The mean radius of the earth is `R`, its angular speed on its own axis is `omega` and the acceleration due to gravity at earth's surface is `g`. The cube of the radius of the orbit of a geo-stationary satellite will be

Options

A. `R^(2)g//omega`
B. `R^(2)omega^(2)//g`
C. `Rg//omega^(2)`
D. `R^(2)g//omega^(2)`

Explanation

To find the cube of the radius of the orbit of a geo-stationary satellite, we can follow these steps:

### Step 1: Understand the relationship between gravitational force and centripetal force
For a satellite in orbit, the gravitational force acting on it provides the necessary centripetal force for circular motion. The gravitational force \( F_g \) is given by:
\[ F_g = \frac{GMm}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( r \) is the distance from the center of the Earth to the satellite.

The centripetal force \( F_c \) required to keep the satellite in circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( v \) is the orbital velocity of the satellite.

### Step 2: Set the gravitational force equal to the centripetal force
Setting \( F_g = F_c \):
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{r^2} = \frac{v^2}{r} \]
Multiplying both sides by \( r \):
\[ \frac{GM}{r} = v^2 \]

### Step 3: Relate orbital velocity to the angular speed
The orbital velocity \( v \) can also be expressed in terms of the angular speed \( \omega \):
\[ v = \omega r \]
Substituting this into the equation from Step 2 gives:
\[ \frac{GM}{r} = (\omega r)^2 \]
This simplifies to:
\[ \frac{GM}{r} = \omega^2 r^2 \]
Rearranging gives:
\[ GM = \omega^2 r^3 \]

### Step 4: Relate \( GM \) to the acceleration due to gravity \( g \)
The acceleration due to gravity at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
where \( R \) is the radius of the Earth. Rearranging this gives:
\[ GM = g R^2 \]

### Step 5: Substitute \( GM \) into the equation for \( r^3 \)
Substituting \( GM = g R^2 \) into the equation \( GM = \omega^2 r^3 \):
\[ g R^2 = \omega^2 r^3 \]
Rearranging gives:
\[ r^3 = \frac{g R^2}{\omega^2} \]

### Final Result
Thus, the cube of the radius of the orbit of a geo-stationary satellite is:
\[ r^3 = \frac{g R^2}{\omega^2} \]