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Q. At a certain depth "$d$" below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $3 R$ above earth surface. Where $R$ is Radius of earth (Take $R =6400\, km$ ). The depth $d$ is equal to

JEE MainJEE Main 2023Gravitation

Solution:

$ \frac{ GM }{ R ^2}\left[1-\frac{ d }{ R }\right]=\frac{4 \times GM }{(4 R )^2}$
$ 1-\frac{ d }{ R }=\frac{1}{4} \Rightarrow \frac{ d }{ R }=\frac{3}{4} \Rightarrow d \frac{3}{4} R $
$ d =4800\, km $