Question

A satellite close to the earth is in orbit above the equator with a period of rotation of $1.5\, h$. If it is above a point $P$ on the equator at some time, it will be above $P$ again after time
(1) $1.5\, h$
(2) $1.6\, h$ if it is rotating from west to east
(3) $\frac{24}{17} \,h$ if it is rotating from west to east
(4) $\frac{24}{17} \,h$ if it is rotating from east to west

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (C)

Let $\omega_{0}$ be the angular velocity of earth about its axis
$\omega_{0}=\frac{2 \pi}{T}=\frac{2 \pi}{24}$
Let $\omega$ be the angular velocity of the satellite
$\therefore \omega_{0}=\frac{2 \pi}{1.5}$
For a satellite rotating from west to east (the same as the earth), the relative angular velocity is
$\omega_{1}=\omega-\omega_{0} $
$=\frac{2 \pi}{1.5}-\frac{2 \pi}{24} $
$=2 \pi \times \frac{15}{24}$
The time period of rotation of satellite relative to earth is
$T_{1}=\frac{2 \pi}{\omega_{1}}=\frac{2 \pi \times 24}{2 \pi \times 15} $
$=1.6\, h$
For a satellite rotating from east to west (opposite to the other), the relative velocity of satellite is
$\omega_{2}=\omega+\omega_{0} $
$=\frac{2 \pi}{1.5}+\frac{2 \pi}{24} $
$=2 \pi \times \frac{17}{24}$
The time period of rotation of satellite relative to the earth is
$T_{2}=\frac{2 \pi}{\omega_{2}}=\frac{2 \pi \times 24}{2 \pi \times 17} $
$=\frac{24}{17} h$