A) 4 days
B) 16 days
C) \[2\sqrt{2}\] days
D) 64 days
Correct Answer: C
Solution :
Key Idea: Parking orbit is a geostationary satellites orbit. From Keplers third law of planetary motion, the square of period of revolution (T) is directly proportional to cube of semi-major axis of its elliptical orbit (a), i.e., \[{{T}^{2}}\propto {{a}^{3}}\] Given, \[{{T}_{1}}=1\,day\] (geostationary) \[{{a}_{1}}=a,{{a}_{2}}=2a\] \[\therefore \] \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{a_{1}^{3}}{a_{2}^{3}}\] \[\Rightarrow \] \[T_{2}^{2}=\frac{a_{2}^{3}}{a_{1}^{3}}T_{1}^{2}=\frac{{{(2a)}^{3}}}{{{a}^{3}}}\times 1=8\] \[\Rightarrow \] \[{{T}_{2}}=2\sqrt{2}\,days\]
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