Gravitational fields

AshPhys · Revision cheat sheet · Sheet 1 of 6

9702 A Level Ch.17Chapter 17: Gravitational fieldsLesson 17.1Sections 17.1–17.7A Level only (not AS)

Goal: Use F = GMm/r², g = GM/r² and φ = −GM/r, and analyse circular orbits, including geostationary satellites.

forcevelocity

01The big picture

A gravitational field is a region in which a mass feels a force. Its strength at a point is the force per unit mass on a small test mass placed there, $g = F/m$ , in N kg⁻¹. The force is always attractive.

Two point masses attract with $F = GMm/r^2$ : proportional to each mass and inversely proportional to the square of the distance between their centres. Outside a uniform sphere, the sphere acts as if all its mass were at its centre, so $r$ is always measured from the centre. Dividing by $m$ gives the field of a point mass, $g = GM/r^2$ .

Field lines show the direction of the force on a mass; their spacing shows the strength. Round a planet they are radial and point inwards. Near the surface they are almost parallel and equally spaced: the field is nearly uniform, so $g \approx 9.81$ N kg⁻¹ and $\Delta E_\mathrm{p} = mg\Delta h$ work only for heights much smaller than the radius.

Gravitational potential $\varphi = -GM/r$ is zero at infinity and negative everywhere else: work must be done to pull a mass away. The potential energy of a mass $m$ is $E_\mathrm{p} = m\varphi = -GMm/r$ , and moving it changes its energy by $m\Delta\varphi$ .

For an orbit, gravity provides the centripetal force. So $GMm/r^2 = mv^2/r$ , giving $v = \sqrt{GM/r}$ and $T^2 \propto r^3$ , whatever the satellite’s mass.

Fig. 1 · Radial field of a planet

Field lines point inwards, towards the centre. They spread out with distance: the field gets weaker, as

1/r^2

.

Fig. 2 · Near the surface: uniform

Over heights much smaller than the radius, the field lines are parallel and equally spaced, so

g

is constant.

Gravitational field strength [paraphrase]: The gravitational force per unit mass on a small test mass at that point, $g = F/m$ .
Gravitational potential (syllabus) [syllabus wording]: The work done per unit mass in bringing a small test mass from infinity to the point.
Newton’s law of gravitation [paraphrase]: Two point masses attract each other with a force proportional to the product of their masses and inversely proportional to the square of their separation.
Geostationary orbit (syllabus) [syllabus wording]: A satellite in 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.

Gravitational fields · 9702 A Level Ch.17Sheet 2 of 6

02Deriving the equations

D1 Field strength of a point mass

Put a small test mass $m$ at distance $r$ from the centre of a mass $M$ , and use the definition $g = F/m$ .

Newton’s law of gravitation gives the force on the test mass:
$F = \frac{GMm}{r^{2}}$
Field strength is force per unit mass, so divide by $m$ ; the test mass cancels:
$g = \frac{F}{m} = \frac{GM}{r^{2}}$
At the surface, $r = R$ and $g_0 = GM/R^2$ . Dividing the two gives how $g$ falls with distance (inverse square):
$\frac{g}{g_{0}} = \frac{R^{2}}{r^{2}}$

Result

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

D2 Potential and potential energy

Potential is zero at infinity. A mass coming in from infinity is pulled by the field, so the work done on it per unit mass is negative.

Fig. 3 · Potential against distance

φ rises from −GM/R at the surface towards zero at infinity. The gradient of the graph at any point is the field strength there.

The potential of a point mass:
$\varphi = -\frac{GM}{r}$
The potential energy of a mass $m$ at distance $r$ is $m$ times the potential:
$E_\mathrm{p} = m\varphi = -\frac{GMm}{r}$
Moving from $r_1$ out to $r_2$ raises the potential (towards zero), so it takes energy:
$\Delta E_\mathrm{p} = m\Delta\varphi = GMm\left(\frac{1}{r_{1}} - \frac{1}{r_{2}}\right)$
The field strength is the size of the gradient of the potential–distance graph:
$g = -\frac{\Delta\varphi}{\Delta r}$

Result

\varphi = -\frac{GM}{r} \qquad E_\mathrm{p} = -\frac{GMm}{r}

Gravitational fields · 9702 A Level Ch.17Sheet 3 of 6

02Deriving the equations (continued)

D3 Circular orbits and T² ∝ r³

A satellite of mass $m$ orbits a planet of mass $M$ at radius $r$ . Gravity is the only force on it, so gravity provides the centripetal force.

Fig. 4 · A satellite in a circular orbit

Gravity is the only force on the satellite: it points to the centre of the orbit and provides the centripetal force.

Gravitational force = centripetal force:
$\frac{GMm}{r^{2}} = \frac{mv^{2}}{r}$
Cancel $m$ and one $r$ : the orbital speed does not depend on the satellite’s mass.
$v = \sqrt{\frac{GM}{r}}$
Substitute $v = 2\pi r/T$ and square:
$\frac{4\pi^{2}r^{2}}{T^{2}} = \frac{GM}{r} \;\Longrightarrow\; T^{2} = \frac{4\pi^{2}}{GM}\,r^{3}$
For satellites of the same planet $4\pi^2/GM$ is constant, so $T^2 \propto r^3$ .

Result

v = \sqrt{\frac{GM}{r}} \qquad T^{2} = \frac{4\pi^{2}r^{3}}{GM}

D4 Geostationary orbit

A geostationary satellite stays above the same point on the Equator: it orbits west to east with the Earth’s rotation, with period $T = 24$ h $= 86\,400$ s.

Cross-multiply $T^2 = 4\pi^2 r^3/(GM)$ and divide by $4\pi^2$ :
$r^{3} = \frac{GMT^{2}}{4\pi^{2}} = \frac{6.67\times10^{-11} \times 5.97\times10^{24} \times 86\,400^{2}}{4\pi^{2}} = 7.53\times10^{22}\ \mathrm{m^{3}}$
Take the cube root, then subtract the Earth’s radius for the height:
$r = 4.22\times10^{7}\ \mathrm{m}, \qquad h = r - R = 3.59\times10^{7}\ \mathrm{m}$

Result

r \approx 4.2\times10^{7}\ \mathrm{m} \quad (h \approx 3.6\times10^{7}\ \mathrm{m})

Gravitational fields · 9702 A Level Ch.17Sheet 4 of 6

03Key equations

Field strength (definition)

g = \frac{F}{m}

Newton’s law of gravitation

F = \frac{GMm}{r^{2}}

Field of a point mass

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

r from the centre

Gravitational potential

\varphi = -\frac{GM}{r}

zero at infinity

Potential energy

E_\mathrm{p} = m\varphi = -\frac{GMm}{r}

Change in potential energy

\Delta E_\mathrm{p} = m\,\Delta\varphi

Orbital speed

v = \sqrt{\frac{GM}{r}}

from GMm/r² = mv²/r

Orbital period (T² ∝ r³)

T^{2} = \frac{4\pi^{2}r^{3}}{GM}

Near the surface only

\Delta E_\mathrm{p} = mg\,\Delta h

when Δh ≪ R

Symbols and units

$G$ gravitational constant, 6.67 × 10⁻¹¹ $\mathrm{N\,m^{2}\,kg^{-2}}$
$M$ mass of the planet or star $\mathrm{kg}$
$m$ mass of the test mass or satellite $\mathrm{kg}$
$r$ distance from the centre $\mathrm{m}$
$R$ radius of the planet $\mathrm{m}$
$g$ gravitational field strength $\mathrm{N\,kg^{-1}}$
$\varphi$ gravitational potential $\mathrm{J\,kg^{-1}}$
$E_\mathrm{p}$ gravitational potential energy $\mathrm{J}$
$v$ orbital speed $\mathrm{m\,s^{-1}}$
$T$ orbital period $\mathrm{s}$
$M_\mathrm{E}$ Earth’s mass, 5.97 × 10²⁴ $\mathrm{kg}$
$R_\mathrm{E}$ Earth’s radius, 6.37 × 10⁶ $\mathrm{m}$

04Sample problems with solutions

Sample 1 · Field strength above the surface
The International Space Station orbits 4.1 × 10⁵ m above the Earth’s surface. Calculate the gravitational field strength there. Use the Earth data in the symbols list.

$r$ is measured from the centre of the Earth, not from the surface:
$r = R + h = 6.37\times10^{6} + 4.1\times10^{5} = 6.78\times10^{6}\ \mathrm{m}$
Substitute into $g = GM/r^2$ :
$g = \frac{6.67\times10^{-11} \times 5.97\times10^{24}}{(6.78\times10^{6})^{2}} = 8.66\ \mathrm{N\,kg^{-1}}$
Sense check: a little less than 9.81 N kg⁻¹. Astronauts feel weightless because they are in free fall, not because $g$ is zero.

Answer:

g = 8.7\ \mathrm{N\,kg^{-1}}

Unit check:

\frac{\mathrm{N\,m^{2}\,kg^{-2}} \times \mathrm{kg}}{\mathrm{m^{2}}} = \mathrm{N\,kg^{-1}}\ \checkmark

Gravitational fields · 9702 A Level Ch.17Sheet 5 of 6

04Sample problems with solutions (continued)

Sample 2 · Energy to lift a probe a long way up
A 500 kg probe is lifted from the Earth’s surface to a height equal to the Earth’s radius. Calculate the increase in its gravitational potential energy.

Potential at the surface ( $r = R$ ) and at height $R$ ( $r = 2R$ ):
$\varphi_{R} = -\frac{GM}{R} = -6.25\times10^{7}\ \mathrm{J\,kg^{-1}}, \qquad \varphi_{2R} = -3.13\times10^{7}\ \mathrm{J\,kg^{-1}}$
Change in potential energy:
$\Delta E_\mathrm{p} = m\Delta\varphi = 500 \times 3.13\times10^{7} = 1.56\times10^{10}\ \mathrm{J}$
Why not $mgh$ ? It gives $3.1\times10^{10}$ J, twice too much, because $g$ falls to a quarter of its surface value over this height.

Answer:

\Delta E_\mathrm{p} = 1.6\times10^{10}\ \mathrm{J}

Unit check:

\mathrm{kg} \times \mathrm{J\,kg^{-1}} = \mathrm{J}\ \checkmark

Sample 3 · Speed and period of a satellite
A satellite orbits 2.0 × 10⁶ m above the Earth’s surface. Calculate its orbital speed and its period.

Radius of the orbit, from the centre:
$r = 6.37\times10^{6} + 2.0\times10^{6} = 8.37\times10^{6}\ \mathrm{m}$
Gravity provides the centripetal force, so:
$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67\times10^{-11} \times 5.97\times10^{24}}{8.37\times10^{6}}} = 6.90\times10^{3}\ \mathrm{m\,s^{-1}}$
One orbit is a distance $2\pi r$ :
$T = \frac{2\pi r}{v} = \frac{2\pi \times 8.37\times10^{6}}{6.90\times10^{3}} = 7.62\times10^{3}\ \mathrm{s}$

Answer:

v = 6.9\times10^{3}\ \mathrm{m\,s^{-1}}

,

T = 7.6\times10^{3}

s (about 2.1 hours)

Unit check:

\sqrt{\frac{\mathrm{N\,m^{2}\,kg^{-2}} \times \mathrm{kg}}{\mathrm{m}}} = \sqrt{\mathrm{m^{2}\,s^{-2}}} = \mathrm{m\,s^{-1}}\ \checkmark

Sample 4 · Weighing the Earth with the Moon
The Moon orbits the Earth at a radius of 3.84 × 10⁸ m with a period of 27.3 days. Calculate the mass of the Earth.

Period in seconds:
$T = 27.3 \times 86\,400 = 2.36\times10^{6}\ \mathrm{s}$
$M$ is in the denominator and $T^2$ is the lone term on the other side: swap them in one step (cross-multiplication).
$T^{2} = \frac{4\pi^{2}r^{3}}{GM} \;\Longrightarrow\; M = \frac{4\pi^{2}r^{3}}{GT^{2}}$
Substitute:
$M = \frac{4\pi^{2} \times (3.84\times10^{8})^{3}}{6.67\times10^{-11} \times (2.36\times10^{6})^{2}} = 6.02\times10^{24}\ \mathrm{kg}$

Answer:

M = 6.0\times10^{24}\ \mathrm{kg}

(close to the accepted

5.97\times10^{24}

kg)

Unit check:

\frac{\mathrm{m^{3}}}{\mathrm{N\,m^{2}\,kg^{-2}} \times \mathrm{s^{2}}} = \frac{\mathrm{m^{3}\,kg^{2}}}{\mathrm{kg\,m^{3}}} = \mathrm{kg}\ \checkmark

Gravitational fields · 9702 A Level Ch.17Sheet 6 of 6

05Practice problems

P1
Calculate the gravitational force between two 70 kg people whose centres are 1.0 m apart.
Final answer
3.3 × 10⁻⁷ N
Answer: 3.3 × 10⁻⁷ N
P2
The Moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m. Calculate the field strength at its surface.
Final answer
1.6 N kg⁻¹
Answer: 1.6 N kg⁻¹
P3
A planet’s field strength at distance r is g. What is it at distance 3r?
Final answer
g/9
Answer: g/9
P4
At what height above the Earth’s surface is g half its surface value?
Final answer
2.6 × 10⁶ m
Answer: 2.6 × 10⁶ m
P5
Calculate the gravitational potential at the Earth’s surface.
Final answer
−6.3 × 10⁷ J kg⁻¹
Answer: −6.3 × 10⁷ J kg⁻¹
P6
The potential at a point is −4.0 × 10⁷ J kg⁻¹. Calculate the potential energy of a 200 kg satellite there.
Final answer
−8.0 × 10⁹ J
Answer: −8.0 × 10⁹ J
P7
Calculate the speed of a satellite in a circular orbit of radius 7.0 × 10⁶ m round the Earth.
Final answer
7.5 × 10³ m s⁻¹
Answer: 7.5 × 10³ m s⁻¹
P8
Two satellites orbit the same planet; one orbit has 4 times the radius of the other. Calculate the ratio of their periods.
Final answer
8
Answer: 8
P9
Phobos orbits Mars at a radius of 9.38 × 10⁶ m with a period of 7.65 h. Calculate the mass of Mars.
Final answer
6.4 × 10²³ kg
Answer: 6.4 × 10²³ kg
P10
Calculate the height above the Earth’s surface of a geostationary satellite.
Final answer
3.6 × 10⁷ m
Answer: 3.6 × 10⁷ m

06Other useful information

Common mistakes

✗ Wrong: Use the height above the surface as r.
✓ Right: r is the distance from the centre: r = R + h.
✗ Wrong: Gravitational potential is positive, like height.
✓ Right: It is zero at infinity and negative everywhere else; it rises towards zero as you move away.
✗ Wrong: Use mgh for any change in height.
✓ Right: Only when g is constant (h ≪ R). Otherwise use ΔEp = mΔφ.
✗ Wrong: A heavier satellite must orbit faster.
✓ Right: m cancels: v = √(GM/r) depends only on M and r.
✗ Wrong: Astronauts float because there is no gravity in orbit.
✓ Right: g is still about 8.7 N kg⁻¹ at the ISS; they are in free fall with the station.

Exam tips

Write r = R + h at the start of every calculation above a planet.
Orbits: start from GMm/r² = mv²/r (or mrω²), then cancel m.
Ratio questions: use g ∝ 1/r², φ ∝ −1/r and T² ∝ r³; no constants needed.
Convert periods to seconds: 24 h = 86 400 s.

Explore

What the syllabus asks

9702 topic 13, Gravitational fields: field strength as force per unit mass, field lines, F = Gm₁m₂/r², g = GM/r², φ = −GM/r, Ep = −GMm/r, circular orbits and geostationary orbits. Coursebook Chapter 17.
A Level only: not part of AS (AS uses ΔEp = mgΔh in a uniform field).
Outside a uniform sphere, treat its mass as a point mass at its centre.
Gravitational and electric fields are compared in Chapter 22.