First slide

Gravitaional potential energy

Question

A body is projected upwards with a velocity of 4 x 11.2km / s from the surface of earth. What will be the velocity of the body when it escapes the gravitational pull of earth ?

Moderate

A

11.2km / s
B

2 x 11.2km/s
C

3 x 11.2km/s
D

$\sqrt{15} \times 11 \cdot 2 km / s$

Solution

Initial kinetic energy
$= \frac{1}{2} {mv}^{2} = \frac{1}{2} m (4 \times 11 \cdot 2)^{2} = 16 \times \frac{1}{2} {mv}_{e}^{2}$
Final kinetic energy should be $15 \times \frac{1}{2} {mv}_{e}^{2}$
Hence $\frac{1}{2} {mv}^{' 2} = 15 \times \frac{1}{2} {mv}_{e}^{2}$
$\begin{array}{l} v^{' 2} = 15 \times {v_{e}}^{2} or v^{'} = \sqrt{15} \cdot v_{e} \\ ∴ v^{'} = \sqrt{15} \times 11 \cdot 2 km / s \end{array}$

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