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

Options

A. `2R`
B. `(R )/sqrtr(3)`
C. `(R )/(2)`
D. `sqrt(2)R`

Answer

A

Explanation

`g=(GM)/(r+h)^(2)` acceleration due to gravity at height h
? `(g)/(g)=(GM)/(R^(2)).(R^(2))/(R+h)^(2)=g(r )/(R+h)^(2)`
`rarr 1/9=(R )/(R+n)^(2) rarr (R )/(R+h)=1/3`
`rarr 3R =R +h rarr 2R =h`