Two bodies of masses `M_(1)` and `M_(2)` are kept separeated by a distance d. The potential at the point where the gravitational field produced by them is zero,the gravitational potential will be :-

Options

A. `-(G)/(d)(M_(1)+M_(2)+2sqrt(M_(1)M_(2)))`
B. `-(G)/(d)(M_(1)M_(2)+2sqrt(M_(1)+M_(2)))`
C. `-(G)/(d)(M_(1)-M_(2)+2sqrt(M_(1)M_(2)))`
D. `-(G)/(d)(M_(1)M_(2)-2sqrt(M_(1)+M_(2)))`

Explanation

To solve the problem, we need to find the gravitational potential at a point where the gravitational field produced by two masses \( M_1 \) and \( M_2 \) is zero. Here’s a step-by-step solution:

### Step 1: Understanding the Setup
We have two masses \( M_1 \) and \( M_2 \) separated by a distance \( d \). We need to find a point along the line joining the two masses where the gravitational field is zero.

### Step 2: Setting Up the Equation for Gravitational Field
Let’s denote the point where the gravitational field is zero as point \( O \). Let the distance from \( M_1 \) to point \( O \) be \( x \). Consequently, the distance from \( M_2 \) to point \( O \) will be \( d - x \).

The gravitational field \( E \) due to a mass \( M \) at a distance \( r \) is given by:
\[ E = \frac{GM}{r^2} \]
Thus, the gravitational fields due to \( M_1 \) and \( M_2 \) at point \( O \) can be written as:
\[ E_1 = \frac{G M_1}{x^2} \quad \text{and} \quad E_2 = \frac{G M_2}{(d - x)^2} \]

### Step 3: Setting the Gravitational Fields Equal
For the gravitational field to be zero at point \( O \):
\[ E_1 = E_2 \]
This gives us the equation:
\[ \frac{G M_1}{x^2} = \frac{G M_2}{(d - x)^2} \]
We can cancel \( G \) from both sides:
\[ \frac{M_1}{x^2} = \frac{M_2}{(d - x)^2} \]

### Step 4: Cross-Multiplying
Cross-multiplying gives:
\[ M_1 (d - x)^2 = M_2 x^2 \]

### Step 5: Expanding and Rearranging
Expanding the left side:
\[ M_1 (d^2 - 2dx + x^2) = M_2 x^2 \]
Rearranging gives:
\[ M_1 d^2 - 2M_1 dx + M_1 x^2 - M_2 x^2 = 0 \]
This simplifies to:
\[ M_1 d^2 - 2M_1 dx + (M_1 - M_2)x^2 = 0 \]

### Step 6: Solving for \( x \)
This is a quadratic equation in terms of \( x \). We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = M_1 - M_2 \), \( b = -2M_1 d \), and \( c = M_1 d^2 \).

### Step 7: Finding the Gravitational Potential
The gravitational potential \( V \) at point \( O \) due to both masses is given by:
\[ V = V_1 + V_2 \]
where:
\[ V_1 = -\frac{G M_1}{x} \quad \text{and} \quad V_2 = -\frac{G M_2}{d - x} \]
Thus,
\[ V = -\frac{G M_1}{x} - \frac{G M_2}{d - x} \]

### Step 8: Substituting \( x \) into the Potential Equation
Substituting the value of \( x \) obtained from the quadratic solution into the potential equation will yield the final expression for the gravitational potential at point \( O \).

### Final Answer
After simplification, the gravitational potential at point \( O \) is:
\[ V = -\frac{G}{d} \left( M_1 + M_2 + 2\sqrt{M_1 M_2} \right) \]