The radius of a planet is n times the radius of earth (R). A satellite revolves around it in a circle of radius 4nR with angular velocity `omega`. The acceleration due to gravity on planet's surface is

Options

A. `R omega^(2)`
B. 16 `R omega^(2)`
C. `32 n R omega^(2)`
D. `64 n R omega^(2)`

Explanation

To find the acceleration due to gravity on a planet whose radius is \( n \) times the radius of Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the formula for acceleration due to gravity (g)**:
The acceleration due to gravity at the surface of a planet is given by the formula:
\[ g = \frac{G \cdot M}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

2. **Define the radius of the planet**:
Given that the radius of the planet is \( n \) times the radius of Earth (\( R \)), we can express the radius of the planet as:
\[ R_p = nR \]

3. **Determine the mass of the planet**:
The mass of the planet can be expressed in terms of its density (\( \rho \)) and volume. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass of the planet is:
\[ M_p = \rho \cdot V = \rho \cdot \frac{4}{3} \pi (nR)^3 = \rho \cdot \frac{4}{3} \pi n^3 R^3 \]

4. **Substitute the mass into the gravity formula**:
Now substituting \( M_p \) into the formula for \( g \):
\[ g = \frac{G \cdot \left(\rho \cdot \frac{4}{3} \pi n^3 R^3\right)}{(nR)^2} \]

5. **Simplify the expression**:
Simplifying the expression:
\[ g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi n^3 R^3}{n^2 R^2} \]
\[ g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi n^3 R}{n^2} \]
\[ g = \frac{4}{3} \cdot \frac{G \cdot \rho \cdot n}{3} \]

6. **Relate to the satellite's motion**:
We know that the satellite is revolving around the planet at a radius of \( 4nR \) with angular velocity \( \omega \). The centripetal acceleration \( a \) for the satellite is given by:
\[ a = \omega^2 r \]
where \( r = 4nR \). Thus,
\[ a = \omega^2 (4nR) = 4nR \omega^2 \]

7. **Set gravitational force equal to centripetal force**:
The gravitational force acting on the satellite can also be expressed as:
\[ F = \frac{G \cdot M_p \cdot m_s}{(4nR)^2} \]
Setting the gravitational force equal to the centripetal force gives:
\[ \frac{G \cdot M_p \cdot m_s}{(4nR)^2} = m_s \cdot 4nR \omega^2 \]
Canceling \( m_s \) and rearranging gives:
\[ G \cdot M_p = 16n^2R^2 \omega^2 \]

8. **Substituting for \( M_p \)**:
Substitute \( M_p = \rho \cdot \frac{4}{3} \pi n^3 R^3 \) into the equation:
\[ G \cdot \left(\rho \cdot \frac{4}{3} \pi n^3 R^3\right) = 16n^2R^2 \omega^2 \]

9. **Final expression for \( g \)**:
Now substituting back into the formula for \( g \):
\[ g = \frac{G \cdot M_p}{R_p^2} = \frac{G \cdot \left(\rho \cdot \frac{4}{3} \pi n^3 R^3\right)}{(nR)^2} \]
Simplifying gives:
\[ g = \frac{64 \omega^2 nR}{n^2} = 64 \omega^2 R \]

### Final Answer:
The acceleration due to gravity on the planet is:
\[ g = 64 \omega^2 nR \]