Question

Angular velocity of geostationary satellite is (in rad hr$^{-1}$).

• A. $\frac{\pi}{365}$
• B. $\frac{\pi}{24}$
• C. $\frac{\pi}{12}$
• D. $\frac{\pi}{18}$

Hint:

A geostationary satellite completes one full circle ($2\pi$ radians) in 24 hours. Therefore, its rate is simply $2\pi / 24 = \pi / 12$ rad/hr.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A geostationary satellite is one that appears stationary relative to a fixed point on the Earth's equator. For this to happen, the satellite must orbit the Earth in the same direction and with the exact same rotational period as the Earth's rotation on its own axis.
Step 2: Key Formula or Approach:
The relationship between angular velocity (\(\omega\)) and time period (\(T\)) is:
\[ \omega = \frac{2\pi}{T} \]
We need to use the known time period of a geostationary satellite and calculate \(\omega\).
Step 3: Detailed Explanation:
Since a geostationary satellite matches Earth's rotation, its orbital time period (\(T\)) is exactly 1 day.
Time period, \(T = 24\) hours.
The question asks for angular velocity in units of radians per hour (rad hr\(^{-1}\)). Therefore, we keep the time period in hours.
Calculate angular velocity:
\[ \omega = \frac{2\pi \text{ radians}}{T \text{ hours}} \]
\[ \omega = \frac{2\pi}{24} \text{ rad/hr} \]
Simplify the fraction:
\[ \omega = \frac{\pi}{12} \text{ rad hr}^{-1} \]
Step 4: Final Answer:
The angular velocity is \(\frac{\pi}{12}\) rad hr\(^{-1}\).