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 \( m \) and \( 2m \) where the gravitational field intensity is zero, we can follow these steps:

### Step-by-Step Solution:

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

2. **Define the point where the gravitational field 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 mass \( M \) at a distance \( r \) is given by:
\[ E = \frac{GM}{r^2} \]
where \( G \) is the gravitational constant.

4. **Calculate the gravitational field intensity at point O**:
- The gravitational field intensity due to mass \( m \) at point O (distance \( x \)):
\[ E_1 = \frac{Gm}{x^2} \]
- The gravitational field intensity due to mass \( 2m \) at point O (distance \( d - x \)):
\[ E_2 = \frac{G(2m)}{(d - x)^2} \]

5. **Set the gravitational field intensities equal to each other**: Since the gravitational field intensity is zero at point O, we set \( E_1 \) equal to \( E_2 \):
\[ \frac{Gm}{x^2} = \frac{G(2m)}{(d - x)^2} \]

6. **Cancel out common terms**: The gravitational constant \( G \) and mass \( m \) can be canceled from both sides:
\[ \frac{1}{x^2} = \frac{2}{(d - x)^2} \]

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

8. **Expand and rearrange the equation**:
\[ d^2 - 2dx + x^2 = 2x^2 \]
\[ d^2 - 2dx - x^2 = 0 \]

9. **Rearranging gives us a quadratic equation**:
\[ x^2 + 2dx - d^2 = 0 \]

10. **Use the quadratic formula to solve for \( x \)**:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 2d \), and \( c = -d^2 \):
\[ x = \frac{-2d \pm \sqrt{(2d)^2 - 4(1)(-d^2)}}{2(1)} \]
\[ x = \frac{-2d \pm \sqrt{4d^2 + 4d^2}}{2} \]
\[ x = \frac{-2d \pm \sqrt{8d^2}}{2} \]
\[ x = \frac{-2d \pm 2d\sqrt{2}}{2} \]
\[ x = -d + d\sqrt{2} \quad \text{(discarding the negative root as distance cannot be negative)} \]
\[ x = d(\sqrt{2} - 1) \]

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