Write g as GM/R² and Change Only the Quantity That Changes
Gravity questions mostly ask how g, or the weight of a body, changes when the mass, the radius, the height or the depth changes. Write g = GM/R² once and scale it.
Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.
Draw every push or pull as an arrow
Balanced arrows: no change · unbalanced: acceleration
Arrows show direction. Balanced arrows give no change in motion; an unbalanced resultant points toward the acceleration.
Newton's law and the surface value of g
The force between two masses is F = G m1 m² / r², and by Newton's third law the force of the Earth on the Moon equals the force of the Moon on the Earth, whatever their masses, a ratio of 1 : 1.
Mars, with 0.5 of Earth's diameter and 0.1 of its mass, has g ratio 0.1 / (0.5)² = 0.4.
If the radius shrinks by 1% with the mass fixed, g rises by about 2%, because g is proportional to R⁻².
Height and depth
At height h above the surface, g' = g R²/(R + h)², which is about g (1 - 2h/R) for small h.
Inside a uniform Earth, g' = g (1 - d/R) at depth d, falling linearly to zero at the centre.
Plotted against distance from the centre, g rises linearly inside, peaks at the surface and then falls as 1/r².
Rotation, balances and other worlds
Rotation reduces the effective g everywhere except at the poles, so if the Earth stopped rotating, g would increase at every place other than the poles. A spring balance measures weight, which varies with g, while a physical (beam) balance compares masses, so it reads the same at the poles and equator. The Moon's g is one sixth of the Earth's and it has no atmosphere, so a hydrogen balloon does not float up there but falls with acceleration g/6, and a stone dropped from the same height takes sqrt(6) times as long to land. For a 60 kg passenger flying from Earth to Mars the weight is 600 N at Earth's surface and 240 N at Mars's surface, with a much smaller net pull for most of the journey, since both pulls weaken with distance.
Adding gravitational forces as vectors
The force on a mass from several others is a vector sum. For three 1 kg masses at (0, 0), (0, 0.2 m) and (0.2 m, 0), each neighbour pulls the mass at the origin with G m²/r² = 6.67 x 10⁻¹¹ / 0.04 = 1.67 x 10⁻⁹ N, one along +y and one along +x, so the net force is 1.67 x 10⁻⁹ (i + j) N.
OlympIQ exam lens
- ✓g = GM/R²; for diameter D, g = 4GM/D²; M = gR²/G
- ✓Above the surface g' = gR²/(R + h)² ≈ g(1 - 2h/R); below it g' = g(1 - d/R)
- ✓Third law: the pull of A on B equals the pull of B on A
- ✓Rotation lowers effective g except at the poles; a spring balance shows weight, a beam balance compares mass
- ✓Sum forces from several masses as vectors
Do not apply g' = g(1 - 2h/R) when h is comparable to R. For large heights use g' = gR²/(R + h)²; the linear approximation fails, and at h = R the value is g/4, not zero.
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