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(C) The escape velocity $v$ at a distance $r$ from the center of the Earth is given by the formula $v = \sqrt{\frac{2GM}{r}}$.For the surface of the E

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$1/\sqrt{2}$

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(C) The escape velocity $v$ at a distance $r$ from the center of the Earth is given by the formula $v = \sqrt{\frac{2GM}{r}}$.For the surface of the Earth,the distance $r = R$,so the escape velocity is $V_{E} = \sqrt{\frac{2GM}{R}}$.For a space station at a height $h = R$ above the surface,the distance from the center of the Earth is $r = R + h = R + R = 2R$.The escape velocity $v_{s}$ at the space station is $v_{s} = \sqrt{\frac{2GM}{2R}} = \sqrt{\frac{GM}{R}}$.Comparing $v_{s}$ with $V_{E}$:$v_{s} = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM}{R}} = \frac{1}{\sqrt{2}} V_{E}$.Thus,the escape velocity on the space station is $\frac{1}{\sqrt{2}}$ times $V_{E}$.

$A$ space station is at a height equal to the radius of the Earth. If $V_{E}$ is the escape velocity on the surface of the Earth,the escape velocity on the space station is __ times $V_{E}$.

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