Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is:

Options

A. `-(4Gm)/(r )`
B. `-(6Gm)/(r )`
C. `-(9Gm)/(r )`
D. zero

Explanation

To solve the problem, we need to find the gravitational potential at a point on the line joining two masses \( m \) and \( 4m \), where the gravitational field is zero. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have two masses:
- Mass \( m \) is at point A.
- Mass \( 4m \) is at point B.
The distance between them is \( r \).

### Step 2: Identify the Point Where the Gravitational Field is Zero
Let’s denote the point where the gravitational field is zero as point P. We need to find the distance \( x \) from mass \( m \) to point P. Consequently, the distance from mass \( 4m \) to point P will be \( r - x \).

### Step 3: Set Up the Equation for Gravitational Fields
The gravitational field \( E \) due to mass \( m \) at point P is given by:
\[ E_m = \frac{Gm}{x^2} \]
The gravitational field \( E \) due to mass \( 4m \) at point P is given by:
\[ E_{4m} = \frac{G(4m)}{(r - x)^2} \]
Since the gravitational field at point P is zero, we can set these two fields equal to each other:
\[ \frac{Gm}{x^2} = \frac{G(4m)}{(r - x)^2} \]

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

### Step 5: Expand and Rearrange the Equation
Expanding the left side:
\[ r^2 - 2rx + x^2 = 4x^2 \]
Rearranging gives:
\[ r^2 - 2rx - 3x^2 = 0 \]

### Step 6: Solve the Quadratic Equation
This is a quadratic equation in \( x \):
\[ 3x^2 + 2rx - r^2 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = 2r \), and \( c = -r^2 \).
\[ x = \frac{-2r \pm \sqrt{(2r)^2 - 4 \cdot 3 \cdot (-r^2)}}{2 \cdot 3} \]
\[ x = \frac{-2r \pm \sqrt{4r^2 + 12r^2}}{6} \]
\[ x = \frac{-2r \pm \sqrt{16r^2}}{6} \]
\[ x = \frac{-2r \pm 4r}{6} \]
This gives two potential solutions:
1. \( x = \frac{2r}{6} = \frac{r}{3} \) (valid)
2. \( x = \frac{-6r}{6} = -r \) (not valid)

### Step 7: Calculate the Gravitational Potential
Now, we know \( x = \frac{r}{3} \) and \( r - x = \frac{2r}{3} \).

The gravitational potential \( V \) at point P due to mass \( m \) is:
\[ V_m = -\frac{Gm}{\frac{r}{3}} = -\frac{3Gm}{r} \]
The gravitational potential \( V \) at point P due to mass \( 4m \) is:
\[ V_{4m} = -\frac{G(4m)}{\frac{2r}{3}} = -\frac{6Gm}{r} \]

### Step 8: Total Gravitational Potential
The total gravitational potential \( V \) at point P is:
\[ V = V_m + V_{4m} = -\frac{3Gm}{r} - \frac{6Gm}{r} = -\frac{9Gm}{r} \]

### Final Answer
Thus, the gravitational potential at the point where the gravitational field is zero is:
\[ \boxed{-\frac{9Gm}{r}} \]