Two small bodies of masses 2.00 kg and 4.00 kg are kept at rest at a separation of 2.0 m. Where should a particle of mass 0.10 kg be placed to experience no net gravitational force from these bodies? The particle is placed at this point. What is the gravitational potential energy of the system of three particle with usual reference level?

Explanation

To solve the problem, we need to determine where to place a mass of 0.10 kg (let's call it m3) between two other masses (m1 = 2.00 kg and m2 = 4.00 kg) that are 2.0 m apart, such that m3 experiences no net gravitational force. After finding the position, we will calculate the gravitational potential energy of the system.

### Step-by-Step Solution:

1. **Identify the positions of the masses**:
- Let m1 (2.00 kg) be at position 0 m.
- Let m2 (4.00 kg) be at position 2.00 m.
- We need to find the position of m3 (0.10 kg) at a distance x from m1.

2. **Set up the equations for gravitational forces**:
- The gravitational force on m3 due to m1 (F1) is given by:
\[ F1 = \frac{G \cdot m1 \cdot m3}{x^2} \]
- The gravitational force on m3 due to m2 (F2) is given by:
\[ F2 = \frac{G \cdot m2 \cdot m3}{(2 - x)^2} \]

3. **Set the forces equal to each other**:
- For m3 to experience no net gravitational force, F1 must equal F2:
\[ \frac{G \cdot m1 \cdot m3}{x^2} = \frac{G \cdot m2 \cdot m3}{(2 - x)^2} \]
- Since G and m3 are common on both sides, they can be canceled out:
\[ \frac{m1}{x^2} = \frac{m2}{(2 - x)^2} \]

4. **Substitute the values of m1 and m2**:
- Substituting m1 = 2 kg and m2 = 4 kg:
\[ \frac{2}{x^2} = \frac{4}{(2 - x)^2} \]

5. **Cross-multiply to solve for x**:
- Cross-multiplying gives:
\[ 2(2 - x)^2 = 4x^2 \]
- Expanding and simplifying:
\[ 2(4 - 4x + x^2) = 4x^2 \]
\[ 8 - 8x + 2x^2 = 4x^2 \]
\[ 0 = 2x^2 + 8x - 8 \]
\[ 0 = x^2 + 4x - 4 \]

6. **Use the quadratic formula to solve for x**:
- Applying the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- Here, \(a = 1\), \(b = 4\), and \(c = -4\):
\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ x = \frac{-4 \pm \sqrt{16 + 16}}{2} \]
\[ x = \frac{-4 \pm \sqrt{32}}{2} \]
\[ x = \frac{-4 \pm 4\sqrt{2}}{2} \]
\[ x = -2 \pm 2\sqrt{2} \]
- Since x must be positive and less than 2, we take:
\[ x = -2 + 2\sqrt{2} \approx 0.83 \text{ m} \]

7. **Calculate the gravitational potential energy**:
- The gravitational potential energy (U) between two masses is given by:
\[ U = -\frac{G \cdot m1 \cdot m2}{r} \]
- Calculate U for each pair:
- For m1 and m2:
\[ U_{12} = -\frac{G \cdot 2 \cdot 4}{2} = -4G \]
- For m1 and m3:
\[ U_{13} = -\frac{G \cdot 2 \cdot 0.1}{0.83} \approx -0.24G \]
- For m2 and m3:
\[ U_{23} = -\frac{G \cdot 4 \cdot 0.1}{2 - 0.83} \approx -0.34G \]

8. **Sum the potential energies**:
- The total gravitational potential energy of the system is:
\[ U = U_{12} + U_{13} + U_{23} = -4G - 0.24G - 0.34G = -4.58G \]

### Final Answer:
The gravitational potential energy of the system is approximately:
\[ U \approx -4.58G \text{ joules} \]