The escape velocity from the surface of the Earth is about 11.2 km/s. If the radius of Earth were doubled (mass unchanged), the new escape velocity would be approximately:
Solution
$v_e = \sqrt{2GM/R}$, so $v_e \propto 1/\sqrt{R}$. Doubling $R$: $v_e' = 11.2/\sqrt{2} \approx 7.9$ km/s
About this question
Subject: Physics · Chapter: Gravitation · Topic: Escape Velocity
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