A) \[\frac{1}{2}\]lunar month
B) \[\frac{2}{3}\] lunar month
C) \[{{2}^{3/2}}\] lunar month
D) \[{{2}^{3/2}}\]lunar month
Correct Answer: C
Solution :
The period of revolution of a satellite at a height h from the surface of earth is given by \[T=2\pi \sqrt{\frac{{{({{R}_{e}}+h)}^{3}}}{gR_{e}^{2}}}\] Given, \[{{T}_{m}}=1\]lunar month, \[\therefore \] \[{{T}_{sat}}=2\pi \sqrt{\frac{{{\left( R+\frac{h}{2} \right)}^{2}}}{g{{R}^{2}}}}\] \[\Rightarrow \] \[{{T}_{sat}}\approx \frac{1}{{{2}^{3/2}}}\] \[{{T}_{moon}}={{2}^{-3/2}}lunar\text{ }month\]
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