If `g` is acceleration due to gravity on the surface of the earth, having radius `R`, the height at which the acceleration due to gravity reduces to `g//2` is

Options

A. R/2
B. `sqrt2R`
C. `R//sqrt2`
D. `(sqrt2-1)R`

Answer

d

Explanation

`(g_(2))/(g_(1))=((R)/(R+h))^(2)`
`(g)/(2g)=((R)/(R+h))^(2)`
`(1)/(sqrt2)=(R)/(R+h)`
`therefore" "R+h=sqrt2R`
`h=sqrt2R-R=R(sqrt2-1).`