Answers

2015-09-05T15:32:01+05:30
Gravity at a depth d is given by

g(d) = GM_e \frac{R_e-d}{R_e^3}

given that gravity at depth d is 1/4th the value of g on surface. So

 \frac{g(d)}{g(0)}= \frac{1}{4}\\ \\ \Rightarrow  \frac{GM_e \frac{R_e-d}{R_e^3} }{\frac{GM_e}{R_e^2} } = \frac{1}{4} \\ \\ \Rightarrow  \frac{R_e-d}{R_e^3} \times R_e^2 =  \frac{1}{4}  \\ \\ \Rightarrow  \frac{R_e-d}{R_e}=  \frac{1}{4}  \\ \\ \Rightarrow  4R_e-4d = R_e\\ \\ \Rightarrow d = \frac{3}{4}R_e

So at a depth of 3R/4, gravity will be 1/4th of the value of g on surface.
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