If different planets have the same density but different radii then the acceleration due to gravity (g) on the surface of the planet will depend on its radius (R) as

Options

A. `g prop 1/R^2`
B. `g prop 1/R`
C. `g prop R`
D. `g prop R^2`

Answer

To solve the problem of how the acceleration due to gravity (g) on the surface of a planet depends on its radius (R) when different planets have the same density, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Gravitational Acceleration**:
The acceleration due to gravity (g) on 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 in Terms of Density**:
The mass \( M \) of a planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho \cdot V \]
For a spherical planet, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, we can write:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

3. **Substitute Mass into the Gravitational Formula**:
Substituting the expression for mass \( M \) into the formula for \( g \):
\[ g = \frac{G \cdot \left(\rho \cdot \frac{4}{3} \pi R^3\right)}{R^2} \]

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

5. **Conclusion**:
From the final equation, we see that the acceleration due to gravity \( g \) is directly proportional to the radius \( R \) of the planet when the density \( \rho \) is constant:
\[ g \propto R \]

### Final Answer:
The acceleration due to gravity (g) on the surface of the planet will depend on its radius (R) as:
\[ g \propto R \]