Two particles of masses 'm' and '9m' are separated by a distance 'r'. At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is (G = Universal constant of gravitation)

Options

A. `-(4Gm)/r`
B. `-(8Gm)/r`
C. `-(16Gm)/r`
D. `-(32Gm)/r`

Answer

To solve the problem step by step, we need to find the gravitational potential at a point where the gravitational field due to two masses \( m \) and \( 9m \) is zero. The two masses are separated by a distance \( r \).

### Step 1: Understanding the Setup
We have two masses:
- Mass \( m \) located at point A
- Mass \( 9m \) located at point B
The distance between them is \( r \). We need to find a point on the line joining them where the gravitational field is zero.

### Step 2: Define the Point of Interest
Let’s denote the point where the gravitational field is zero as point P. Let the distance from mass \( m \) to point P be \( x \). Consequently, the distance from mass \( 9m \) to point P will be \( r - x \).

### Step 3: Setting Up the Gravitational Fields
The gravitational field \( g \) due to a mass \( M \) at a distance \( d \) is given by:
\[ g = \frac{GM}{d^2} \]
where \( G \) is the universal gravitational constant.

At point P, the gravitational field due to mass \( m \) (let's call it \( g_1 \)) and the gravitational field due to mass \( 9m \) (let's call it \( g_2 \)) must be equal in magnitude but opposite in direction for the net gravitational field to be zero:
\[ g_1 = g_2 \]
Substituting the expressions for \( g_1 \) and \( g_2 \):
\[ \frac{Gm}{x^2} = \frac{G(9m)}{(r - x)^2} \]

### Step 4: Simplifying the Equation
We can cancel \( G \) and \( m \) from both sides:
\[ \frac{1}{x^2} = \frac{9}{(r - x)^2} \]
Cross-multiplying gives:
\[ (r - x)^2 = 9x^2 \]

### Step 5: Expanding and Rearranging
Expanding the left side:
\[ r^2 - 2rx + x^2 = 9x^2 \]
Rearranging gives:
\[ r^2 - 2rx - 8x^2 = 0 \]

### Step 6: Solving the Quadratic Equation
This is a quadratic equation in \( x \):
\[ 8x^2 + 2rx - r^2 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 8 \), \( b = 2r \), and \( c = -r^2 \).
\[ x = \frac{-2r \pm \sqrt{(2r)^2 - 4 \cdot 8 \cdot (-r^2)}}{2 \cdot 8} \]
\[ x = \frac{-2r \pm \sqrt{4r^2 + 32r^2}}{16} \]
\[ x = \frac{-2r \pm \sqrt{36r^2}}{16} \]
\[ x = \frac{-2r \pm 6r}{16} \]

### Step 7: Finding the Positive Solution
Calculating the two potential solutions:
1. \( x = \frac{4r}{16} = \frac{r}{4} \) (valid)
2. \( x = \frac{-8r}{16} = -\frac{r}{2} \) (not valid as distance cannot be negative)

Thus, \( x = \frac{r}{4} \).

### Step 8: Finding the Distance from Mass \( 9m \)
The distance from mass \( 9m \) to point P is:
\[ r - x = r - \frac{r}{4} = \frac{3r}{4} \]

### Step 9: Calculating the Gravitational Potential
The gravitational potential \( V \) at point P due to both masses is given by:
\[ V = V_m + V_{9m} \]
Where:
\[ V_m = -\frac{Gm}{x} = -\frac{Gm}{\frac{r}{4}} = -\frac{4Gm}{r} \]
\[ V_{9m} = -\frac{G(9m)}{r - x} = -\frac{G(9m)}{\frac{3r}{4}} = -\frac{12Gm}{r} \]

### Step 10: Summing the Potentials
Now, summing these:
\[ V = -\frac{4Gm}{r} - \frac{12Gm}{r} = -\frac{16Gm}{r} \]

### Final Answer
Thus, the gravitational potential at the point where the gravitational field is zero is:
\[ \boxed{-\frac{16Gm}{r}} \]