Discussion
An artificial satellite revolves about the earth at height H (H \(\ll\) R, R = radius of earth). The orbital period, for which a man in the satellite will be in state of weightlessness is given as :
Answer: B
The period is \(2\pi\sqrt{R/g}\).
For a satellite close to the earth (\(H \ll R\)), gravity supplies the centripetal force, and \(GM = gR^{2}\):
\[ \begin{aligned} \frac{mv^{2}}{R} &= mg \Rightarrow v = \sqrt{gR} \\ T &= \frac{2\pi R}{v} = 2\pi\sqrt{\frac{R}{g}} \end{aligned} \]
In such an orbit the man and the satellite fall freely together, so he is weightless. This is about 84 minutes. The option \(2\pi\sqrt{g/R}\) has the ratio inverted and is not even a time, and the \(1/2\pi\) options are frequencies rather than periods.
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