Question

A body stretches a spring by a particular length at the earth's surface at equator. At what height above the south pole will it stretch the same spring by the same length ? Assume the earth to be spherical.

Answer 1

At equator $g^{\prime }=g-{\omega }^{2}R$
Let at 'h' height above south pole, the acceleration due to gravity is same.
then , here $g^{\prime }=g(1-\frac{2h}{R})\therefore g-{\omega }^{2}r=g(1-\left(\frac{2h}{R}\right))or1-\frac{{\omega }^{2}R}{2g}=1-\frac{2h}{R}orh=\frac{{\omega }^2{R}^2}{2g}=\frac{{(7.3\times {10}^{-5})}^2\times {(6400\times {10}^{-3})}^2}{2\times 9.81}=\frac{(7.3{)}^2\times (64{)}^2}{19.62}11125m=10$ km(approx)