Particles of masses `m_1` and `m_2` are at a fixed distance apart . If the gravitational field strength at `m_1` and `m_2` are `vecI_1` and `vecI_2` respectively . Then ,

Options

A. `m_1 vecI_1 + m_2 vecI_2=0`
B. `m_1 vecI_2+m_2 vecI_1=0`
C. `m_1 vecI_1 - m_2vecI_2=0`
D. `m_1vecI_2-m_2vecI_1=0`

Answer

To solve the problem, we need to find the relationship between the gravitational field strengths at two masses \( m_1 \) and \( m_2 \), denoted as \( \vec{I}_1 \) and \( \vec{I}_2 \) respectively.

### Step-by-Step Solution:

1. **Understanding Gravitational Field Strength**:
The gravitational field strength \( \vec{I} \) at a point due to a mass \( m \) is given by the formula:
\[ \vec{I} = \frac{G \cdot m}{r^2} \hat{r} \]
where \( G \) is the gravitational constant, \( r \) is the distance from the mass to the point where the field is being measured, and \( \hat{r} \) is the unit vector pointing away from the mass.

2. **Gravitational Field Strength at \( m_1 \)**:
The gravitational field strength \( \vec{I}_1 \) at mass \( m_1 \) due to mass \( m_2 \) is:
\[ \vec{I}_1 = \frac{G \cdot m_2}{r^2} \hat{r}_{12} \]
where \( \hat{r}_{12} \) is the unit vector pointing from \( m_2 \) to \( m_1 \).

3. **Gravitational Field Strength at \( m_2 \)**:
Similarly, the gravitational field strength \( \vec{I}_2 \) at mass \( m_2 \) due to mass \( m_1 \) is:
\[ \vec{I}_2 = \frac{G \cdot m_1}{r^2} (-\hat{r}_{12}) \]
Here, the negative sign indicates that the gravitational field due to \( m_1 \) at \( m_2 \) is directed towards \( m_1 \).

4. **Multiplying by Masses**:
Now, we multiply both gravitational field strengths by their respective masses:
- For \( m_1 \):
\[ m_1 \vec{I}_1 = m_1 \left(\frac{G \cdot m_2}{r^2} \hat{r}_{12}\right) = \frac{G \cdot m_1 \cdot m_2}{r^2} \hat{r}_{12} \]
- For \( m_2 \):
\[ m_2 \vec{I}_2 = m_2 \left(\frac{G \cdot m_1}{r^2} (-\hat{r}_{12})\right) = -\frac{G \cdot m_1 \cdot m_2}{r^2} \hat{r}_{12} \]

5. **Adding the Two Expressions**:
Now, we add the two expressions:
\[ m_1 \vec{I}_1 + m_2 \vec{I}_2 = \frac{G \cdot m_1 \cdot m_2}{r^2} \hat{r}_{12} - \frac{G \cdot m_1 \cdot m_2}{r^2} \hat{r}_{12} = 0 \]

6. **Conclusion**:
Therefore, we conclude that:
\[ m_1 \vec{I}_1 + m_2 \vec{I}_2 = 0 \]

### Final Answer:
The relation is:
\[ m_1 \vec{I}_1 + m_2 \vec{I}_2 = 0 \]