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Question

Physics Question on Gravitation

If the radius of earth of RR then the height h at which the value of gg becomes one-fourth, will be

A

R8\frac{R}{8}

B

3R8\frac{3R}{8}

C

3R4\frac{3R}{4}

D

R2\frac{R}{2}

Answer

3R8\frac{3R}{8}

Explanation

Solution

The value of acceleration due to gravity at a height hh above the surface of the earth is given by
g=g(1+hR)2g^{'}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}
where RR is radius of earth. When li is negligible compared to RR, we have
g=g(1+hR)2=g(12hR)g^{'}=g\left(1+\frac{h}{R}\right)^{-2}=g\left(1-\frac{2 h}{R}\right)
Given g=g4 g^{'}=\frac{g}{4}
g4=g(12hR)\frac{g}{4}=g\left(1-\frac{2 h}{R}\right)
14=12hR\Rightarrow \frac{1}{4}=1-\frac{2 h}{R}
2hR=34\Rightarrow\frac{2 h}{R}=\frac{3}{4}
h=3R8\Rightarrow h=\frac{3 R}{8}Note : The value of acceleration due to gravity decreases on going above or below the surface of earth.