Why do objects weigh less at the equator than at the poles?
They feel centrifugal acceleration due to the earth’s rotation. a = v^2 / r = omega^2 r = (2 pi / T)^2 r , where r is the radius of the earth (at the equator) and T is one sidereal day. You can compare this to g, gravitational acceleration. It’s less than a percent, so you wouldn’t notice much. Moreover, because of the centrifugal force, the earth is fatter in the middle, so you are a bit further out, so gravity is a wee bit less. You can model this by assuming that the earth’s surface is an equipotential. But you have not only the spherically symmetric gravitational potential (-GM / r), but also the centrifugal potential: -1/2 omega^2 R^2 = -1/2 omega^2 r^2 cos(lattitude)^2 Their sum is constant at the earth’s surface: GM/r + 1/2 (2 pi / T)^2 r^2 cos(lattitude)^2 = constant = GM / Rpolar GM (r – Rpolar) = 2 pi^2 Rpolar T^2 r^3 cos(lat)^2 To first order, you can assume that the r^3 on the right is pretty close to Rpolar^3, so r – Rpolar = 2 pi^2 Rpolar^4 T^2 cos(lat)^2 / GM This relati
