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 find the radius of a planet where the acceleration due to gravity at its surface is equal to that at the surface of the Earth, and its density is 1.5 times that of Earth, we can follow these steps:

### Step-by-Step Solution

1. **Understand the formula for acceleration due to gravity (g)**:
The acceleration due to gravity 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. **Express mass in terms of density**:
The mass \( M \) of the planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho V \]
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, we can write:
\[ M = \rho \left(\frac{4}{3} \pi R^3\right) \]

3. **Substituting mass into the gravity formula**:
Substituting this expression for mass into the formula for gravity, we have:
\[ g = \frac{G \rho \left(\frac{4}{3} \pi R^3\right)}{R^2} \]
Simplifying this gives:
\[ g = \frac{4}{3} \pi G \rho R \]

4. **Setting the gravity of the planet equal to that of Earth**:
Let \( g_E \) be the acceleration due to gravity at the surface of Earth, which can be expressed as:
\[ g_E = \frac{4}{3} \pi G \rho_E R_E \]
where \( \rho_E \) is the density of Earth and \( R_E \) is the radius of Earth. Since the problem states that the acceleration due to gravity on the planet is equal to that on Earth:
\[ g = g_E \]
Therefore:
\[ \frac{4}{3} \pi G \rho R = \frac{4}{3} \pi G \rho_E R_E \]

5. **Relating the densities**:
Given that the density of the planet \( \rho \) is 1.5 times the density of Earth:
\[ \rho = 1.5 \rho_E \]
Substituting this into the equation gives:
\[ \frac{4}{3} \pi G (1.5 \rho_E) R = \frac{4}{3} \pi G \rho_E R_E \]

6. **Canceling common terms**:
The \( \frac{4}{3} \pi G \) terms cancel out, leading to:
\[ 1.5 \rho_E R = \rho_E R_E \]

7. **Solving for the radius of the planet**:
Dividing both sides by \( \rho_E \) (assuming \( \rho_E \neq 0 \)):
\[ 1.5 R = R_E \]
Thus, we can solve for \( R \):
\[ R = \frac{R_E}{1.5} = \frac{2}{3} R_E \]

### Final Result
The radius of the planet \( R \) is:
\[ R = \frac{2}{3} R_E \]