Question

In the case of earth, mean radius is $R$, acceleration due to gravity on the surface is $g$, angular speed about its own axis is $\omega$. What will be the radius of the orbit of a geostationary satellite?

• A. $\left[\dfrac{\omega^2}{gR^2}\right]^{1/2}$
• B. $\left[\dfrac{gR^2}{\omega^2}\right]^{1/3}$
• C. $\left[\dfrac{gR^2}{\omega^2}\right]^{1/2}$
• D. $\left[\dfrac{\omega^2}{gR^2}\right]^{1/3}$

Hint:

Always replace $GM$ using $gR^2$ when Earth-related quantities are given.

Answer 1

The Correct Option is B

Step 1: Write condition for geostationary orbit.
For a geostationary satellite, gravitational force provides the required centripetal force: \[ \dfrac{GMm}{r^2} = m\omega^2 r \] Step 2: Express $GM$ in terms of $g$ and $R$.
We know that: \[ g = \dfrac{GM}{R^2} \Rightarrow GM = gR^2 \] Step 3: Substitute value of $GM$.
\[ \dfrac{gR^2}{r^2} = \omega^2 r \] Step 4: Solve for orbital radius $r$.
\[ r^3 = \dfrac{gR^2}{\omega^2} \] \[ r = \left(\dfrac{gR^2}{\omega^2}\right)^{1/3} \]