The mean radius of earth is R, its angular speed on its own axis is `omega` and the acceleration due to gravity at earth's surface is g. What will be the radius of the orbit of a geostationary satellite
(a) `((R^(2)g)/(omega^(2)))^(1//3)` (b) `((Rg)/(omega^(2)))^(1//3)` (c) `((R^(2)omega^(2))/(g))^(1//3)` (d) `((R^(2)g)/(omega))^(1//3)`

Options

A. `((R^(2)g)/(omega^(2)))^(1//3)`
B. `((Rg)/(omega^(2)))^(1//3)`
C. `((R^(2)omega^(2))/(g))^(1//3)`
D. `((R^(2)g)/(omega))^(1//3)`

Answer

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

### Step 1: Understand the relationship between gravitational force and centripetal force
A geostationary satellite must have a centripetal force equal to the gravitational force acting on it. The gravitational force \( F_g \) acting on the satellite is given by:

\[ F_g = \frac{G M m}{r^2} \]

where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, 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 orbit is given by:

\[ F_c = \frac{m v^2}{r} \]

where \( v \) is the orbital velocity of the satellite.

### Step 2: Set the gravitational force equal to the centripetal force
For a geostationary satellite, we set \( F_g = F_c \):

\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \]

### Step 3: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):

\[ \frac{G M}{r^2} = \frac{v^2}{r} \]

Multiplying both sides by \( r^2 \):

\[ G M = v^2 r \]

### Step 4: Express the orbital velocity in terms of 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 gives:

\[ G M = (\omega r)^2 r \]

### Step 5: Rearrange the equation
This simplifies to:

\[ G M = \omega^2 r^3 \]

### Step 6: Solve for \( r \)
Rearranging gives:

\[ r^3 = \frac{G M}{\omega^2} \]

Taking the cube root of both sides, we find:

\[ r = \left(\frac{G M}{\omega^2}\right)^{1/3} \]

### Step 7: Relate \( G M \) to \( g \) and \( R \)
We know that at the surface of the Earth, the acceleration due to gravity \( g \) 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 \( r \)
Substituting this into our equation for \( r \):

\[ r = \left(\frac{g R^2}{\omega^2}\right)^{1/3} \]

### Final Answer
Thus, the radius of the orbit of a geostationary satellite is:

\[ r = \left(\frac{g R^2}{\omega^2}\right)^{1/3} \]

This corresponds to option (b):

\[ \boxed{\left(\frac{Rg}{\omega^2}\right)^{1/3}} \]