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Question

The escape velocity on earth is 11.2 km s−1. What is the escape velocity on a planet of radius four times the radius of earth and one fourth of the density of earth?

A
11.2 km s1
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B
22.4 km s1
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C
44.8 km s1
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D
26.2 km s1
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Solution

The correct option is B 22.4 km s1
The potential energy is given by GMr

Escape velocity scales as the square root of the potential(since kinetic energy is quadratic with speed). Double the radius gives eight times the mass . Since the surface is at double distance from the center now the potential energy has increased by 4 times . Thus the escape velocity is twice as big. Hence the escape velocity will be 22.4 km s1

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