Question

The radius of the orbit of a geostationary satellite is (mean radius of the earth is \( R \), angular velocity about an axis is \( \omega \), and acceleration due to gravity on earth's surface is \( g \))

• A. \( \left( \frac{gR^2}{\omega^2} \right)^{1/3} \)
• B. \( \frac{gR^2}{\omega^2} \)
• C. \( \left( \frac{gR^2}{\omega^2} \right)^{2/3} \)
• D. \( \left( \frac{gR^2}{\omega^2} \right)^{1/2} \)

Hint:

For a geostationary satellite, the radius is found by equating the gravitational and centrifugal forces. Always remember to convert the given quantities (like gravitational constant) in terms of known values like \( g \) and \( R \).

Answer 1

The Correct Option is A

Step 1: Understanding the motion of a geostationary satellite.
The radius of a geostationary satellite is derived from the balance between the gravitational force and the centrifugal force due to its orbital motion. The force of gravity is given by: \[ F = \frac{GMm}{r^2} \] where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the radius of the orbit. The centrifugal force is: \[ F = m \omega^2 r \] Setting these two forces equal gives: \[ \frac{GMm}{r^2} = m \omega^2 r \] Simplifying and solving for \( r \) yields: \[ r = \left( \frac{GM}{\omega^2} \right)^{1/3} \] where \( G \) is related to \( g \) by \( g = \frac{GM}{R^2} \). Substituting this gives the final result: \[ r = \left( \frac{gR^2}{\omega^2} \right)^{1/3} \] Step 2: Conclusion.
Thus, the correct answer is (A) \( \left( \frac{gR^2}{\omega^2} \right)^{1/3} \).