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 :

Options

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

Answer

D

Explanation

(d) we know that (g')/g) = (R^2)/((R +h)^2`
`:. (g//9)/(g) = [(R )/(R +h)]^2 :. h =2R`