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 on its surface. We'll denote the following:

- \( g \): acceleration due to gravity
- \( \rho_e \): density of Earth
- \( \rho_p \): density of the planet
- \( R_e \): radius of Earth
- \( R_p \): radius of the planet
- \( G \): universal gravitational constant

### Step-by-Step Solution:

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

2. **Expressing 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:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass becomes:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

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

4. **Setting Up the Equation for Earth and the Planet**:
For Earth, we have:
\[ g_e = \frac{4}{3} G \rho_e \pi R_e \]
For the planet, we have:
\[ g_p = \frac{4}{3} G \rho_p \pi R_p \]
Given that \( g_e = g_p \), we can set these two equations equal to each other:
\[ \frac{4}{3} G \rho_e \pi R_e = \frac{4}{3} G \rho_p \pi R_p \]

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

6. **Substituting the Density of the Planet**:
We know from the problem that the density of the planet \( \rho_p \) is twice that of Earth:
\[ \rho_p = 2 \rho_e \]
Substituting this into the equation gives:
\[ \rho_e R_e = (2 \rho_e) R_p \]

7. **Simplifying the Equation**:
Dividing both sides by \( \rho_e \) (assuming \( \rho_e \neq 0 \)):
\[ R_e = 2 R_p \]
Rearranging gives:
\[ R_p = \frac{R_e}{2} \]

8. **Final Answer**:
Thus, the radius of the planet in terms of Earth's radius \( R \) is:
\[ R_p = \frac{R}{2} \]

### Summary:
The radius of the planet is half the radius of Earth.