Acceleration due to gravity

Near the earth, the pull on a mass m is GM_Em/R_E². Divide by m and you get the acceleration it gives, the acceleration due to gravity: g = GM_E/R_E² ≈ 9.8 m s⁻². It does not depend on the falling body's mass, which is why all bodies fall alike (without air).

Turn it round and we can weigh the earth: M_E = gR_E²/G ≈ 6 × 10²⁴ kg. Since M = density × (4/3)πR³, we can also write g = (4/3)πGρR.

A body's weight is mg. Its mass never changes, but its weight changes with g. Example: on Mars (mass 1/9 of the earth's, radius 1/2), g_Mars = (1/9) ÷ (1/2)² × g = (4/9)g. A pack weighing 900 N on earth weighs 400 N there.

Going up: at a height h, the distance from the centre is R_E + h, so g(h) = GM_E/(R_E + h)². For heights small compared with R_E, this is about g(1 – 2h/R_E). Example: g falls by 36% (to 64%, so (R_E + h)/R_E = 10/8) at h = R_E/4 = 1600 km.

Going down: at a depth d, the outer shell you have passed pulls you neither way (last lesson), and only the inner sphere of radius R_E – d pulls. In a uniform earth this gives g(d) = g(1 – d/R_E). Example: 16 km down a deep mine, g is less by 16/6400 = 0.25%.

So g is largest at the surface and falls both above it and below it, reaching zero at the centre.

g also varies across the surface. The earth bulges at the equator, so the poles are nearer the centre and g is a little larger at the poles. The earth's spin lowers the felt g at the equator a little more, and not at all at the poles.

One caution for energy: V = mgh works only for small heights, where g hardly changes. Over big distances use the exact form, ΔV = –GMm(1/r₂ – 1/r₁).