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The height at which the acceleration due...

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

2R

B

`R/sqrt2`

C

`R/2`

D

`sqrt2R`

Text Solution

Verified by Experts

The correct Answer is:
A
`g'=(GM)/((R+h)^2)`, acceleration due to gravity at height h
`=g/9=(GM)/R^2 R^2/((R+h)^2)=g(R/(R+h))^2 rArr 1/9=(R/(R+h))^2 rArr R/(R+h) =1/3`
`rArr ` 3R = R+h `rArr ` 2R=h
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