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

Options

A. 2R
B. `R/sqrt2`
C. `R/2`
D. `sqrt2R`

Answer

A

Explanation

`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