Gravitation

JEE Main level, single correct, medium

66 of 223

Question 66 of 223

At a certain depth dd below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height 3R3R above earth surface. Where RR is Radius of earth (Take R=6400kmR=6400km). The depth dd is equal to

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Answer: (D) 4800km4800km

The magnitude of acceleration due to gravity g1{g}_{1} at a depth of dd below the surface of Earth is
g1=GMR3(R−d)=GMR2(1−dR){g}_{1}=G\dfrac{M}{{R}^{3}}\left(R-d\right)=G\dfrac{M}{{R}^{2}}\left(1-\dfrac{d}{R}\right) ....(1)
The magnitude of acceleration due to gravity g2{g}_{2} at a height dd above the surface of Earth is
g2=GM(R+d)2{g}_{2}=G\dfrac{M}{{\left(R+d\right)}^{2}} ....(2)
Given here, g1=4g2{g}_{1}=4{g}_{2}
GMR2(1−dR)=4GM(R+3R)2\dfrac{GM}{{R}^{2}}\left(1-\dfrac{d}{R}\right)=\dfrac{4GM}{{\left(R+3R\right)}^{2}}
⇒GMR2[1−dR]=4GM(4R)2\Rightarrow \dfrac{GM}{{R}^{2}}\left[1-\dfrac{d}{R}\right]=\dfrac{4GM}{{\left(4R\right)}^{2}}
⇒1−dR=14⇒dR=34⇒d=34R\Rightarrow 1-\dfrac{d}{R}=\dfrac{1}{4} \Rightarrow \dfrac{d}{R}=\dfrac{3}{4} \Rightarrow d=\dfrac{3}{4}R
⇒d=4800km\Rightarrow d=4800km

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