The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is `R`, the radius of the planet would be

Options

A. `2R`
B. `4R`
C. `(1)/(4)R`
D. `(1)/(2)R`

Answer

To solve the problem, we need to find the radius of a newly discovered planet given that its density is twice that of Earth and that the acceleration due to gravity at the surface of the planet is equal to that at the surface of the Earth.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Density of the planet, \( \rho_P = 2 \rho_E \) (where \( \rho_E \) is the density of Earth).
- Acceleration due to gravity at the surface of the planet, \( g_P = g_E \) (where \( g_E \) is the acceleration due to gravity on Earth).
- Radius of Earth, \( R_E = R \).

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

3. **Write the Equations for Both Planets**:
For Earth:
\[ g_E = \frac{G M_E}{R_E^2} \]
For the new planet:
\[ g_P = \frac{G M_P}{R_P^2} \]

4. **Set the Two Equations Equal**:
Since \( g_P = g_E \):
\[ \frac{G M_P}{R_P^2} = \frac{G M_E}{R_E^2} \]
We can cancel \( G \) from both sides:
\[ \frac{M_P}{R_P^2} = \frac{M_E}{R_E^2} \]

5. **Express Mass in Terms of Density**:
The mass of a planet can be expressed in terms of its density and volume:
\[ M = \rho \cdot V = \rho \cdot \left(\frac{4}{3} \pi R^3\right) \]
Therefore, for the new planet:
\[ M_P = \rho_P \cdot \left(\frac{4}{3} \pi R_P^3\right) \]
And for Earth:
\[ M_E = \rho_E \cdot \left(\frac{4}{3} \pi R_E^3\right) \]

6. **Substitute Mass into the Equation**:
Substitute \( M_P \) and \( M_E \) into the equation:
\[ \frac{\rho_P \cdot \left(\frac{4}{3} \pi R_P^3\right)}{R_P^2} = \frac{\rho_E \cdot \left(\frac{4}{3} \pi R_E^3\right)}{R_E^2} \]
Simplifying gives:
\[ \rho_P \cdot R_P = \rho_E \cdot R_E \]

7. **Substitute the Density Relation**:
Since \( \rho_P = 2 \rho_E \):
\[ 2 \rho_E \cdot R_P = \rho_E \cdot R \]
Dividing both sides by \( \rho_E \) (assuming \( \rho_E \neq 0 \)):
\[ 2 R_P = R \]

8. **Solve for \( R_P \)**:
\[ R_P = \frac{R}{2} \]

### Final Answer:
The radius of the newly discovered planet is \( \frac{R}{2} \).