Question

A rocket has landed on Planet X , which has half
the radius of Earth. An astronaut onboard the
rocket weighs twice as much on Planet X as on
Earth. If the escape velocity for the rocket taking
off from Earth is V , then its escape velocity on
Planet X is? The answer is V but why? So far I have calculated that the acceleration due to gravity on planet X is two times the acceleration due to gravity on planet Earth.

Answer 1

420 blaze it \[e=mc^2\]

Answer 2

what we know about the planet (p) and earth (e):
\[g_p = 2 g_e \ and \ r_p = \frac{r_e}{2}\]
we also know that gravity at the surface is calculated from\[mg = \frac{G \ M \ m}{r^2} \ which \ can \ be \ restated \ as \ G = \frac{g \ r^2}{M} \]
comparing the planet and earth
\[\frac{2 g_e \frac{r^2_e}{4}}{M_p} = \frac{g_e \ r^2_e}{M_e} \ so \ M_p = \frac{1}{2}M_e\]

so the escape velovity on the planet in earth terms is
\[V_{escape} = \sqrt{\frac{2 G (\frac{1}{2} M_e)}{\frac{r_e}{2}}} = \sqrt{\frac{2G M_e}{r_e}}\]
ie same result as for earth