Question

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 equator at some time, it will be above P again after time

• A. $1.5$ hours.
• B. $1.6$ hours if it is rotating from west to east.
• C. $24/17$ hours if it is rotating from west to east.
• D. $24/17$ hours if it is rotating from east to west.

Answer 1

Answer: (B), (D)

$1.6$ hours if it is rotating from west to east.
$24/17$ hours if it is rotating from east to west.

The angular velocity of earth about its axis is ${\omega }_0=\frac{2\pi }{24}$
The angular velocity of satellite is $\omega =\frac{2\pi }{1.5}$.
For a satellite rotating from west to east (the same as the earth), the relative angular velocity, ${\omega }_1=\omega -{\omega }_0=2\pi (\frac{1}{1.5}-\frac{1}{24})=2\pi \left(0.625\right)$
Time period of rotation relative to the earth $=\frac{2\pi }{{\omega }_1}=\frac{1}{0.625}=1.6hour$
For a satellite rotating from east to west (opposite to the earth), the relative angular velocity, ${\omega }_2=\omega +{\omega }_0=2\pi (\frac{1}{1.5}+\frac{1}{24})=2\pi \left(\frac{17}{24}\right)$
Now time period of rotation relative to the earth $=\frac{2\pi }{{\omega }_2}=\frac{1}{17/24}=24/17hour$