What must the separation be between a 5.2 kg particle and a 2.4 kg particle for their gravitational attraction to have a magnitude of `2.3 xx 10^(-12)` N?

Answer

To find the separation between a 5.2 kg particle and a 2.4 kg particle for their gravitational attraction to have a magnitude of \(2.3 \times 10^{-12}\) N, we can use Newton's law of gravitation. The formula for gravitational force is given by:

\[ F_g = \frac{G \cdot M_1 \cdot M_2}{R^2} \]

Where:
- \(F_g\) is the gravitational force,
- \(G\) is the gravitational constant (\(6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\)),
- \(M_1\) is the mass of the first particle (5.2 kg),
- \(M_2\) is the mass of the second particle (2.4 kg),
- \(R\) is the separation between the two masses.

### Step-by-Step Solution:

1. **Identify the known values**:
- \(M_1 = 5.2 \, \text{kg}\)
- \(M_2 = 2.4 \, \text{kg}\)
- \(F_g = 2.3 \times 10^{-12} \, \text{N}\)
- \(G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\)

2. **Rearrange the formula to solve for \(R\)**:
\[ R^2 = \frac{G \cdot M_1 \cdot M_2}{F_g} \]
Therefore,
\[ R = \sqrt{\frac{G \cdot M_1 \cdot M_2}{F_g}} \]

3. **Substitute the known values into the equation**:
\[ R = \sqrt{\frac{(6.67 \times 10^{-11}) \cdot (5.2) \cdot (2.4)}{2.3 \times 10^{-12}}} \]

4. **Calculate the numerator**:
\[ (6.67 \times 10^{-11}) \cdot (5.2) \cdot (2.4) = 8.33976 \times 10^{-10} \]

5. **Calculate \(R^2\)**:
\[ R^2 = \frac{8.33976 \times 10^{-10}}{2.3 \times 10^{-12}} = 3.628 \times 10^{2} \]

6. **Take the square root to find \(R\)**:
\[ R = \sqrt{3.628 \times 10^{2}} \approx 19.03 \, \text{m} \]

7. **Final answer**:
The separation \(R\) must be approximately \(19 \, \text{m}\).