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Learn9702Paper 4 · A Level Structured QuestionsGravitational fieldsGravitational potential and orbits
Gravitational fields · Subtopic 2 of 2

Gravitational Potential and Orbits

Gravitational potential is the work done per unit mass bringing a small mass from infinity: φ=−GMr\varphi=-\dfrac{GM}r. It is zero at infinity and negative everywhere else, because gravity is attractive. A mass mm has potential energy Ep=−GMmrE_p=-\dfrac{GMm}r. For a circular orbit, gravity provides the centripetal force, giving v2=GMrv^2=\dfrac{GM}r and T2=4π2GMr3T^2=\dfrac{4\pi^2}{GM}r^3. A geostationary satellite orbits above the equator with a period of 2424 hours.

Cambridge 9702 Paper 4Edexcel YPH11 WPH15AQA 7408 Paper 2≈ 40 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Define gravitational potential and explain why it is negative.
  • Use φ=−GMr\varphi=-\dfrac{GM}r and ΔEp=mΔφ\Delta E_p=m\Delta\varphi.
  • Derive T2∝r3T^2\propto r^3 by equating gravitational and centripetal forces.
  • Describe a geostationary orbit and calculate its radius.
Pause and think

A satellite in a low orbit fires its thrusters and moves to a higher circular orbit. Afterwards, is it moving faster or slower?

Section 2 of 14

Core idea

Start hereClimbing out of a well

Escaping Earth is like climbing out of a deep well that gets shallower the higher you go. Gravitational potential measures how deep in the well you are, with zero at the top (infinity) and negative values everywhere below.

The big idea

ϕ=−GMr\phi=-\frac{GM}r: work per unit mass to bring a mass from infinity. Ep=mϕE_p=m\phi. Escape speed 2GMR\sqrt{\frac{2GM}R}.

1. The potential well

Potential around the Earth: a "well"
12345678−70−60−50−40−30−20−100distance from centre (Earth radii)φ (MJ/kg)φ = −GM/rsurface −62.5 MJ/kg
Energy per kg needed to escape from the surface: 62.5 MJ. The gradient of the curve at any point is the field strength gg.
Potential from the field
  1. g=−dϕdrg=-\frac{d\phi}{dr}
    Field strength is minus the potential gradient (taking outward as positive).
  2. g=−GMr2⇒dϕdr=GMr2g=-\frac{GM}{r^2}\Rightarrow\frac{d\phi}{dr}=\frac{GM}{r^2}
    The field points inwards.
  3. ϕ=−GMr+C\phi=-\frac{GM}r+C
    Integrate.
  4. ϕ(∞)=0⇒C=0\phi(\infty)=0\Rightarrow C=0
    Zero of potential defined at infinity.
Negative everywhere; rises towards zero far away.
Worked exampleSurface potential and escape speed

Earth's surface.

Goal
ϕ\phi; escape speed.
Strategy
KE must climb to ϕ=0\phi=0.
  1. ϕ=−GMR=−6.25×107 J kg−1\phi=-\frac{GM}R=-6.25\times10^7\text{ J kg}^{-1}
    Formula.
  2. 12mv2=GMmR\tfrac12mv^2=\frac{GMm}R
    Just reaches infinity with zero speed.
  3. v=2GMR=11.2 km s−1v=\sqrt{\frac{2GM}R}=11.2\text{ km s}^{-1}
    mm cancels.
Answer−6.25×107-6.25\times10^7 J kg−1^{-1}; 11.211.2 km s−1^{-1}.
✓ Check2×\sqrt2\times the low-orbit speed (7.97.9 km s−1^{-1}) ✓.
Worked exampleLifting a satellite to 2R

1000 kg.

Goal
Energy needed (ignore orbital KE).
Strategy
mΔϕm\Delta\phi.
  1. Δϕ=−GM2R+GMR=GM2R=3.13×107\Delta\phi=-\frac{GM}{2R}+\frac{GM}R=\frac{GM}{2R}=3.13\times10^7
    J kg−1^{-1}.
  2. ΔE=3.13×1010 J\Delta E=3.13\times10^{10}\text{ J}
    × 1000 kg.
Answer3.13×10103.13\times10^{10} J.
✓ Checkmgh=1000(9.81)(6.37×106)=6.25×1010mgh=1000(9.81)(6.37\times10^6)=6.25\times10^{10} J would double it: gg isn't constant ✓.

2. Kepler's third law

Kepler's third law: T² against r³ is a straight line
12345678123456780r³ (×10²² m³)T² (×10¹⁰ s²)T² = (4π²/GM) r³
For all satellites of the same planet, T2r3\dfrac{T^2}{r^3} is constant, so the gradient gives the planet's mass.
Common misconceptionUsing mghmgh far out Only valid when h≪Rh\ll R.
Pause and think

Can gravitational potential be positive?

Summary

  • ϕ=−GM/r\phi=-GM/r.
  • Escape v=2GM/Rv=\sqrt{2GM/R}.
Key words
Gravitational potential
Work per unit mass from infinity.
HistoryClarke's orbit. In 19451945 science-fiction writer Arthur C. Clarke published a paper proposing three geostationary satellites to relay radio signals around the world. Twenty years later, Intelsat 1 did exactly that. The geostationary belt is still called the Clarke orbit.
Fun factEscape speed. To escape Earth completely from the surface, you need 12mv2=GMmR\tfrac12mv^2=\dfrac{GMm}R, giving v=2GMR≈11.2v=\sqrt{\dfrac{2GM}R}\approx11.2 km s⁻¹, about 3333 times the speed of sound. Escape speed doesn't depend on the mass of the rocket.
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