A remote-sensing satellite of earth revolves in a circular orbit at a height of $0.25 \times 10^{6} \mathrm{~m}$ above the surface of earth. If earth’s radius is $6.38 \times 10^{6} \mathrm{~m}$ and $\mathrm{g}=9.8 \mathrm{~ms}^{-2}$, then the orbital speed of the satellite is: (1) $6.67 \mathrm{~km} \mathrm{~s}^{-1}$ (2) $7.76 \mathrm{~km} \mathrm{~s}^{-1}$ (3) $8.56 \mathrm{~km} \mathrm{~s}^{-1}$ (4) $9.13 \mathrm{~km} \mathrm{~s}^{-1}$

contexto.120

3 years ago

The correct Solution is (2)
$
\begin{aligned}
& V_{0}=\sqrt{\frac{G M}{r}}=\sqrt{\frac{G M}{\overline{R^{2}}}, \frac{R^{2}}{r}} \\
=& \sqrt{\frac{9.8 \times 6.38 \times 6.38}{6.63 \times 10^{6}}}=\sqrt{60 \times 10^{6}} \mathrm{~m} / \mathrm{sec} \\
=& 7.76 \mathrm{~km} / \mathrm{sec}
\end{aligned}
$