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Escape velocity on a planet is ve. If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes(A) 4ve(B) 2ve(C) ve(D) 0.5 ve |
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Answer» Correct option is: (B) \(2v_e\) Escape Velocity \(v_e = \sqrt {2gRe}\) \(\frac{V_{e1}}{V_{e2}} \propto \sqrt \frac{4R_1}{R_1}\) \(V_{escape \ 2} = 2 \ V_{escape \ 1}\) Correct option is: (B) 2ve |
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