The height at which the acceleration due to gravity becomes `g/9` (where g = acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is .........

Options

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

Answer

B

Explanation

`implies` Gravitational acceleration at a height from the earth .
`g/9 = (GM_e)/((R+h)^2)`
`:. (GM_e) /(9R^2) =(GM_e)/((R+h)^2) " "[ :. g = (GM_e)/R^2]`
`:. 9R^2 = (R+h)^2`
`:. 3R = R + h `
`:. h = 2R`