary satellite (r=4.23\times10^{7}\,\mathrm{m}) moves at only 3.1\,\mathrm{km\,s^{-1}}.

2026 Annual I, Set B, Q.2 (iv) OR3 marks

Find field strength g at a height of 1500\,\mathrm{km} above surface of Earth.

Solution

Given

g = 9.8\,\mathrm{m\,s^{-2}} surface value (paper's data)
R_e = 6.4\times10^{6}\,\mathrm{m}
h = 1500\,\mathrm{km}=1.5\times10^{6}\,\mathrm{m}

To find

Formula (1 mark). The height is not small compared with R_e, so use the exact relation.
g_h=g\left(\dfrac{R_e}{R_e+h}\right)^{2}
Substitution (1 mark).
g_h=9.8\left(\dfrac{6.4\times10^{6}}{6.4\times10^{6}+1.5\times10^{6}}\right)^{2}=9.8\left(\dfrac{6.4}{7.9}\right)^{2}
Answer with unit (1 mark).
g_h=9.8\times0.656=6.43\,\mathrm{m\,s^{-2}}

g_h\approx6.4\,\mathrm{m\,s^{-2}}

2026 Annual I, Set B, Q.2 (xii)1.5+1.5 = 3 marks

What factors determine the strength of gravitational field g around a planet?

Solution

The field strength at a distance r from the centre of a planet of mass M is

g=\dfrac{GM}{r^{2}}

Factor 1: mass of the planet (1.5 marks). g\propto M. A more massive planet has a stronger field; for example Jupiter, with 316 times the Earth's mass, has g\approx24.8\,\mathrm{m\,s^{-2}} at its surface.

Factor 2: distance from the planet's centre (1.5 marks). g\propto1/r^{2}. At the surface r equals the planet's radius R, so for the same mass a smaller planet has a stronger surface field. Above the surface, g_h=GM/(R+h)^{2} falls with height; at h=R it is only g/4.

g does not depend on the mass of the body placed in the field.

2026 Annual I, Set B, Q.32+4+1 = 7 marks

Briefly describe gravitational potential and derive the expression \varphi=-\dfrac{GM}{r} at a distance r from the centre of Earth. Make a sketch.

Solution

Gravitational potential (2 marks)

The gravitational potential \varphi (or V) at a point is the work done per unit mass in bringing a small test mass from infinity to that point. It equals the gravitational potential energy per unit mass, \varphi=U/m. It is a scalar with SI unit \mathrm{J\,kg^{-1}}. Its zero is at infinity, so it is negative at every finite distance from the Earth, and it increases (becomes less negative) with distance.

Derivation (4 marks)

Here M is the mass of the Earth (M=M_E). The force changes with distance, so the path is split into small steps and the work in each step is added.

Derivation Derivation of \varphi=-\dfrac{GM_E}{r}

Aim. Find the work done per unit mass in bringing a small test mass m from infinity to a point at distance r from the centre of the Earth.

Assumptions.

The Earth is a uniform sphere of mass M_E and radius R_E; it acts as a point mass at its centre O.
The mass m is moved slowly, so its kinetic energy does not change and the work done on it by the agent goes into potential energy.
U=0 at infinity.

The path from the surface (point 0, distance R_E from the centre O) to a very distant point n is divided into small steps \Delta r; in each step the force is taken as constant.

By the work and energy principle, when m is moved slowly the work done on it by the external agent equals the change in its potential energy.
\Delta P.E.=W_{net}\qquad(1)
The force changes with distance, so W=Fd cannot be used for the whole path at once. Divide the path between the surface (point 0) and a very distant point n into n small steps of length \Delta r, so small that the force is almost constant in each. The total work is the sum of the works in the steps.
W_{net}=W_1+W_2+W_3+\dots+W_{n-1}+W_n\qquad(2)
Take the step between point 1 (distance r_1) and point 0 (distance R_E). To bring m in slowly, the agent pulls outward with a force F_{av} equal to the pull of gravity, while m moves inward through \Delta r. Force and displacement are opposite: \theta=180^\circ and \cos180^\circ=-1.
W_1=F_{av}\,\Delta r\cos180^\circ=-F_{av}\,\Delta r\qquad(3)
The average force in this step follows from Newton's law, using the average distance r_{av} of the step.
F_{av}=\dfrac{GM_Em}{r_{av}^{2}}\qquad(4)
The average distance is neither R_E nor r_1 but their mean. From the figure \Delta r=r_1-R_E, so r_1=R_E+\Delta r.
r_{av}=\dfrac{R_E+r_1}{2}=\dfrac{2R_E+\Delta r}{2}=R_E+\dfrac{\Delta r}{2}\qquad(5\text{ to }7)
Square the average distance.
r_{av}^{2}=\left(R_E+\dfrac{\Delta r}{2}\right)^{2}=R_E^{2}+R_E\,\Delta r+\dfrac{(\Delta r)^{2}}{4}
\Delta r is very small, so the term (\Delta r)^{2}/4 is negligible compared with the others. Then put \Delta r=r_1-R_E.
r_{av}^{2}\approx R_E^{2}+R_E\,\Delta r=R_E^{2}+R_E(r_1-R_E)=R_Er_1\qquad(8,\ 9)
Put Eq. (9) in Eq. (4).
F_{av}=\dfrac{GM_Em}{R_Er_1}\qquad(10)
Put Eq. (10) and \Delta r=r_1-R_E (Eq. 11) in Eq. (3), then split the fraction.
W_1=-\dfrac{GM_Em}{R_Er_1}\,(r_1-R_E)=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_1}\right)\qquad(12)
The same working applies to every other step.
\begin{aligned}W_2&=-GM_Em\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\\ W_3&=-GM_Em\left(\dfrac{1}{r_2}-\dfrac{1}{r_3}\right)\\ &\ \ \vdots\\ W_{n-1}&=-GM_Em\left(\dfrac{1}{r_{n-2}}-\dfrac{1}{r_{n-1}}\right)\\ W_n&=-GM_Em\left(\dfrac{1}{r_{n-1}}-\dfrac{1}{r_n}\right)\end{aligned}\qquad(13\text{ to }16)
Add all the steps as in Eq. (2). Every middle term appears once with + and once with -, so they cancel in pairs; only the first and last terms remain.
\begin{aligned}W_{net}&=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_1}+\dfrac{1}{r_1}-\dfrac{1}{r_2}+\dots+\dfrac{1}{r_{n-1}}-\dfrac{1}{r_n}\right)\\ &=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_n}\right)\end{aligned}\qquad(17)
By Eq. (1), this is the change in potential energy when m is brought from point n to the surface.
\Delta P.E.=U_{\text{surface}}-U_n=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_n}\right)\qquad(18)
Now take point n at infinity: r_n=\infty, so 1/r_n=0, and U_n=0 (our zero). The work done in bringing m from infinity to the surface is then
U=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{\infty}\right)=-\dfrac{GM_Em}{R_E}
The same steps, started from any point P at distance r from the centre instead of the surface, give the general result.
U=-\dfrac{GM_Em}{r}\qquad(15.9)
By definition the potential is the work done per unit mass, so divide by m. For a unit mass (m=1\,\mathrm{kg}) the work itself is the potential.
\varphi=\dfrac{U}{m}=\dfrac{-GM_Em/r}{m}=-\dfrac{GM_E}{r}

Result\varphi=-\dfrac{GM_E}{r}

Sketch (1 mark)

The labelled diagram of the path (above) is the sketch that goes with the derivation. The graph of \varphi against r shows the result:

Sketch: \varphi (\mathrm{MJ\,kg^{-1}}) against distance from the centre

\varphi is most negative (-62.5\,\mathrm{MJ\,kg^{-1}}) at the surface and rises towards zero as r increases.

2026 Annual I, Set D, MCQ 31 mark

Two objects attract each other with a gravitational force F. If mass of one object is doubled and distance between their centers remains same, then gravitational force is:

A2FAnswer
B4F
CF/2
DF

Why

F=G\dfrac{m_1m_2}{r^{2}} is directly proportional to each mass. Doubling one mass (same distance) doubles the force: F'=G\dfrac{(2m_1)m_2}{r^{2}}=2F. B (4F) would need both masses doubled.

2026 Annual I, Set D, MCQ 41 mark

The gravitational field strength at a point depends on:

AShape of the test body
BVelocity of the body
CMass of the test body
DMass causing the fieldAnswer

Why

g=\dfrac{F}{m}=\dfrac{GM}{r^{2}}: the test mass m cancels, so g depends only on the mass M causing the field and the distance r from it. The shape, velocity and mass of the test body do not matter.

2026 Annual I, Set D, Q.2 (i)1+2 = 3 marks

How will the orbital velocity of a satellite be affected when it is put into a lower orbit (closer to the Earth)? Explain mathematically.

Solution

Effect (1 mark). The orbital velocity increases: a satellite in a lower orbit must move faster.

Mathematical explanation (2 marks). For a circular orbit of radius r, gravity provides the centripetal force:

\dfrac{GM_Em}{r^{2}}=\dfrac{mv_o^{2}}{r}\quad\Rightarrow\quad v_o=\sqrt{\dfrac{GM_E}{r}}\quad\Rightarrow\quad v_o\propto\dfrac{1}{\sqrt r}

In a lower orbit r is smaller and the Earth's pull is stronger, so a larger speed is needed to keep the satellite on its circle. If r is reduced to one quarter, v_o doubles. The fastest circular orbit is the one just above the surface: v_o=\sqrt{gR_E}=7.9\,\mathrm{km\,s^{-1}}.

2026 Annual I, Set D, Q.2 (iv) OR3 marks

What will be the value of gravitational field strength g at 35\,700\,\mathrm{km}, where a satellite is orbiting the earth?

Solution

35\,700\,\mathrm{km} is the well-known height of communication (geostationary) satellites above the Earth's surface (Table 15.1), so we take it as the height h above the surface. This is Assignment 15.2 of the book.

Given

g = 9.8\,\mathrm{m\,s^{-2}}
R_e = 6.4\times10^{6}\,\mathrm{m}
h = 35\,700\,\mathrm{km}=3.57\times10^{7}\,\mathrm{m}

To find

Formula (1 mark).
g_h=g\left(\dfrac{R_e}{R_e+h}\right)^{2}
Substitution (1 mark). R_e+h=4.21\times10^{7}\,\mathrm{m}.
g_h=9.8\left(\dfrac{6.4\times10^{6}}{4.21\times10^{7}}\right)^{2}=9.8\times(0.152)^{2}
Answer with unit (1 mark).
g_h=0.23\,\mathrm{m\,s^{-2}}

g_h\approx0.23\,\mathrm{m\,s^{-2}}

2026 Annual I, Set D, Q.2 (xi)3 marks

Why the value of g does not change significantly near the Earth's surface? (Show mathematically)

Solution

At height h above the surface the field strength is

g_h=g\left(\dfrac{R_E}{R_E+h}\right)^{2}=g\left(1+\dfrac{h}{R_E}\right)^{-2}\approx g\left(1-\dfrac{2h}{R_E}\right)\qquad(h\ll R_E)

using the binomial approximation. So the fractional decrease in g is only 2h/R_E. The Earth's radius is very large (R_E=6.4\times10^{6}\,\mathrm{m}), so for ordinary heights this is tiny:

h=1\,\mathrm{km}:\quad\dfrac{2h}{R_E}=\dfrac{2\times1000}{6.4\times10^{6}}=3.1\times10^{-4}\ (0.03\%)

Even at the top of Mount Everest (h\approx8.8\,\mathrm{km}) g falls by less than 0.3\%, from 9.80 to about 9.77\,\mathrm{m\,s^{-2}}. Hence near the surface g is practically constant.

2026 Annual I, Set D, Q.32+4+1 = 7 marks

Briefly describe gravitational potential and derive the expression \varphi=-\dfrac{GM}{r} at a distance r from the center of Earth. Make a sketch.

Solution

Gravitational potential (2 marks)

Derivation (4 marks)

Here M is the mass of the Earth (M=M_E). The force changes with distance, so the path is split into small steps and the work in each step is added.

Derivation Derivation of \varphi=-\dfrac{GM_E}{r}

Aim. Find the work done per unit mass in bringing a small test mass m from infinity to a point at distance r from the centre of the Earth.

Assumptions.

The Earth is a uniform sphere of mass M_E and radius R_E; it acts as a point mass at its centre O.
The mass m is moved slowly, so its kinetic energy does not change and the work done on it by the agent goes into potential energy.
U=0 at infinity.

The path from the surface (point 0, distance R_E from the centre O) to a very distant point n is divided into small steps \Delta r; in each step the force is taken as constant.

By the work and energy principle, when m is moved slowly the work done on it by the external agent equals the change in its potential energy.
\Delta P.E.=W_{net}\qquad(1)
The force changes with distance, so W=Fd cannot be used for the whole path at once. Divide the path between the surface (point 0) and a very distant point n into n small steps of length \Delta r, so small that the force is almost constant in each. The total work is the sum of the works in the steps.
W_{net}=W_1+W_2+W_3+\dots+W_{n-1}+W_n\qquad(2)
Take the step between point 1 (distance r_1) and point 0 (distance R_E). To bring m in slowly, the agent pulls outward with a force F_{av} equal to the pull of gravity, while m moves inward through \Delta r. Force and displacement are opposite: \theta=180^\circ and \cos180^\circ=-1.
W_1=F_{av}\,\Delta r\cos180^\circ=-F_{av}\,\Delta r\qquad(3)
The average force in this step follows from Newton's law, using the average distance r_{av} of the step.
F_{av}=\dfrac{GM_Em}{r_{av}^{2}}\qquad(4)
The average distance is neither R_E nor r_1 but their mean. From the figure \Delta r=r_1-R_E, so r_1=R_E+\Delta r.
r_{av}=\dfrac{R_E+r_1}{2}=\dfrac{2R_E+\Delta r}{2}=R_E+\dfrac{\Delta r}{2}\qquad(5\text{ to }7)
Square the average distance.
r_{av}^{2}=\left(R_E+\dfrac{\Delta r}{2}\right)^{2}=R_E^{2}+R_E\,\Delta r+\dfrac{(\Delta r)^{2}}{4}
\Delta r is very small, so the term (\Delta r)^{2}/4 is negligible compared with the others. Then put \Delta r=r_1-R_E.
r_{av}^{2}\approx R_E^{2}+R_E\,\Delta r=R_E^{2}+R_E(r_1-R_E)=R_Er_1\qquad(8,\ 9)
Put Eq. (9) in Eq. (4).
F_{av}=\dfrac{GM_Em}{R_Er_1}\qquad(10)
Put Eq. (10) and \Delta r=r_1-R_E (Eq. 11) in Eq. (3), then split the fraction.
W_1=-\dfrac{GM_Em}{R_Er_1}\,(r_1-R_E)=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_1}\right)\qquad(12)
The same working applies to every other step.
\begin{aligned}W_2&=-GM_Em\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\\ W_3&=-GM_Em\left(\dfrac{1}{r_2}-\dfrac{1}{r_3}\right)\\ &\ \ \vdots\\ W_{n-1}&=-GM_Em\left(\dfrac{1}{r_{n-2}}-\dfrac{1}{r_{n-1}}\right)\\ W_n&=-GM_Em\left(\dfrac{1}{r_{n-1}}-\dfrac{1}{r_n}\right)\end{aligned}\qquad(13\text{ to }16)
Add all the steps as in Eq. (2). Every middle term appears once with + and once with -, so they cancel in pairs; only the first and last terms remain.
\begin{aligned}W_{net}&=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_1}+\dfrac{1}{r_1}-\dfrac{1}{r_2}+\dots+\dfrac{1}{r_{n-1}}-\dfrac{1}{r_n}\right)\\ &=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_n}\right)\end{aligned}\qquad(17)
By Eq. (1), this is the change in potential energy when m is brought from point n to the surface.
\Delta P.E.=U_{\text{surface}}-U_n=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_n}\right)\qquad(18)
Now take point n at infinity: r_n=\infty, so 1/r_n=0, and U_n=0 (our zero). The work done in bringing m from infinity to the surface is then
U=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{\infty}\right)=-\dfrac{GM_Em}{R_E}
The same steps, started from any point P at distance r from the centre instead of the surface, give the general result.
U=-\dfrac{GM_Em}{r}\qquad(15.9)
By definition the potential is the work done per unit mass, so divide by m. For a unit mass (m=1\,\mathrm{kg}) the work itself is the potential.
\varphi=\dfrac{U}{m}=\dfrac{-GM_Em/r}{m}=-\dfrac{GM_E}{r}

Result\varphi=-\dfrac{GM_E}{r}

Sketch (1 mark)

The labelled diagram of the path (above) is the sketch that goes with the derivation. The graph of \varphi against r shows the result:

Sketch: \varphi (\mathrm{MJ\,kg^{-1}}) against distance from the centre

\varphi is most negative (-62.5\,\mathrm{MJ\,kg^{-1}}) at the surface and rises towards zero as r increases.

Exam focus

Describe gravitational potential and derive \varphi=-\dfrac{GM}{r} by the step method, with a sketch. (long)
Q.3 (7 marks) in both 2026 new-syllabus sets (B and D), and comprehensive question 5 of the book.
Go to answer
Derive the orbital velocity v_o=\sqrt{GM_E/r} and explain how it changes when the orbit is raised or lowered. (short)
Q.2 (i) in both 2026 sets (higher orbit in Set B, lower orbit in Set D); also comprehensive question 3.
Go to answer
Derive g=GM_E/R_E^{2} and g_h=g\left(\dfrac{R_E}{R_E+h}\right)^{2}; discuss the variation of g with altitude. (long)
Comprehensive question 2; its results were examined as numericals in both 2026 sets.
Go to answer
Numericals on g at a height (1500\,\mathrm{km}, 35\,700\,\mathrm{km}). (numerical)
Both 2026 sets had one as the Q.2 (iv) OR part; also book exercise MCQ 3 and Assignment 15.2.
Go to answer
Why is g nearly constant near the Earth's surface? Show mathematically. (short)
2026 Set D Q.2 (xi), and one of the unit's SLOs.
Go to answer
What factors determine the gravitational field strength around a planet? (short)
2026 Set B Q.2 (xii); same as short question 5.
Go to answer
Geostationary satellites: definition, conditions, and the calculation of r, v_o and h. (long)
Comprehensive question 4; the numbers (4.23\times10^{7}\,\mathrm{m}, 3.07\,\mathrm{km\,s^{-1}}, 3 satellites) feed several MCQs.
Go to answer
State Newton's law of gravitation and show it is consistent with Newton's third law. (long)
Comprehensive question 1; a standard definition-plus-explanation question.
Go to answer
One-line ratio MCQs: force with changed masses or distance; g at h=R_E. (mcq)
Four of the six 2026 unit 15 MCQs were of this type (Set B MCQs 1 and 2, Set D MCQs 3 and 4).
Go to answer
Potential energy numericals: U=-GMm/r and \Delta U between two orbits. (numerical)
Book numericals 8 to 10 test the exact formula instead of mgh.
Go to answer
Meaning of negative potential energy and the link U=mV. (short)
Short questions 9 and 10; the SLO asks you to justify how potential leads to the potential energy of two point masses.
Go to answer