Suppose that the acceleration of a free fall at the surface of a distant planet was found to be equal to that at the surface of the earth. If the diameter of the planet were twice the diameter of the earth, then the ratio of mean density of the planet to that of the earth would be

Options

A. `4:1`
B. `2:1`
C. `1:1`
D. `1:2`

Explanation

To solve the problem, we need to find the ratio of the mean density of a distant planet to that of the Earth, given that the acceleration due to gravity on both bodies is the same and that the diameter of the planet is twice that of Earth.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the acceleration due to gravity \( g \) 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. **Setting Up the Equations**:
For Earth:
\[ g_e = \frac{G \cdot M_e}{R_e^2} \]
For the distant planet:
\[ g_p = \frac{G \cdot M_p}{R_p^2} \]
Since \( g_e = g_p \), we have:
\[ \frac{G \cdot M_e}{R_e^2} = \frac{G \cdot M_p}{R_p^2} \]
We can cancel \( G \) from both sides:
\[ \frac{M_e}{R_e^2} = \frac{M_p}{R_p^2} \]

3. **Expressing Mass in Terms of Density**:
The mass of a planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho \cdot V \]
The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, we can write:
\[ M_e = \rho_e \cdot \frac{4}{3} \pi R_e^3 \]
\[ M_p = \rho_p \cdot \frac{4}{3} \pi R_p^3 \]

4. **Substituting Mass into the Equation**:
Substituting these expressions into the equation we derived:
\[ \frac{\rho_e \cdot \frac{4}{3} \pi R_e^3}{R_e^2} = \frac{\rho_p \cdot \frac{4}{3} \pi R_p^3}{R_p^2} \]
This simplifies to:
\[ \rho_e \cdot R_e = \rho_p \cdot R_p \]

5. **Using the Given Information**:
We are given that the diameter of the planet is twice that of Earth. Therefore, the radius of the planet \( R_p \) is:
\[ R_p = 2R_e \]

6. **Substituting Radius into the Density Equation**:
Substituting \( R_p = 2R_e \) into the density equation:
\[ \rho_e \cdot R_e = \rho_p \cdot (2R_e) \]
Dividing both sides by \( R_e \) (assuming \( R_e \neq 0 \)):
\[ \rho_e = 2\rho_p \]

7. **Finding the Ratio of Densities**:
Rearranging gives us:
\[ \frac{\rho_p}{\rho_e} = \frac{1}{2} \]
Therefore, the ratio of the mean density of the planet to that of Earth is:
\[ \frac{\rho_p}{\rho_e} = 1:2 \]

### Final Answer:
The ratio of the mean density of the planet to that of the Earth is \( 1:2 \).