First slide

Gravitaional potential energy

Question

A body is projected up with a velocity equal to 3/4th of the escape velocity from the surface of earth. The height it reaches is (R=Radius of earth)

Moderate

A

$10R / 9$
B

$9 R / 7$
C

$9 R / 8$
D

$10 R / 3$

Solution

From the law of conservation of energy
$\frac{1}{2} m v^{2} = \frac{G M m}{R} - \frac{G M m}{R + h}$
$\frac{v^{2}}{2} = \frac{G M}{R (R + h)} (h)$
$\frac{1}{2} {(\frac{3}{4} \sqrt{\frac{2 G M}{R}})}^{2} = \frac{G M}{R} \frac{h}{(R + h)}$
$\frac{9}{16} \frac{2 G M}{R} \frac{1}{2} = \frac{G M}{R} \frac{h}{R + h}$
$9 R + 9 h = 16 h$
$9 R = 7 h$
$h = \frac{9 R}{7}$

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