A body of mass 'm' is raised to a height ' $10 R$ ' from the surface of the earth, where 'R' is the radius of the earth. Suppose it is released now. The speed with which it will strike the surface of the earth is nearly
[ $g$ = acceleration due to gravity]

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TestHub · Accepted Answer

By the principle of conservation of mechanical energy:

$K_{i} + U_{i} = K_{f} + U_{f}$

The initial height from the surface is $10 R$ , so the initial distance from the center of the earth is $r_{i} = R + 10 R = 11 R$ . The final distance is $r_{f} = R$ .

$0 - \frac{GM m}{11 R} = \frac{1}{2} m v^{2} - \frac{GM m}{R}$

$\frac{1}{2} m v^{2} = \frac{GM m}{R} - \frac{GM m}{11 R}$

$\frac{1}{2} v^{2} = \frac{10 GM}{11 R}$

$v = \frac{20 GM}{11 R}$

Using the relation $g = \frac{GM}{R ^{2}}$ , we substitute $GM = g R^{2}$ :

$v = \frac{20 g R ^{2}}{11 R} = \frac{20}{11} g R$

Answer: $\frac{20 g R}{11}$

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