The radius of a planet is `R`. A satellite revolves around it in a circle of radius `r` with angular velocity `omega_(0)`. The acceleration due to the gravity on planet's surface is

Options

A. `(r^(3)omega_(0))/R`
B. `(r^(3)omega_(0)^(2))/(R^(2))`
C. `(r^(3)omega_(0)^(2))/R`
D. `(r^(3)omega_(0)^(2))/(R^(2))`

Explanation

To find the acceleration due to gravity on the surface of a planet with radius \( R \), given that a satellite revolves around it in a circular orbit of radius \( r \) with angular velocity \( \omega_0 \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Satellite
The satellite is in circular motion around the planet, which means it experiences a gravitational force pulling it towards the planet and a centripetal force required to keep it in circular motion.

### Step 2: Write the Gravitational Force Equation
The gravitational force \( F \) between the planet and the satellite is given by Newton's law of gravitation:
\[ F = \frac{G M m}{r^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the planet,
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the planet to the satellite.

### Step 3: Write the Centripetal Force Equation
The centripetal force \( F_c \) required to keep the satellite in circular motion is given by:
\[ F_c = \frac{m v^2}{r} \]
where \( v \) is the linear velocity of the satellite.

### Step 4: Relate Linear Velocity to Angular Velocity
For circular motion, the linear velocity \( v \) can be expressed in terms of angular velocity \( \omega_0 \):
\[ v = \omega_0 r \]
Substituting this into the centripetal force equation gives:
\[ F_c = \frac{m (\omega_0 r)^2}{r} = m \omega_0^2 r \]

### Step 5: Set Gravitational Force Equal to Centripetal Force
Since the gravitational force provides the necessary centripetal force for the satellite, we can set the two forces equal to each other:
\[ \frac{G M m}{r^2} = m \omega_0^2 r \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M}{r^2} = \omega_0^2 r \]

### Step 6: Solve for \( G M \)
Rearranging the equation gives:
\[ G M = \omega_0^2 r^3 \]

### Step 7: Relate \( G \) to Acceleration Due to Gravity
The acceleration due to gravity \( g \) at the surface of the planet is given by:
\[ g = \frac{G M}{R^2} \]
Substituting \( G M \) from the previous step into this equation gives:
\[ g = \frac{\omega_0^2 r^3}{R^2} \]

### Final Result
Thus, the acceleration due to gravity on the surface of the planet is:
\[ g = \frac{\omega_0^2 r^3}{R^2} \]