Question:
A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet?
[Given ; Mass of planet = \(8\times10^{22}\) kg, Radius of planet = \(2\times10^{6}\) m, Gravitational constant \(G=6.67\times10^{-11}\ \mathrm{Nm^2/kg^2}\)]
A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet?
[Given ; Mass of planet = \(8\times10^{22}\) kg, Radius of planet = \(2\times10^{6}\) m, Gravitational constant \(G=6.67\times10^{-11}\ \mathrm{Nm^2/kg^2}\)]
[Given ; Mass of planet = \(8\times10^{22}\) kg, Radius of planet = \(2\times10^{6}\) m, Gravitational constant \(G=6.67\times10^{-11}\ \mathrm{Nm^2/kg^2}\)]
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Orbital radius = R + h; use \(T=2\pi\sqrt{r^3/GM}\) and divide 24 h by T.
Updated On: Nov 5, 2026
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- 11
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The Correct Option is D
Solution and Explanation
Orbital radius \(r=2\times10^6+0.02\times10^6=2.02\times10^6\) m.
\(GM=6.67\times10^{-11}\times8\times10^{22}=5.336\times10^{12}\).
\(r^3=(2.02)^3\times10^{18}=8.242\times10^{18}\).
\(T=2\pi\sqrt{\frac{r^3}{GM}}=2\pi\sqrt{1.5446\times10^{6}}=2\pi\times1242.8\approx7.81\times10^3\) s.
Number of revolutions \(=\frac{24\times3600}{7809}=\frac{86400}{7809}\approx11.06\).
Complete revolutions = 11.
Answer: (4) 11
\(GM=6.67\times10^{-11}\times8\times10^{22}=5.336\times10^{12}\).
\(r^3=(2.02)^3\times10^{18}=8.242\times10^{18}\).
\(T=2\pi\sqrt{\frac{r^3}{GM}}=2\pi\sqrt{1.5446\times10^{6}}=2\pi\times1242.8\approx7.81\times10^3\) s.
Number of revolutions \(=\frac{24\times3600}{7809}=\frac{86400}{7809}\approx11.06\).
Complete revolutions = 11.
Answer: (4) 11
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