Let the acceleration due to gravity be `g_(1)` at a height `h` above the earth's surface `g_(2)` at a depth `d` below the earth's surface. If `g_(1)=g_(2),h lt lt R` and `d lt lt R` then

Options

A. `h = (d)/(2)`
B. h = 2d
C. h = 3d
D. h = d

Answer

A

Explanation

Let .g. be acceleration due to gravity on earth.s surface `g = (GM)/(R^(2))`
Acceleration due to gravity at depth .d. from the earth,s surface
`g. = g(1-(d)/(R )) " "because d lt lt R`
Acceleration due to gravity at height .h., from the earth.s surface
`g" = (gR^(2))/((R+h)^(2)) = g[1+(h)/(R)]^(-2) = g(1-(2h)/(R )), because h lt lt R`
given, `rArr g.. = g.`
`g(1-(2h)/(R )) = g(1-(d)/(R ))`
`2h = d rArr h = (d)/(2)`