A satellite is orbiting around the earth in the equatorial plane rotating from west to east as the earth does. If ωₑ be the angular speed of the earth and ωₛ be that of satellite, then the satellite will repeatedly appear at the some location after a time t =

The correct answer is 2π/(ωₛ-ωₑ).

Let us understand the relative motion of the satellite with respect to the Earth.
Both the Earth and the satellite are rotating in the same direction, which is from west to east in the equatorial plane.
The angular speed of the Earth is given as _ωₑ.
The angular speed of the satellite is given as _ωₛ.

Since both are rotating in the same direction, the relative angular velocity of the satellite with respect to an observer on the Earth is:
${\omega }_{\mathrm{rel}}={\omega }_s-{\omega }_e$
Here, we assume the satellite moves faster than the Earth's rotation (_ωₛ > _ωₑ) to make consecutive appearances.

The time interval t after which the satellite will repeatedly appear at the same location above the Earth corresponds to the time it takes for the satellite to complete one full relative rotation (an angle of 2π radians) with respect to the Earth.
Using the relation between time, angular displacement, and relative angular velocity:
$t=\frac{2\pi }{{\omega }_{\mathrm{rel}}}$
Substituting the value of ${\omega }_{\mathrm{rel}}$:
$t=\frac{2\pi }{{\omega }_s-{\omega }_e}$
Thus, the satellite will repeatedly appear at the same location after a time interval of 2π/(ωₛ-ωₑ).