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Question

Asked in: JEE-2009

The height at which the acceleration due to gravity becomes g/9 ( where g = the acceleration due to gravity on the surface of the earth ) in terms of R, the radius of the earth is

A.

3r

B.

\frac{R}{\sqrt{2}}

C.

R/2

D.

\sqrt{2}R

Answers (1)

best_answer

assume at height h from the surface of earth the gravity is g/9. so by applying gravity formula

\frac{GM}{9R^2}=\frac{GM}{(R+h)^2}

9R^2=(R+h)^2

3R=R+h

2R=h

this is right way to solve the above question

Posted by

neetan kumar lalotra

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