The change in the value of g a height h about the surface of the earth is the same as a depth d below the surface of earth- When both d and h are much smaller than the radius of earth- then which one of the following is correct-

Question

Toppr Admin · Accepted Answer

Acceleration due to gravity at a height $h$ , $g_{h}^{'} = g (1 - \frac{2 h}{R})$
Acceleration due to gravity at a depth $d$ , $g_{d}^{'} = g (1 - \frac{d}{R})$
$∴ g (1 - \frac{2 h}{R})$ $= g (1 - \frac{d}{R})$
or $\frac{2 h}{R} = \frac{d}{R}$
or $d = 2 h$ .

The change in the value of g at a height h about the surface of the earth is the same as at a depth d below the surface of earth. When both d and h are much smaller than the radius of earth, then which one of the following is correct?

Acceleration due to gravity at a height h, g′h=g(1−2hR) Acceleration due to gravity at a depth d, g′d=g(1−dR)∴ g(1−2hR) =g(1−dR)or 2hR=dR or d=2h.

The change in the value of $g$ at a height $h$ about the surface of the earth is the same as at a depth $d$ below the surface of earth. When both $d$ and $h$ are much smaller than the radius of earth, then which one of the following is correct?

Acceleration due to gravity at a height $h$ , $g_{h}^{'} = g (1 - \frac{2 h}{R})$
Acceleration due to gravity at a depth $d$ , $g_{d}^{'} = g (1 - \frac{d}{R})$
$∴ g (1 - \frac{2 h}{R})$ $= g (1 - \frac{d}{R})$
or $\frac{2 h}{R} = \frac{d}{R}$
or $d = 2 h$ .