A) 1
B) 2
C) \[\sqrt{\frac{1}{2}}\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
The velocity of the satellite is\[\sqrt{\frac{GM}{r}}.\] \[\therefore \] \[\frac{{{T}_{A}}}{{{T}_{B}}}=\frac{\frac{1}{2}{{m}_{A}}v_{A}^{2}}{\frac{1}{2}{{m}_{B}}v_{B}^{2}}=\frac{1}{2}\times \frac{2R}{R}=1\]You need to login to perform this action.
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