Acceleration due to gravity at surface of a planet is equal to that at surface of earth and density is 1.5 times that of earth. If radius is R. radius of planet is

Options

A. `3/2 R`
B. `2/3 R`
C. `9/4 R`
D. `4/9` R

Answer

To solve the problem, we need to find the radius of a planet given that the acceleration due to gravity at its surface is equal to that at the surface of Earth, and that its density is 1.5 times that of Earth.

### Step-by-Step Solution:

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

2. **Setting up the equation for the planet and Earth**:
Since the acceleration due to gravity on the planet is equal to that on Earth, we can write:
\[ \frac{GM_p}{R_p^2} = \frac{GM_e}{R_e^2} \]
where \( M_p \) and \( R_p \) are the mass and radius of the planet, and \( M_e \) and \( R_e \) are the mass and radius of the Earth.

3. **Expressing mass in terms of density**:
The mass of the planet can also be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho V \]
The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass of the planet can be expressed as:
\[ M_p = \rho_p \cdot \frac{4}{3} \pi R_p^3 \]
Given that the density of the planet is 1.5 times that of Earth (\( \rho_p = 1.5 \rho_e \)), we can write:
\[ M_p = 1.5 \rho_e \cdot \frac{4}{3} \pi R_p^3 \]

4. **Setting up the mass ratio**:
The mass of the Earth can be expressed similarly:
\[ M_e = \rho_e \cdot \frac{4}{3} \pi R_e^3 \]
Now, substituting the expressions for \( M_p \) and \( M_e \) into the gravity equation gives:
\[ \frac{1.5 \rho_e \cdot \frac{4}{3} \pi R_p^3}{R_p^2} = \frac{\rho_e \cdot \frac{4}{3} \pi R_e^3}{R_e^2} \]

5. **Canceling common terms**:
The \( \frac{4}{3} \pi \) and \( \rho_e \) terms cancel out, leading to:
\[ \frac{1.5 R_p^3}{R_p^2} = \frac{R_e^3}{R_e^2} \]
Simplifying gives:
\[ 1.5 R_p = R_e \]

6. **Expressing radius in terms of Earth's radius**:
Given that the radius of the Earth \( R_e = R \), we can substitute:
\[ 1.5 R_p = R \]
Thus, the radius of the planet \( R_p \) can be expressed as:
\[ R_p = \frac{R}{1.5} = \frac{2R}{3} \]

### Final Answer:
The radius of the planet is \( \frac{2R}{3} \).