A satellite `P` is revolving around the earth at a height `h`=radius of earth `(R)` above equator. Another satellite `Q` is at a height `2h` revolving in oppisite direction. At an instant the two are at same vertical line passing through centres of sphere. Find the least time of after which again they are in this situation.

Answer

`(2piR^(3//2)(6sqrt(6)))/(sqrt(GM)(2sqrt(2)+3sqrt(3)))`

Explanation

`omega = sqrt((Gm)/(r^(3))) rArr omega_(1) = sqrt((Gm)/((R+R)^(3))) rArr omega_(1) = sqrt((Gm)/((2R)^(3)))`
`omega_(2) = sqrt((Gm)/((R+2R)^(3))) rArr omega_(2) = sqrt((Gm)/((3R)^(3))) rArr omega_("rel")=omega_(1)+omega_(2)`
`omega_(2) = (sqrt(Gm))/(R^(3//2))[(1)/(2sqrt(2))+(1)/(3sqrt(3))] rArr omega_("rel") = (sqrt(Gm))/(R^(3//2))xx((3sqrt(3)+2sqrt(2)))/(6sqrt(6))`
`T = (2pi)/(omega_("rel")) rArr T = (2piR^(3//2)xx6sqrt(6))/(sqrt(Gm)(3sqrt(3)+2sqrt(2)))`