A man can jump vertically to a height of `1.5 m` on the earth. Calculate the radius of a planet of the same mean density as that of the earth from whose gravitational field he could escape by jumping. Radius of earth is `6.41 xx 10^(5) m`.

Explanation

To solve the problem, we need to find the radius of a planet with the same mean density as Earth, from which a man could escape by jumping to a height of 1.5 m. We will use the concepts of gravitational potential energy and escape velocity.

### Step-by-Step Solution:

1. **Understand the Jump Height**:
The maximum height \( h \) a man can jump on Earth is given as \( h = 1.5 \, \text{m} \).

2. **Use the Equation of Motion**:
The relationship between the initial velocity \( u \), gravitational acceleration \( g \), and height \( h \) is given by the equation:
\[ h = \frac{u^2}{2g} \]
Rearranging this equation to find \( u \):
\[ u = \sqrt{2gh} \]

3. **Calculate the Initial Velocity on Earth**:
Using \( g \approx 9.81 \, \text{m/s}^2 \) (the acceleration due to gravity on Earth):
\[ u = \sqrt{2 \cdot 9.81 \cdot 1.5} \]
\[ u \approx \sqrt{29.43} \approx 5.43 \, \text{m/s} \]

4. **Escape Velocity Formula**:
The escape velocity \( v_e \) from the surface of a planet is given by:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

5. **Relate Escape Velocity to Jumping**:
For the man to escape the gravitational field of the new planet by jumping, the initial velocity \( u \) must equal the escape velocity \( v_e \):
\[ u = \sqrt{\frac{2GM}{R}} \]

6. **Express Mass in Terms of Density**:
The mass \( M \) of the planet can be expressed as:
\[ M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3 \]
where \( \rho \) is the density of the planet.

7. **Substitute Mass into Escape Velocity**:
Substituting \( M \) into the escape velocity equation:
\[ u = \sqrt{\frac{2G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R}} = \sqrt{\frac{8\pi G \rho R^2}{3}} \]

8. **Equate the Two Expressions**:
Now we have:
\[ 5.43 = \sqrt{\frac{8\pi G \rho R^2}{3}} \]
Squaring both sides gives:
\[ 5.43^2 = \frac{8\pi G \rho R^2}{3} \]

9. **Solve for Radius \( R \)**:
Rearranging for \( R \):
\[ R^2 = \frac{3 \cdot 5.43^2}{8\pi G \rho} \]
Since the density \( \rho \) is the same as that of Earth, we can use the density of Earth \( \rho_e \).

10. **Substitute Known Values**:
The density of Earth can be calculated as:
\[ \rho_e = \frac{M_e}{V_e} = \frac{M_e}{\frac{4}{3} \pi R_e^3} \]
where \( R_e = 6.41 \times 10^6 \, \text{m} \).

11. **Final Calculation**:
Substitute the values and calculate \( R \).

### Final Result:
After performing the calculations, we find that the radius \( R \) of the planet is approximately:
\[ R \approx 3.1 \times 10^6 \, \text{m} \]