Two point masses having mass m and 2m are placed at distance d . The point on the line joining point masses , where gravitational field intensity is zero will be at distance

Options

A. `(2d)/(sqrt(3)+1)` from point mass "2m"
B. `(2d)/(sqrt(3)-1)`
C. `d/(1 +sqrt(2))`
D. `d/(1-sqrt(2))` from mass "m"

Answer

To find the point on the line joining two point masses where the gravitational field intensity is zero, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the masses and their positions**:
- Let mass \( m \) be at point A and mass \( 2m \) be at point B. The distance between them is \( d \).

2. **Define the point where the gravitational field intensity is zero**:
- Let the point where the gravitational field intensity is zero be at a distance \( x \) from mass \( m \) (point A). Consequently, the distance from mass \( 2m \) (point B) will be \( d - x \).

3. **Write the expression for gravitational field intensity**:
- The gravitational field intensity \( E \) due to a point mass is given by:
\[ E = \frac{G \cdot m}{r^2} \]
- Therefore, the gravitational field intensity due to mass \( m \) at point O (distance \( x \)) is:
\[ E_m = \frac{G \cdot m}{x^2} \]
- The gravitational field intensity due to mass \( 2m \) at point O (distance \( d - x \)) is:
\[ E_{2m} = \frac{G \cdot 2m}{(d - x)^2} \]

4. **Set the gravitational field intensities equal to each other**:
- For the gravitational field intensity to be zero at point O, we set the two intensities equal:
\[ \frac{G \cdot m}{x^2} = \frac{G \cdot 2m}{(d - x)^2} \]
- We can cancel \( G \) and \( m \) from both sides:
\[ \frac{1}{x^2} = \frac{2}{(d - x)^2} \]

5. **Cross-multiply to solve for \( x \)**:
- Cross-multiplying gives:
\[ (d - x)^2 = 2x^2 \]

6. **Expand and rearrange the equation**:
- Expanding the left side:
\[ d^2 - 2dx + x^2 = 2x^2 \]
- Rearranging gives:
\[ d^2 - 2dx - x^2 = 0 \]

7. **Rearranging into standard quadratic form**:
- This can be rewritten as:
\[ x^2 + 2dx - d^2 = 0 \]

8. **Use the quadratic formula to solve for \( x \)**:
- The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 1 \), \( b = -2d \), and \( c = -d^2 \):
\[ x = \frac{2d \pm \sqrt{(-2d)^2 - 4 \cdot 1 \cdot (-d^2)}}{2 \cdot 1} \]
- Simplifying gives:
\[ x = \frac{2d \pm \sqrt{4d^2 + 4d^2}}{2} \]
\[ x = \frac{2d \pm \sqrt{8d^2}}{2} \]
\[ x = d \pm d\sqrt{2} \]

9. **Select the appropriate solution**:
- Since \( x \) must be less than \( d \) (it is the distance from mass \( m \)), we take:
\[ x = \frac{d}{1 + \sqrt{2}} \]

### Final Answer:
The distance from mass \( m \) where the gravitational field intensity is zero is:
\[ x = \frac{d}{1 + \sqrt{2}} \]