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. `1/4R`
D. `1/2R`

Explanation

To find the radius of the newly discovered planet in relation to the radius of Earth, we can follow these steps:

### Step 1: Understanding the Given Information
We know that:
- The density of the planet (ρ_p) is twice that of Earth (ρ_e), so:
\[ \rho_p = 2 \rho_e \]
- The acceleration due to gravity at the surface of the planet (g_p) is equal to that at the surface of Earth (g_e), so:
\[ g_p = g_e \]

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

### Step 3: Express Mass in Terms of Density
The mass (M) of a planet can be expressed in terms of its density (ρ) and volume (V):
\[ M = \rho \cdot V \]
For a spherical planet, the volume (V) is:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass can be rewritten as:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

### Step 4: Substitute Mass in the Gravity Formula
Substituting the expression for mass into the gravity formula gives:
\[ g = \frac{G \cdot \left(\rho \cdot \frac{4}{3} \pi R^3\right)}{R^2} \]
This simplifies to:
\[ g = \frac{4}{3} \pi G \rho R \]

### Step 5: Write the Equations for Earth and the Planet
For Earth:
\[ g_e = \frac{4}{3} \pi G \rho_e R_e \]
For the planet:
\[ g_p = \frac{4}{3} \pi G \rho_p R_p \]

### Step 6: Set the Two Equations Equal
Since \( g_p = g_e \):
\[ \frac{4}{3} \pi G \rho_e R_e = \frac{4}{3} \pi G \rho_p R_p \]
We can cancel out the common terms:
\[ \rho_e R_e = \rho_p R_p \]

### Step 7: Substitute Density of the Planet
Substituting \( \rho_p = 2 \rho_e \):
\[ \rho_e R_e = (2 \rho_e) R_p \]
Dividing both sides by \( \rho_e \) (assuming \( \rho_e \neq 0 \)):
\[ R_e = 2 R_p \]

### Step 8: Solve for the Radius of the Planet
Rearranging gives:
\[ R_p = \frac{R_e}{2} \]

### Conclusion
Thus, the radius of the newly discovered planet in relation to the radius of Earth is:
\[ R_p = \frac{R}{2} \]