First slide

Acceleration due to gravity and its variation

Question

If the change in the value of g at a height h above the surface of the earth is the same as at a depth x below it, then (both x and h being much smaller than the radius of the earth)

Moderate

A

$x = h$
B

$x = 2 h$
C

$x = \frac{h}{2}$
D

$x = h^{2}$

Solution

The value of g at the height h from the surface of earth
$g^{'} = g (1 - \frac{2 h}{R})$
The value of g at depth x below the surface of earth
$g^{'} = g (1 - \frac{x}{R})$
These two are given equal, hence $(1 - \frac{2 h}{R}) = (1 - \frac{x}{R})$
On solving, we get x = 2h

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