The acceleration due to gravity at the surface of earth `(g_(e))` and acceleration due to gravity at the surface of a planet `(g_(p))` are equal. The density of the planet is three times than that of earth. The ratio of radius of earth to the radius of planet will be.

Options

A. `1:3`
B. `1:9`
C. `3:1`
D. `9:1`

Explanation

To solve the problem, we need to find the ratio of the radius of the Earth to the radius of the planet given that the acceleration due to gravity at the surface of both is equal and the density of the planet is three times that of Earth.

### Step-by-Step Solution:

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

2. **Expressing Mass in Terms of Density**:
The mass \( M \) of a body can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho \cdot V \]
For a sphere, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass can be rewritten as:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

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

4. **Setting Up the Equations for Earth and the Planet**:
Let \( g_e \) be the acceleration due to gravity at the surface of the Earth and \( g_p \) at the surface of the planet. We know that:
\[ g_e = \frac{4}{3} \pi G \rho_e R_e \]
\[ g_p = \frac{4}{3} \pi G \rho_p R_p \]
Given that \( g_e = g_p \), we can equate the two expressions:
\[ \frac{4}{3} \pi G \rho_e R_e = \frac{4}{3} \pi G \rho_p R_p \]

5. **Canceling Common Terms**:
We can cancel \( \frac{4}{3} \pi G \) from both sides:
\[ \rho_e R_e = \rho_p R_p \]

6. **Substituting the Density of the Planet**:
We are given that the density of the planet \( \rho_p \) is three times that of Earth:
\[ \rho_p = 3 \rho_e \]
Substituting this into the equation gives:
\[ \rho_e R_e = (3 \rho_e) R_p \]

7. **Simplifying the Equation**:
Dividing both sides by \( \rho_e \) (assuming \( \rho_e \neq 0 \)):
\[ R_e = 3 R_p \]

8. **Finding the Ratio of Radii**:
Rearranging gives us the ratio of the radius of Earth to the radius of the planet:
\[ \frac{R_e}{R_p} = 3 \]

### Final Answer:
The ratio of the radius of the Earth to the radius of the planet is:
\[ \frac{R_e}{R_p} = 3:1 \]