Gravitation

JEE Main level, single correct, medium

85 of 223

Question 85 of 223

The value of the acceleration due to gravity is g1{g}_{1} at a height h=R2h=\dfrac{R}{2} (R=R= radius of the earth) from the surface of the earth. It is again equal to g1{g}_{1} at a depth dd below the surface the earth. The ratio (dR)\left(\dfrac{d}{R}\right) equals:

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Answer: (B) 59\dfrac{5}{9}

Given that
gn=gd{g}_{n}={g}_{d}
GM(R+h)2=GMR3(R−d)\dfrac{GM}{(R+h{)}^{2}}=\dfrac{GM}{{R}^{3}}(R-d)
GM(R+R/2)2=GMR3(R−d)\dfrac{GM}{(R+R/2{)}^{2}}=\dfrac{GM}{{R}^{3}}(R-d)
4GM9R2=GMR2(1−dR)\dfrac{4GM}{9{R}^{2}}=\dfrac{GM}{{R}^{2}}\left(1-\dfrac{d}{R}\right)
49=1−dR\dfrac{4}{9}=1-\dfrac{d}{R}
dR=1−49=59\dfrac{d}{R}=1-\dfrac{4}{9}=\dfrac{5}{9}
d=59Rd=\dfrac{5}{9}R

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