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

Options

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

Answer

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

### Step 1: Understand the forces acting on the satellite
A geostationary satellite is in a circular orbit around the Earth. The two main forces acting on it are:
1. Gravitational force (Fg) acting towards the center of the Earth.
2. Centripetal force (Fc) required to keep the satellite in circular motion.

### Step 2: Write the equations for the forces
The gravitational force acting on the satellite can be expressed as:
\[ F_g = \frac{G M m}{r^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the Earth to the satellite.

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

### Step 3: Relate the orbital speed to angular speed
The orbital speed \( v \) can be related to the angular speed \( \omega \) by:
\[ v = r \omega \]
Substituting this into the centripetal force equation gives:
\[ F_c = \frac{m (r \omega)^2}{r} = m r \omega^2 \]

### Step 4: Set the gravitational force equal to the centripetal force
For a geostationary satellite, the gravitational force equals the centripetal force:
\[ \frac{G M m}{r^2} = m r \omega^2 \]

### Step 5: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M}{r^2} = r \omega^2 \]

### Step 6: Rearrange the equation to find \( r^3 \)
Multiplying both sides by \( r^2 \) gives:
\[ G M = r^3 \omega^2 \]
Now, rearranging this gives:
\[ r^3 = \frac{G M}{\omega^2} \]

### Step 7: Express \( G M \) in terms of \( g \) and \( R \)
We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{G M}{R^2} \]
From this, we can express \( G M \) as:
\[ G M = g R^2 \]

### Step 8: Substitute \( G M \) back into the equation for \( r^3 \)
Substituting \( G M \) into the equation for \( r^3 \) gives:
\[ r^3 = \frac{g R^2}{\omega^2} \]

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