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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/sqrt3`

C

`R/2`

D

`sqrt(2R)`

Text Solution

Verified by Experts

The correct Answer is:
A
Acceleration due to gravity at height h.
`rArr 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 therefore h=2R`
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