OxfordAQA International AS and A-level Physics: Gravitational potential and orbits of planets and satellites (9630)

This guide follows the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for exams from May/June 2020 onwards. It covers sections 3.7.3 (Gravitational potential) and 3.7.4 (Orbits of planets and satellites). Section 3.7 is marked International A-level only, so none of it is part of the International AS qualification. Use the course hub and the printable checklist to track your progress.

Companion pages: the revision notes and the practice questions. Not sure where you stand? Take a free diagnostic first. You should already know 3.7.1 and 3.7.2 (Newton’s law and field strength).

Data for the examples, as a question would supply them: G = 6.67 × 10⁻¹¹ N m² kg⁻²; Jupiter’s mass is 1.90 × 10²⁷ kg; the Earth has mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m; the Moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m.

What 3.7.3 and 3.7.4 ask you to do

Specification section	What you must be able to do
3.7.3	Define gravitational potential, with zero at infinity, and potential difference
3.7.3	Use the work done in moving a mass, ΔW = mΔV
3.7.3	Use V = −GM/r in a radial field
3.7.3	Explain that no work is done moving a mass along an equipotential surface
3.7.3	Sketch and interpret graphs of g and V against r; use g = −ΔV/Δr
3.7.4	Relate orbital period and speed to the radius of a circular orbit
3.7.4	Discuss the energy of an orbiting satellite
3.7.4	Explain the significance of a geosynchronous orbit

The specification prints the work-done relation with a lower-case v. It means the potential difference, so this guide writes ΔW = mΔV.

Gravitational potential (3.7.3)

Definition

The gravitational potential V at a point is the work done per unit mass, by an external agent, in moving a small test mass at constant speed from infinity to that point. Infinity is the zero of potential. Gravity is attractive: the field itself does the work as the mass falls in, so the external agent’s work is negative and the potential everywhere else is below zero. The unit is J kg⁻¹.

A point mass M (or any spherically symmetric body, treated as if its mass were at the centre) gives a radial field with

V = −GM/r

where r is the distance from the centre. Potential is a scalar. Closer to the mass, V is more negative. As r → ∞, V → 0 from below, so V increases as you move away.

Potential difference and work done

The potential difference ΔV between two points is the work done per unit mass in moving a mass between them. Moving a mass m through a potential difference ΔV needs

ΔW = mΔV

with ΔV = V(final) − V(initial). If ΔV is positive (moving away from the mass), ΔW is positive: someone must do work against the field. If it is negative, the field does the work and the mass gains kinetic energy if it is free to move.

The potential energy of mass m at distance r is mV = −GMm/r. ΔW = mΔV works for any size of change, unlike mgh, which assumes g stays constant.

Worked example 1: raising a capsule

A 450 kg capsule is moved from a circular path of radius 7.00 × 10⁶ m to one of radius 1.20 × 10⁷ m about the centre of the Earth. Find the work that must be done on it.

GM = 6.67e-11 x 5.97e24 = 3.982 x 10^14 N m^2 kg^-1
V(initial) = -3.982e14 / 7.00e6 = -5.69 x 10^7 J/kg
V(final)   = -3.982e14 / 1.20e7 = -3.32 x 10^7 J/kg
ΔV = -3.318e7 - (-5.689e7) = +2.37 x 10^7 J/kg
ΔW = mΔV = 450 x 2.370e7 = 1.07 x 10^10 J

Equipotential surfaces

Around a point mass, equipotential surfaces are spheres centred on the mass. The field is always at right angles to an equipotential, so moving along one has ΔV = 0 and therefore ΔW = 0. No work is done, whatever path you take along the surface. Equipotentials drawn at equal steps of V are closer together where the field is stronger.

Graphs of V and g against r

For r larger than the radius of the body:

Quantity	Formula	Shape of graph against r
g	GM/r²	Starts high and falls, never reaching zero (an inverse square curve, positive magnitude)
V	−GM/r	Starts at a large negative value and rises towards zero from below, never reaching it

Both graphs flatten as r grows.

The link g = −ΔV/Δr

The field strength is the negative gradient of the potential graph:

g = −ΔV/Δr

Check it with V = −GM/r: dV/dr = +GM/r², so g = −GM/r², and the minus sign shows that g points towards the mass. Its magnitude is GM/r², as in 3.7.2, and equals the positive gradient of the V–r curve.

Worked example 2: field strength from a potential graph

Potential values near the Moon, with GM = 4.90 × 10¹² N m² kg⁻¹ from the data:

r / 10⁶ m	2.0	2.4	2.6	3.0
V / 10⁶ J kg⁻¹	−2.451	−2.043	−1.885	−1.634

Estimate g at r = 2.5 × 10⁶ m.

Use the narrow chord from 2.4e6 to 2.6e6 m:
ΔV = -1.885e6 - (-2.043e6) = +1.58 x 10^5 J/kg
Δr = 0.2e6 m
|g| = 1.58e5 / 2.0e5 = 0.79 N/kg
Check: GM/r^2 = 4.90e12 / (2.5e6)^2 = 0.78 N/kg

A narrow chord gives a good estimate. A wide chord, 2.0 to 3.0 × 10⁶ m, gives 0.82 N kg⁻¹ because the gradient changes along the interval.

Worked example 3: a lander leaving the Moon

A 2200 kg lander lifts from the Moon’s surface to a point 100 km above it. How much work must the engines do against gravity (ignore fuel burnt and the Moon’s rotation)?

GM(Moon) = 4.902e12
V(surface) = -4.902e12 / 1.74e6 = -2.8175 x 10^6 J/kg
r(final) = 1.74e6 + 1.00e5 = 1.84e6 m
V(final) = -4.902e12 / 1.84e6 = -2.6644 x 10^6 J/kg
ΔV = +1.531 x 10^5 J/kg
ΔW = 2200 x 1.531e5 = 3.37 x 10^8 J

Using mgh with the surface value 1.62 N kg⁻¹ gives 3.56 × 10⁸ J, about 6% too high, because g falls as the lander rises. The full method is the one to trust.

Orbits of planets and satellites (3.7.4)

Speed and period of a circular orbit

Take a satellite, mass m, moving in a circle of radius r around a body of mass M. Gravity provides the centripetal force:

GMm/r² = mv²/r

so v² = GM/r and v = √(GM/r).

The satellite mass cancels, so larger orbits are slower. The period is T = 2πr/v, which gives

T² = (4π²/GM) r³

so T² is proportional to r³ (Kepler’s third law for circular orbits). Put ratios to work: if r doubles, T increases by a factor 2^1.5 = 2.83 and v falls by a factor of 1/√2 = 0.707.

Worked example 4: a low Earth orbit

A satellite orbits 600 km above the Earth’s surface. Find its speed and period.

r = 6.37e6 + 6.0e5 = 6.97 x 10^6 m
v = sqrt(3.982e14 / 6.97e6) = 7.56 x 10^3 m/s
T = 2 pi r / v = 2 pi x 6.97e6 / 7558 = 5.79 x 10^3 s  (96.6 minutes)

Worked example 5: a moon of Jupiter

A moon orbits Jupiter at a radius of 4.22 × 10⁸ m. Find its orbital period in days.

GM(Jupiter) = 6.67e-11 x 1.90e27 = 1.267 x 10^17
T = 2 pi sqrt(r^3 / GM) = 2 pi sqrt((4.22e8)^3 / 1.267e17)
  = 1.53 x 10^5 s = 1.77 days

Energy of an orbiting satellite

Use v² = GM/r to write the kinetic energy:

• Kinetic energy: Ek = ½mv² = GMm/(2r)
• Potential energy: Ep = −GMm/r
• Total energy: E = Ek + Ep = −GMm/(2r)

So Ep = −2Ek and E = −Ek. The total energy is negative, which shows the satellite is bound. A bigger orbit has a total energy closer to zero. Ep rises with r while Ek falls.

Worked example 6: energy and changing orbit

A 520 kg satellite is in the 6.97 × 10⁶ m orbit from example 4. (a) Find Ek, Ep and E. (b) Find the energy needed to move it to a circular orbit of radius 1.00 × 10⁷ m.

(a) Ek = 3.982e14 x 520 / (2 x 6.97e6) = 1.49 x 10^10 J
    Ep = -2 x Ek = -2.97 x 10^10 J
    E  = -1.49 x 10^10 J
(b) E(new) = -3.982e14 x 520 / (2 x 1.00e7) = -1.035 x 10^10 J
    ΔE = -1.035e10 - (-1.485e10) = +4.50 x 10^9 J

The satellite ends up moving more slowly (6.31 × 10³ m s⁻¹) even though it gained energy. The extra energy goes into potential energy, and the kinetic energy decreases.

Worked example 7: air resistance

Thin air at the top of the atmosphere acts on a 520 kg satellite and it drops from r = 7.00 × 10⁶ m to r = 6.80 × 10⁶ m.

E at 7.00e6: -3.982e14 x 520 / (2 x 7.00e6) = -1.479 x 10^10 J
E at 6.80e6: -3.982e14 x 520 / (2 x 6.80e6) = -1.523 x 10^10 J
Total energy change = -4.35 x 10^8 J
v: 7.54 x 10^3 -> 7.65 x 10^3 m/s

Drag removes mechanical energy as thermal energy, so the total energy becomes more negative, yet the satellite speeds up. Falling to a lower orbit converts potential energy to kinetic energy. The kinetic energy gained is half the potential energy lost, and the rest is dissipated by drag.

Geosynchronous orbits

A geosynchronous orbit has a period equal to the Earth’s rotation period (one sidereal day, 23 h 56 min, or 86 164 s). A satellite in a geosynchronous orbit that is also above the equator and moves in the same direction as the Earth’s rotation stays above the same point on the equator. This is a geostationary orbit. The significance is that a fixed dish on the ground can keep pointing at it, which suits communications and broadcasting.

Worked example 8: the geosynchronous radius

r^3 = GM T^2 / (4 pi^2) = 3.982e14 x (86164)^2 / (4 pi^2)
r = 4.21 x 10^7 m
Height above the surface = 4.215e7 - 6.37e6 = 3.58 x 10^7 m
v = 2 pi r / T = 3.07 x 10^3 m/s

Every geosynchronous satellite of the Earth has this radius, whatever its mass.

Common errors

• Writing V = +GM/r or forgetting that the potential is negative, so that ΔV comes out with the wrong sign.
• Using the height above the surface as r. In V = −GM/r and in T² ∝ r³ the distance is measured from the centre.
• Using mgh for large changes in height.
• Saying “energy increases when the orbit gets smaller”. The total energy becomes more negative (decreases).
• Saying that a geosynchronous satellite must be stationary. It is only fixed above one point if it is also equatorial.

Next steps

Try the practice questions, then condense with the revision notes.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for exams from May/June 2020 onwards, published by OxfordAQA. Sections 3.7.3 and 3.7.4.