The escape velocity at the surface of Earth is approximately 11 km/s.What is the mass, in units of M_E (the mass of the Earth), of a planet with twice the radius of Earth for which the escape speed is twice that for Earth? A)2 M_E B)4 M_E C)8 M_E D)1/2 M_E E)1/4 M_E

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Quizplus · Accepted Answer

The escape velocity $v_e$ from a planet is given by the formula $v_e = \sqrt{\frac{2GM}{r}}$, where $G$ is the gravitational constant, $M$ is the mass of the planet, and $r$ is the radius of the planet. If the escape velocity is doubled ($2v_e$) and the radius is also doubled ($2r$), we can set up the equation $\sqrt{\frac{2G(2M)}{2r}} = 2\sqrt{\frac{2GM}{r}}$. Simplifying this equation, we find that $M$ must be increased by a factor of 8 to maintain the condition, hence the mass of the planet would be $8 M_E$.

The Escape Velocity at the Surface of Earth Is Approximately