A) 5 h
B) 4 h
C) 3h
D) 12 h
Correct Answer: B
Solution :
From Keplers third law of planetary motion \[{{T}^{2}}\propto {{R}^{3}}\] \[\therefore \] \[\frac{T_{2}^{2}}{T_{1}^{2}}=\frac{R_{2}^{3}}{R_{1}^{3}}\] or \[\frac{T_{2}^{2}}{{{(24)}^{2}}}={{\left( \frac{6400+6400}{36000+6400} \right)}^{2}}\] or \[T_{2}^{2}={{(24)}^{2}}\times {{\left( \frac{16}{53} \right)}^{3}}\] \[\Rightarrow \] \[{{T}_{2}}=4h\]
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