Question

A satellite is seen after every $6$ hours over the equator. It is known that it rotates opposite to that of earth’s direction. Then the angular velocity (in radians per hour) of the satellite about the centre of earth will be:

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{8}$

Answer 1

Answer: (C)

Let ${\omega }_E=$ angular velocity of earth, ${\omega }_S=$ angular velocity of satellite.

Then $6=\frac{2\pi }{({\omega }_E+{\omega }_S)}$

$\Rightarrow {\omega }_E+{\omega }_S=\frac{\pi }{3}$
$\Rightarrow {\omega }_S=\frac{\pi }{3}-\frac{\pi }{12}$

$(\because {\omega }_E=\frac{2\pi }{24}=\frac{\pi }{12}$ rad per hour)

${\omega }_S=\frac{\pi }{4}$ radian per hour