A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as that on the surface of earth. Its radius in terms of earth's radius `R` will be

Options

A. `R//4`
B. `R//2`
C. `R//3`
D. `R//8`

Explanation

To solve the problem, we need to find the radius of a planet that has twice the density of Earth but the same acceleration due to gravity at its surface as Earth. Let's denote the following:

- \( \rho_e \): Density of Earth
- \( R_e \): Radius of Earth
- \( g_e \): Acceleration due to gravity on Earth
- \( \rho_p \): Density of the planet
- \( R_p \): Radius of the planet
- \( g_p \): Acceleration due to gravity on the planet

### Step-by-Step Solution:

1. **Understanding the Relationship for Gravity**:
The acceleration due to gravity \( g \) on the surface of a planet can be expressed as:
\[ g = \frac{G \cdot M}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

2. **Expressing Mass in Terms of Density**:
The mass \( M \) of the planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho \cdot V = \rho \cdot \left(\frac{4}{3} \pi R^3\right) \]
Thus, we can rewrite the expression for gravity as:
\[ g = \frac{G \cdot \rho \cdot \left(\frac{4}{3} \pi R^3\right)}{R^2} = \frac{4}{3} \pi G \cdot \rho \cdot R \]

3. **Applying the Given Conditions**:
For Earth, we have:
\[ g_e = \frac{4}{3} \pi G \cdot \rho_e \cdot R_e \]
For the planet, since it has twice the density of Earth:
\[ \rho_p = 2 \rho_e \]
Therefore, the acceleration due to gravity on the planet is:
\[ g_p = \frac{4}{3} \pi G \cdot \rho_p \cdot R_p = \frac{4}{3} \pi G \cdot (2 \rho_e) \cdot R_p = \frac{8}{3} \pi G \cdot \rho_e \cdot R_p \]

4. **Setting the Accelerations Equal**:
Given that \( g_e = g_p \):
\[ \frac{4}{3} \pi G \cdot \rho_e \cdot R_e = \frac{8}{3} \pi G \cdot \rho_e \cdot R_p \]
We can cancel \( \frac{4}{3} \pi G \cdot \rho_e \) from both sides (assuming \( \rho_e \neq 0 \)):
\[ R_e = 2 R_p \]

5. **Solving for \( R_p \)**:
Rearranging the equation gives:
\[ R_p = \frac{R_e}{2} \]

### Final Answer:
The radius of the planet in terms of Earth's radius \( R \) is:
\[ R_p = \frac{R}{2} \]