A) 90 min
B) \[90\times \sqrt{8}\text{ }min\]
C) 270 min
D) 720 min
Correct Answer: D
Solution :
The period of revolution of a satellite depends only upon its height above earths surface, for a satellite very close to the surface of the earth, period of revolution is \[{{T}_{1}}=2\pi \sqrt{\frac{R}{g}}\] ? (i) For a height h above the earths surface, \[{{T}_{2}}=\frac{2\pi }{R}\sqrt{\frac{{{(R+h)}^{3}}}{g}}\] ?.. (ii) Given, \[h=3R\] \[\therefore \]\[{{T}_{2}}=\frac{2\pi }{R}=\sqrt{\frac{{{(R+3R)}^{3}}}{g}}=2\pi \times 8\sqrt{\frac{R}{g}}\] ?. (iii) From Eqs. (i) and (iii), we have \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{1}{8}\] \[\Rightarrow \] \[{{T}_{2}}=90\times 8=720\text{ }\min \]
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