The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to

Options

A. `rho R`
B. `rho R^(-1)`
C. `rho^(-1)R^(-2)`
D. `rhoR^(2)`

Explanation

To solve the problem of determining how the acceleration due to gravity (g) near the surface of a planet is proportional to its radius (R) and density (d), we can follow these steps:

### Step-by-Step Solution:

1. **Understand 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 universal gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

2. **Express mass (M) in terms of density (d)**:
The mass of the planet can be expressed using its density and volume. The volume \( V \) of a sphere (the shape of the planet) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass \( M \) can be expressed as:
\[ M = d \cdot V = d \cdot \frac{4}{3} \pi R^3 \]

3. **Substitute the expression for mass (M) into the formula for g**:
Now, substitute the expression for \( M \) into the formula for \( g \):
\[ g = \frac{G \cdot \left(d \cdot \frac{4}{3} \pi R^3\right)}{R^2} \]

4. **Simplify the equation**:
Simplifying the equation gives:
\[ g = \frac{G \cdot 4 \pi d}{3} \cdot \frac{R^3}{R^2} \]
This simplifies to:
\[ g = \frac{G \cdot 4 \pi d}{3} \cdot R \]

5. **Identify the proportional relationship**:
From the simplified equation, we can see that:
\[ g \propto R \cdot d \]
This means that the acceleration due to gravity is directly proportional to both the radius \( R \) and the density \( d \) of the planet.

### Conclusion:
Thus, the acceleration due to gravity near the surface of a planet of radius \( R \) and density \( d \) is proportional to \( R \cdot d \).