Discussion
If the acceleration due to gravity on the surface of a planet of mass m and radius r is g, then the escape velocity of a body from the surface of the planet is
Answer: A
The escape velocity is \(\sqrt{2gr}\).
Escape needs kinetic energy enough to bring the gravitational potential energy up to zero:
\[ \begin{aligned} \tfrac{1}{2}mv_{e}^{2} &= \frac{GMm}{r} \\ v_{e} &= \sqrt{\frac{2GM}{r}} \end{aligned} \]
On the surface \(g = \frac{GM}{r^{2}}\), so \(GM = gr^{2}\) and \(v_{e} = \sqrt{2gr}\). (The planet's mass is called m in the question; the result does not depend on the escaping body's mass.)
\(\sqrt{gr}\) is the orbital speed just above the surface, smaller by \(\sqrt{2}\). \(\sqrt{2g/r}\) has the wrong units, and \(gr^{2}\) and \(2gr^{2}\) are not even speeds.
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