Question

State expression for acceleration due to gravity at depth 'd' and altitude 'h'. Hence show that ratio is equal to $\frac{R-d}{R-2h}$, assuming that h < R, where R is radius of the earth.

Answer 1

At depth ${}^{\prime }{d}^{\prime }$

$\begin{array}{c}F=\frac{GM\left(R-d\right)m}{R^3}\\ =m.\frac{GM}{R^2}\left(1-\frac{d}{R}\right)\\ =m.g_{eff}\\ \therefore g_{eff}=g\left(1-\frac{d}{R}\right)\end{array}$

At a height ${}^{\prime }{h}^{\prime }$

$\begin{array}{c}F=\frac{GMm}{{\left(R+h\right)}^2}\\ =m.\left\{\frac{GM}{{\left(R+h\right)}^2}\right\}\\ \Rightarrow g_{eff}=\frac{GM}{{\left(R+h\right)}^2}\\ =\frac{GM}{R^2}.\frac{1}{{\left(1+\frac{h}{R}\right)}^2}\\ =\frac{GM}{R^2}\left(1-\frac{2h}{R}\right)\\ =g\left(1-\frac{2h}{R}\right)\end{array}$

$\therefore$ Ratio is $\frac{R-d}{R-2h}$.