First slide

Gravitaional potential energy

Question

The escape velocity of a body on the surface of the earth is 11.2 km./s. If the earth's mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes

Moderate

A

5.6 km/s
B

11.2km/s
C

22. 4 km/s
D

44.8 km/s

Solution

Escape velocity $v_{e} = \sqrt{(\frac{2 {GM}_{e}}{R_{e}})}$
or $11 \cdot 2 = \sqrt{(\frac{2 {GM}_{e}}{R_{e}})}$
Let the new escape velocity becomes $v_{e}^{'}$ Now
$\begin{aligned} V_{e}^{'} = \sqrt{(\frac{2 G \times (2 M_{e})}{(R_{e} / 2)})} = 2 \sqrt{(\frac{2 {GM}_{e}}{R_{e}})} \\ = 2 \times 11 \cdot 2 = 22 \cdot 4 km / s \end{aligned}$

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