Gravitation

JEE Main level, single correct, medium

50 of 223

Question 50 of 223

The height in terms of radius of the earth (R)(R), at which the acceleration due to gravity becomes g9\dfrac{g}{9}, where gg is acceleration due to gravity on earth's surface, is ______.

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Answer: (C) 2R2R

The acceleration due to gravity at a height hh above the surface of the earth is given by

g′=g(1+hR)2g' = \dfrac{g}{\left(1 + \dfrac{h}{R}\right)^2}

Given that g′=g9g' = \dfrac{g}{9}, we get

g9=g(1+hR)2\dfrac{g}{9} = \dfrac{g}{\left(1 + \dfrac{h}{R}\right)^2}

(1+hR)2=9\left(1 + \dfrac{h}{R}\right)^2 = 9

1+hR=31 + \dfrac{h}{R} = 3

hR=2\dfrac{h}{R} = 2

h=2Rh = 2R

Answer: 2R2R

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