Question

A simple pendulum is taken at a place where its separation from the earth's surface is equal to the radius of the earth. Calculate the time period of small oscillations if the length of the string is 1.0 m. Take $g={\pi }^2$ m/s${}^2$ at the surface of the earth.

Answer 1

At any height h from the surface, the gravitational pull is

$g^{\prime }=\frac{g}{(1+\frac{h}{R}{)}^2}$

Here $H=R$ (as given)

$g^{\prime }=\frac{g}{(1+\frac{R}{R}{)}^2}$

$⟹g^{\prime }=\frac{g}{4}$

So Time period $T=2\pi \sqrt{\frac{l}{g^{\prime }}}$

$T=2\pi \sqrt{\frac{l}{g/4}}$

Also given that $l=1$ and $g={\pi }^2$

$T=2\pi \sqrt{\frac{1}{{\pi }^2/4}}$

$⟹T=4s$