Question

A satellite close to the earth is in orbit above the equator with a period of rotation of $1.5h$. If it is above a point $P$ on the equator at some time, it will be above $P$ again after time
(1) $1.5h$
(2) $1.6h$ 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
$T_1=\frac{2\pi }{{\omega }_1}=\frac{2\pi \times 24}{2\pi \times 15}$
$=1.6h$
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
$T_2=\frac{2\pi }{{\omega }_2}=\frac{2\pi \times 24}{2\pi \times 17}$
$=\frac{24}{17}h$