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(A) The acceleration due to gravity at a height $h$ above the surface is given by $g_h = g \left(1 + \frac{h}{R}\right)^{-2} \approx \frac{g}{(1 + h/R

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(A) The acceleration due to gravity at a height $h$ above the surface is given by $g_h = g \left(1 + \frac{h}{R}\right)^{-2} \approx \frac{g}{(1 + h/R)^2}$.Given $g_h = \frac{g}{4}$,we have $\frac{g}{4} = \frac{g}{(1 + h/R)^2}$.Taking the square root on both sides,$2 = 1 + \frac{h}{R}$,which gives $\frac{h}{R} = 1$,so $h = R$.The acceleration due to gravity at a depth $d$ below the surface is given by $g_d = g \left(1 - \frac{d}{R}\right)$.Given $g_d = \frac{g}{4}$,we have $\frac{g}{4} = g \left(1 - \frac{d}{R}\right)$.Dividing by $g$,$\frac{1}{4} = 1 - \frac{d}{R}$,which gives $\frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4}$,so $d = \frac{3R}{4}$.Now,the ratio $\frac{h}{d} = \frac{R}{3R/4} = \frac{4}{3}$.

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