A) 4.8 h
B) 48\[\sqrt{2}\] h
C) 48 h
D) 24 h
Correct Answer: B
Solution :
According to Kepler's law \[{{T}^{2}}\propto {{R}^{3}}\] or \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{R_{1}^{3}}{R_{2}^{3}}\] For a geostationary satellite \[{{T}_{1}}=24h\] \[{{T}_{1}}=?\] \[{{R}_{1}}=R,\] \[{{R}_{2}}=2R\] \[\therefore \] \[\frac{{{(24)}^{2}}}{T_{2}^{2}}=\frac{{{R}^{3}}}{{{(2R)}^{3}}}\] or \[T_{2}^{2}={{(24)}^{2}}\times 8\] or \[{{T}_{2}}=48\sqrt{2}h\]
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