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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 accleeration 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)//R`

Text Solution

Verified by Experts

The correct Answer is:
A
We know, `g=(GM)/(R^(2))`
`g.=(g)/(9)=(GM)/(9R^(2))" "...(i)`
Also `g.=(GM)/((R+h)^(2))" "...(ii)`
Equating eqns. (i) and (ii), we get
`9R^(2)=(R+h)^(2)`
or `3R=R+h or h=3R-R=2R`
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