A satellite is revolving around the earth in a fixed circular orbit. $V$ is speed of satellite and E is its KE. This satellite will escape from earth’s gravitational influence with a speed of satellite is increased by $Δ V$ and its $K E$ is increased by $Δ E .$ Then

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$Δ V = V, Δ E = \sqrt{2} E$

$Δ V = \frac{V}{\sqrt{2}}, Δ E = \frac{E}{\sqrt{2}}$

$Δ V = 0.41 V, Δ E = E$

$Δ V = \sqrt{2} V, Δ E = E$

answer is B.

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Detailed Solution

$V = \sqrt{\frac{G m}{r}} T E = - K E = - E$

$V_{e} = \sqrt{\frac{2 G m}{r}} K E_{e} = \frac{1}{2} m {V_{e}}^{2} = \frac{1}{2} m 2 {V_{o}}^{2} = m {V_{o}}^{2}$

$V_{e} = \sqrt{2} V_{0} K E_{e} - K E = 2 (\frac{1}{2} m {V_{o}}^{2}) - \frac{1}{2} m {V_{o}}^{2} = \frac{1}{2} m {V_{o}}^{2} = E$

$Δ V = (\sqrt{2} - 1) V_{0} = 0.41 V_{0}$

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