Question

A satellite is orbiting close to the earth above the equator with a period of rotation of 1.5 hours. If it is above a point x on the equator at some time, then after how much time will it be above x again?

Options

• 1.5 hours if it is rotating from east to west.

• 1.6 hours if it is rotating from west to east.

• 1.6 hours if it is rotating east to west.

• 1.8 hours if it is rotating west to east.

Answer 1

1.6 hours if it is rotating from west to east.

Explanation:

Stellite's angle of rotation,

ω_s = `(2pi)/T_s = (2pi)/1.5` rad/hr

ω = `(2pi)/T = (2pi)/4` rad/hr

Considering the relative angular velocity of a satellite rotating from west to east ω_r = ω_s – ω

= `(2pi)/1.5 = (2pi)/24`

= `2pi xx 5/8 = (5pi)/4`

The satellite will revolve around the earth at a relative speed of

T_rel = `(2pi)/(ω_r) = (2pi)/((5pi)/4)` = 1.6 hr

If a satellite rotates from east to west, the relative angular velocity is ω_r = ω_s + ω

= `(2pi)/1.5 + (2pi)/24 = 2pi xx 17/24`

The relative time period of the satellite's revolution with regard to the earth will be

T_rel = `(2pi)/(ω_r) = (2pi)/(2pi xx 17/24) = 24/17` = 1.41 hr