Question

If mean radius of earth is 'R', its angular velocity is '\(\omega\)' and acceleration due to gravity at the surface of the earth is 'g' then the cube of the radius of the orbit of geostationary satellite will be

• A. \(\frac{R^{2}\omega ^{2}}{g}\)
• B. \(\frac{R^{2}g^{2}}{\omega ^{2}}\)
• C. \(\frac{R^{2}g}{\omega }\)
• D. \(\frac{R^{2}g}{\omega ^{2}}\)

Hint:

Equate gravitational force to centripetal force and replace GM using the surface value of g.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
A geostationary satellite has the same angular velocity \(\omega\) as the Earth. Gravity supplies the centripetal force.

Step 2: Equate the Forces
\(\frac{GMm}{r^{2}}=m\omega^{2}r\), so \(r^{3}=\frac{GM}{\omega^{2}}\).

Step 3: Use the Surface Gravity
At the surface \(g=\frac{GM}{R^{2}}\), so \(GM=gR^{2}\).

Step 4: Result
\(r^{3}=\frac{gR^{2}}{\omega^{2}}\), which is option (D).

Final Answer:
The cube of the orbit radius is g R squared divided by omega squared. \[ \boxed{\text{(D) }\dfrac{R^{2}g}{\omega^{2}}} \]