A satellite is orbiting the earth of mass `M` in equatorial plane in a circular orbit having radius `2R` and same sense of rotation as that of the Earth. Find duration of time for which a man standing on the equator will be able tosee the satellite continuously. Assume that the mans can see the satellite when it is above horizon. take Earth's angular velocity `omega`.

Answer

`(2pi)/(3[sqrt((GM)/(8R^(3)))-omega_(e )])`

Explanation

`m(2R)omega_(s)^(2) = (GM_(s)M)/((2R)^(2))`

`omega_(s) = sqrt((GM)/(8R^(3)))`
`omega_(rel)=omega_(s)-omega_("earth") rArr omega_(rel)=sqrt((GM)/(8R^(3)))-omega`
`t = (theta)/(omega_(rel)) rArr T = ((2pi//3))/(sqrt((GM)/(8R^(3))-omega))rArr t = (2pi)/(3(sqrt((GM)/(8R^(3))-omega)))`