A satellite moves estwards very near the surface of the Earth in equitorial plane with speed `(v_(0))`. Another satellite moves at the same height with the same speed in the equatorial plane but westwards. If `R` =radius of the Earth and `omega` be its angular speed of the Earth about its own axis. Then find the approximate difference in the two time period as observed on the earth.

Options

A. `(4piomegaR^(2))/(v_(0)^(2)+R^(2)omega^(2))`
B. `(2piomegaR^(2))/(v_(0)^(2)-R^(2)omega^(2))`
C. `(4piomegaR^(2))/(v_(0)^(2)-R^(2)omega^(2))`
D. `(2piomegaR^(2))/(v_(0)^(2)+R^(2)omega^(2))`

Answer

C

Explanation

`T_(west)=(2piR)/(v_(0)+Romega)` and `T_("east")=(2piR)/(v_(0)-Romega) implies DeltaT=T_("east")`
`implies T_("east")-T_(west)=2piR[(2piR)/(v_(0)^(2)-R^(2)omega^(2))]=(4piomegaR^(2))/(v_(0)^(2)-R^(2)omega^(2))`