Discussion
The orbital velocity \(v_{o}\) of an artificial satellite revolving around the earth at a height R from the surface of the earth in terms of escape velocity \(v_{e}\) from the earth is (R - radius of the earth)
Answer: A
The orbital velocity is \(\frac{v_{e}}{2}\).
At height R above the surface the orbit radius is 2R. The orbital speed there and the escape speed from the surface are
\[ \begin{aligned} v_{o} &= \sqrt{\frac{GM}{2R}} \\ v_{e} &= \sqrt{\frac{2GM}{R}} \end{aligned} \]
\[ \begin{aligned} \frac{v_{o}}{v_{e}} &= \sqrt{\frac{GM/2R}{2GM/R}} \\ &= \sqrt{\frac{1}{4}} = \frac{1}{2} \end{aligned} \]
\(\frac{v_{e}}{\sqrt{2}}\) is the orbital speed just above the surface (orbit radius R), which ignores the height. The options equal to or larger than \(v_{e}\) would let the satellite escape.
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