Discussion
The tangential velocity component 'V' of a spacecraft, which is in a circular orbit of radius 'R' around a spherical Earth (μ = GM → gravitational parameter of Earth) is given by the following expression.
Answer: B
The velocity is \(V = \sqrt{\mu/R}\).
In a circular orbit, gravity provides exactly the centripetal force:
\[ \begin{aligned} \frac{mV^{2}}{R} &= \frac{GMm}{R^{2}} = \frac{\mu m}{R^{2}} \\ V &= \sqrt{\frac{\mu}{R}} \end{aligned} \]
\(\sqrt{\mu/2R}\) is smaller than the circular speed by a factor \(\sqrt{2}\); it equals half the escape speed \(\sqrt{2\mu/R}\). \(\dfrac{2\pi}{\sqrt{\mu}}R^{3/2}\) is the orbital period, a time, not a velocity, and the version with \(R^{2/3}\) is that period with the exponent inverted.
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