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 oscillation if the length of the string is `1.0m`. Take `g = pi^(2) m//s^(2)` at the surface of the earth.

Explanation

At a height R (radis of the earth) the acceleration due to gravity ils
`g'=(GM)/((R+R)^2)=1/4 (GM)/R^2=g/4`
The time period of small oscillations of the simple pendulum is
`T=2pisqrt(l/g')=2pi (sqrt(1.0m)/(1/4xxpi^2ms^-2))=2pi(2/pis)=4s`