A particle is dropped from a height $h$ above the surface of the earth. If the time taken to reach the surface is $t$, what is the expression for $t$ considering the variation of gravity with height $g(r)=\frac{GM}{r^2}$?

Solution

1. The velocity $v$ of a particle falling from height $h$ is given by the conservation of energy: $\frac{1}{2}{v}^2=GM(\frac{1}{r}-\frac{1}{R+h})$, where $r$ is the distance from the center of the Earth.
2. The time $t$ is found by integrating $dt=-\frac{dr}{v}$, resulting in $t={\int }_R^{R+h}\frac{dr}{\sqrt{2GM(\frac{1}{r}-\frac{1}{R+h})}}$.
3. Substituting $r=R+y$ and using the binomial expansion $(1+\frac{y}{R}{)}^{-1/2}\approx 1-\frac{y}{2R}+\frac{3y^2}{8R^2}$ for small $h/R$, the integral evaluates to $t\approx \sqrt{\frac{2h}{g}}(1+\frac{h}{6R})$.
4. Hence the answer is (C).