A simple pendulum has a time peirod `T_(1)` when on the earth's surface & `T_(2)` when taken to a height & above the earth's surface, where R is the radius of the earth. What is the value of `T_(2)//T_(1)`?

Explanation

Let g & g' be the acceleration due to gravity on the earths surface at a height R above the earth's surface.
`g' = g/((1 + n/R)) = g/((1 + (R/R)^(2))) = g/4`
Time period of simple pendulum `T = 2pi sqrt(l/g)`
`(T_2)/(T_1) = sqrt(g/(g')) = sqrt((g)/(g//4)) = 2`.