A satellite is revolving in the circular equatorial orbit of radius `R=2xx10^(4)km` from east to west. Calculate the interval after which it will appear at the same equatorial town. Given that the radius of the earth `=6400km` and `g` (acceleration due to gravity) `=10ms^(-2)`

Explanation

Let `omega` be the actual angular velocity of the satellite from east to west and `omega_(e)` be the angular speed of the earth (west to east)
Then `omega_("relative") =omega-(-omega_(e))=omega+omega_(e)`
By the dynamics of circlar motion
`(GMm)/(R^(2))=momega^(2)R`
or `omega^(2)=(gR_(e)^(2))/(R^(3))` (`:'GM=gR_(e)^(2)`)
`implies omega=sqrt((gR_(e)^(2))/R^(3)) :.omega_(rel)=sqrt((gR_(e)^(2))/R^3)+omega_(e)`
`implies omega_(rel)=sqrt(10xx6.4^(2)xx10^(12))/(2^(3)xx10^(21))+7.27xx10^(-5)`
(`:'omega_(e)=(2pi)/86400=7.27xx10^(-5))`
`implies omega_(rel)=22.6xx10^(-5)+7.27xx110^(-5)=30xx10^(-5)rads^(-1)`
`:. tau=(2pi)/omega_(rel)=(2pi)/(30xx10^(-5))=2.09xx10^(s)=5h 48` min