First slide

Acceleration due to gravity and its variation

Question

The value of g at a certain height h above the free surface of the earth is x/4 where x is the value of g at the surface of the earth. The height h is

Moderate

A

R
B

2R
C

3R
D

4R

Solution

$x = \frac{GM}{R^{2}}$
$Again, \frac{x}{4} = \frac{GM}{(R + h)^{2}} or x = GM {(\frac{2}{R + h})}^{2}$
$∴ \frac{1}{R^{2}} = {(\frac{2}{R + h})}^{2} = \frac{2}{R + h} or R + h = 2 R or h = R$

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