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(B) The orbital velocity $V$ of a satellite at a distance $r$ from the center of a planet is given by $\frac{mV^2}{r} = \frac{GMm}{r^2}$, which simpli

Vedclass · Accepted Answer

$11$

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(B) The orbital velocity $V$ of a satellite at a distance $r$ from the center of a planet is given by $\frac{mV^2}{r} = \frac{GMm}{r^2}$, which simplifies to $V = \sqrt{\frac{GM}{r}}$.Here, $r = R + h = 2 	imes 10^6 \, m + 20 	imes 10^3 \, m = 2.02 	imes 10^6 \, m$.The time period $T_p$ of one revolution is $T_p = \frac{2\pi r}{V} = 2\pi \sqrt{\frac{r^3}{GM}}$.The number of revolutions $n$ in time $T = 24 \, hours = 24 	imes 3600 \, s$ is $n = \frac{T}{T_p} = \frac{T}{2\pi} \sqrt{\frac{GM}{r^3}}$.Substituting the values: $n = \frac{24 	imes 3600}{2 	imes 3.14} \sqrt{\frac{6.67 	imes 10^{-11} 	imes 8 	imes 10^{22}}{(2.02 	imes 10^6)^3}}$.$n = \frac{86400}{6.28} \sqrt{\frac{53.36 	imes 10^{11}}{8.2424 	imes 10^{18}}} = 13758 	imes \sqrt{6.47 	imes 10^{-7}} = 13758 	imes 8.04 	imes 10^{-4} \approx 11.06$.Thus, the number of complete revolutions is $11$.

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