Question

The angular speed of a geostationary satellite (in rad h\(^{-1}\)) is

• A. \(\frac{\pi}{15}\,\text{rad h}^{-1}\)
• B. \(\frac{\pi}{13}\,\text{rad h}^{-1}\)
• C. \(\frac{\pi}{12}\,\text{rad h}^{-1}\)
• D. \(\frac{\pi}{9}\,\text{rad h}^{-1}\)
• E. \(\frac{\pi}{8}\,\text{rad h}^{-1}\)

Hint:

For any periodic motion: \[ \omega=\frac{2\pi}{T} \] A geostationary satellite always has period 24 hours.

Answer 1

The Correct Option is C

A geostationary satellite completes one revolution in 24 hours. So its angular speed is: \[ \omega=\frac{2\pi}{T} \] Here, \[ T=24\text{ h} \] Thus, \[ \omega=\frac{2\pi}{24}=\frac{\pi}{12}\text{ rad h}^{-1} \]
Hence, the correct answer is the option corresponding to: \[ \boxed{\frac{\pi}{12}\text{ rad h}^{-1}} \]