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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//sqrt(2)`

C

R/2

D

`sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`g. = (GM)/(r^(2)) = (GM)/((R+h)^(2))`
`because g. = (g)/(9) " "because g = (GM)/(R^(2))`
`(GM)/((R+h)^(2)) = (1)/(9)(GM)/(R^(2)) rArr 3R = r + h`
`(h)/(R ) = 2` or h = 2R
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