Topic 12 takes the force that drops an apple and uses it to hold a galaxy together. Newton's law of gravitation, field strength, potential, and orbits.
F = GMm/r² · g = GM/r² · V = −GM/r · T² ∝ r³G = 6.67 × 10⁻¹¹ N m² kg⁻² — the weakest of the four fundamental forces by a factor of about 10³⁶
Compare with Topic 7: gravity and electrostatics share the same inverse-square mathematics. But gravity is only ever attractive, and there is no such thing as negative mass — which is why gravitational potential is always negative.
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
Topic 12 · Newton's law
Newton's law of gravitation
F = G M m / r²r is measured from CENTRE to CENTRE, not from the surface
The force is attractive, acts along the line joining the two masses, and forms a Newton's third law pair: the Earth pulls you down with exactly the force you pull the Earth up with.
It is an inverse-square law: double the separation and the force falls to a quarter; treble it and it falls to a ninth.
A uniform sphere behaves, from outside, exactly as if all its mass were concentrated at its centre.
g = F/m = GM/r²gravitational field strength = force per unit mass, in N kg⁻¹ (identical to m s⁻²)
Radial vs uniform: far from a planet the field is radial — lines converge on the centre and g falls as 1/r². Very close to the surface, over a small region, the lines are effectively parallel and evenly spaced, so g is constant — which is why g = 9.81 N kg⁻¹ works for every projectile question in Topic 2.
Calculate
Your turn — Newton's law
1Calculate the gravitational force on a 70 kg person standing on the Earth's surface. (ME = 5.97 × 10²⁴ kg, rE = 6.37 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m² kg⁻².) Give your answer in N to 3 significant figures.
N
Hint: F = GMm/r². GM = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ = 3.98 × 10¹⁴. Multiply by 70, then divide by r² = (6.37 × 10⁶)² = 4.06 × 10¹³.
Calculate
Your turn — field strength
2Using the same data, calculate the gravitational field strength g at the Earth's surface. Give your answer in N kg⁻¹ to 3 significant figures.
N kg⁻¹
Hint: g = GM/r² = 3.98 × 10¹⁴ ÷ 4.06 × 10¹³. (Or simply divide your answer to the last question by 70.)
Topic 12 · potential
Gravitational potential and potential energy
V = − G M / r · Ep = m V = − G M m / rgravitational potential = potential energy per unit mass, in J kg⁻¹
Why negative? By convention, potential is zero at infinity. Because gravity is attractive, you must do work on a mass to move it out to infinity — so anywhere closer than infinity it has less energy than zero. Hence negative, everywhere.
Potential is a scalar — for several masses you simply add the values, with no directions to worry about.
Work done moving a mass between two points: W = mΔV.
Equipotentials are surfaces of constant V — spherical shells around a planet. Moving along one requires no work. They always cross field lines at right angles.
Field strength is the negative gradient of potential: g = −ΔV/Δr.
Two curves, two shapes: g falls as 1/r² but V rises (towards zero) as 1/r. Do not sketch them the same way — a favourite exam question.
Calculate
Your turn — gravitational potential
3Calculate the magnitude of the gravitational potential at the Earth's surface. (GM = 3.98 × 10¹⁴ N m² kg⁻¹, r = 6.37 × 10⁶ m.) Give your answer as a multiple of 10⁷ J kg⁻¹ to 3 significant figures.
× 10⁷ J kg⁻¹
Hint: V = −GM/r = −3.98 × 10¹⁴ ÷ 6.37 × 10⁶ = −6.25 × 10⁷ J kg⁻¹. You are asked for the magnitude.
Topic 12 · orbits
Satellites, orbital speed and Kepler's third law
For a satellite in a circular orbit, gravity supplies the centripetal force — and that single sentence generates the entire theory of orbits:
GMm/r² = mv²/r → v = √(GM/r)the satellite's own mass m cancels — so orbital speed does NOT depend on the satellite's mass
Substitute v = 2πr/T and rearrange:
T² = (4π²/GM) r³ → T² ∝ r³Kepler's third law — and the constant depends only on the mass of the central body
A higher orbit is a slower orbit with a longer period. The ISS at 400 km orbits in about 90 minutes; the Moon, 60 times further out, takes 27 days.
A geostationary satellite must have T = 24 hours, orbit west to east above the equator, giving r ≈ 4.22 × 10⁷ m (about 36 000 km altitude). It stays above one point — ideal for TV and communications.
Total energy in orbit: E = ½mv² + (−GMm/r) = −GMm/2r — negative, meaning the satellite is bound.
Worked example — low Earth orbit
r = 7.00 × 10⁶ m, GM = 3.98 × 10¹⁴.
v = √(3.98 × 10¹⁴ ÷ 7.00 × 10⁶) = √(5.69 × 10⁷) = 7540 m s⁻¹ = 7.54 km s⁻¹
T = 2πr/v = (2π × 7.00 × 10⁶) ÷ 7540 = 5830 s = 97.2 minutes
Calculate
Your turn — orbital speed
4A satellite orbits at a radius of 7.00 × 10⁶ m from the Earth's centre. (GM = 3.98 × 10¹⁴ N m² kg⁻¹.) Calculate its orbital speed. Give your answer in km s⁻¹ to 3 significant figures.
km s⁻¹
Hint: v = √(GM/r) = √(3.98 × 10¹⁴ ÷ 7.00 × 10⁶) = √(5.69 × 10⁷) m s⁻¹. Divide by 1000 for km s⁻¹.
Calculate
Your turn — orbital period
5That satellite travels at 7540 m s⁻¹ at a radius of 7.00 × 10⁶ m. Calculate its orbital period. Give your answer in minutes to 3 significant figures.
minutes
Hint: T = 2πr/v = (2π × 7.00 × 10⁶) ÷ 7540 = 5830 s. Divide by 60.
Calculate
Your turn — geostationary orbit
6Calculate the orbital radius of a geostationary satellite, for which T = 86 400 s. (GM = 3.98 × 10¹⁴ N m² kg⁻¹.) Give your answer as a multiple of 10⁷ m to 3 significant figures.