Gravitation

JEE Main level, single correct, medium

54 of 223

Question 54 of 223

At what distance above and below the surface of the earth a body will have same weight? (Take radius of earth as RR)

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Answer: (D) 5R−R2\dfrac{\sqrt{5}R-R}{2}

The acceleration due to gravity at a height hh above the surface of the Earth is given by
gp=gR2(R+h)2...(1){g}_{p}=\dfrac{g{R}^{2}}{(R+h{)}^{2}}...\left(1\right)
The acceleration due to gravity at a depth hh below the surface of the Earth is given by
gq=g(1−hR)...(2){g}_{q}=g\left(1-\dfrac{h}{R}\right)...\left(2\right)
According to the given problem,
gp=gq...(3){g}_{p}={g}_{q}...\left(3\right)
Equations (1), (2) and (3) imply that
g(1+hR)2=g(1−hR)⇒(1−h2R2)(1+hR)=1...(4)\dfrac{g}{{\left(1+\dfrac{h}{R}\right)}^{2}}=g\left(1-\dfrac{h}{R}\right) \Rightarrow \left(1-\dfrac{{h}^{2}}{{R}^{2}}\right)\left(1+\dfrac{h}{R}\right)=1...\left(4\right)
Take hR=x\dfrac{h}{R}=x
So, from equation (4), it follows that
x3−x+x2=0⇒x2+x−1=0⇒x=5−12⇒h=R2(5−1){x}^{3}-x+{x}^{2}=0 \Rightarrow {x}^{2}+x-1=0 \Rightarrow x=\dfrac{\sqrt{5}-1}{2} \Rightarrow h=\dfrac{R}{2}(\sqrt{5}-1)

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