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Question

(C) The orbital velocity of a satellite at an altitude $h$ is given by $v = \sqrt{\frac{GM}{R+h}}$.For the first satellite,$h = 0$,so the velocity is

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$\sqrt{\frac{2}{3}}v$

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(C) The orbital velocity of a satellite at an altitude $h$ is given by $v = \sqrt{\frac{GM}{R+h}}$.For the first satellite,$h = 0$,so the velocity is $v_1 = \sqrt{\frac{GM}{R}} = v$.For the second satellite,the altitude is $h = \frac{R}{2}$.Substituting this into the formula,we get $v_2 = \sqrt{\frac{GM}{R + \frac{R}{2}}} = \sqrt{\frac{GM}{\frac{3R}{2}}} = \sqrt{\frac{2GM}{3R}}$.We can rewrite this as $v_2 = \sqrt{\frac{2}{3}} \sqrt{\frac{GM}{R}}$.Since $v = \sqrt{\frac{GM}{R}}$,we have $v_2 = \sqrt{\frac{2}{3}}v$.

The orbital velocity of an artificial satellite in a circular orbit just above the Earth's surface is $v$. For a satellite orbiting at an altitude of half of the Earth's radius,the orbital velocity is

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