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Q.

A space ship goes into a circular orbit close to the earth’s surface. What additional velocity must be imparted to the ship so that it is able to escape the gravitational pull of the earth? $(R = 6400 km, g = 9.8 {ms}^{- 2})$

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answer is 3.278.

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Detailed Solution

The orbital velocity in a circular orbit close to the earth is $V = \sqrt{gR}$

The velocity required to escape $V_{e} = \sqrt{2 gR}$

Hence additional velocity required is

$V_{e} - V = (\sqrt{2} - 1) \sqrt{gR}$

$V_{e} - V = 0.414 \times \sqrt{9.8 \times 6400 \times 10^{3}}$

$= 3278.71 m / s$

$V_{e} - V = 3.278 {kms}^{- 1}$

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A space ship goes into a circular orbit close to the earth’s surface. What additional velocity must be imparted to the ship so that it is able to escape the gravitational pull of the earth? (R=6400 km, g=9.8 ms−2)