A satellite close to the earth is in orbit above the equator with a period of rotation of `1.5` hours. If it is above a point `P` on the equtor at some time, it will be above `P` again after time ...........

Explanation

Let `omega_(0)=` the angular velocity of earth above its axis `=(2pi//24)rad//hr`
Let `omega` be the angular velocity of the satellite.
`rArr omega=2pi//1.5`
For a setallite rotating from west to east (same as earth), the relative angular velocity `omega_(1)=omega-omega_(0)`
`:. `Time period of rotation relative to eath `=2pi//omega_(1)=1.6h`
Now, for a satellite rotating from east to west (opposite to earth) the relative angular velocity `omega_(2)=omega+omega_(0)`.
Time period of rotation relative to earth `=2pi//omega_(2)=(24//17)hr`