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Gravitational potential energy = U = \frac{GMm}{R}
Escape velocity = v_{esc} = \sqrt{\frac{2GM}{r}}
Kinetic energy of launch:
\begin{align}
K_0 &= \frac{1}{2}m(2v_{esc})^2 = \frac{1}{2} m 4v_{esc}^2 \\
&= 2 m v_{esc}^2 = 2m \frac{2GM}{r} \\
&= 4 U
\end{align}
Kinetic energy after launch = K_0 - U = 3U
\begin{align}
3U &= 3 \frac{GMm}{r} = \frac{3}{2} \frac{2GM}{r} \\
&= \frac{3}{2} v_{esc}^2 = \frac{1}{2} \left(\sqrt{3}v_{esc}\right)^2
\end{align}
So the answer is D) \sqrt{3}v_{esc}
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Nice job
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I didn't remember the escape velocity formula off the top of my head, so the answer is no.
Me plus looking it up would look like what @SimpleStacker did.
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I enjoy solving puzzles like these... I almost wish mathematical econ was more useful in real life so those skills would be more applicable.
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I didnt want to get too cute with the answer but there arent a lot of details provided, so I do make some assumptions.
  1. There is gravity on the planet (clearly because it has mass and a given escape velocity)
  2. There is matter in outer space that will have gravity
The speed exceeds the escape velocity so greatly that I will say that the rock will land inside some mass in outer space after which the speed of the rock will be:
A) Zero
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@SimpleStacker got it right with his elegant solution
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I was way too late to the game. did that ken game ever end?
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