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Q.

If g is the acceleration due to gravity on the surface of earth, its value at a height equal to double the radius of earth is

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a

$\frac{g}{3}$

b

$\frac{g}{2}$

c

$g$

d

$\frac{g}{9}$

answer is D.

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Detailed Solution

$g_{h} = \frac{g}{{(1 + \frac{h}{R})}^{2}} = \frac{g}{{(1 + \frac{2 R}{R})}^{2}} = \frac{g}{3^{2}} = \frac{g}{9}$

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