The height at which the acceleration due to gravity becomes `g//9` in terms of R the radius of the earth is

Options

A. `(R )/(sqrt(2))`
B. R
C. `sqrt(2)R`
D. 2 R

Answer

D

Explanation

`g=(GM)/(R^(2))` and `g'=(GM)/((R+h)^(2))`
`(g')/(g)=(1)/(9)=(GM)/((R+h)^(2))xx(R^(2))/(GM)=(R^(2))/((R+h)^(2))`
`therefore 9R^(2)=(R+h)^(2) " " therefore 3R=R+h " " therefore h = 2R`