A) \[10\]
B) \[1\]
C) \[\frac{1}{5}\]
D) \[\frac{1}{10}\]
Correct Answer: C
Solution :
Time period of satellite \[T=2\pi \sqrt{\frac{{{({{R}_{e}}+h)}^{2}}}{gR_{e}^{2}}}\] where, \[{{R}_{e}}=\]Radius of earth \[h=\]Height from earth surface Time period does not depend on mass. So, time period of both satellite will be equal. \[\because \] \[{{T}_{A}}={{T}_{B}}\] \[\therefore \] \[{{T}_{A}}={{T}_{B}}=2\pi \sqrt{\frac{{{({{R}_{e}}+h)}^{3}}}{gR_{e}^{2}}}\] Therefore, ratio of time periods \[\frac{{{T}_{A}}}{{{T}_{B}}}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
