Two bodies of masses m and 4 m 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. `-(6 Gm)/(r )`
C. `-(9 Gm)/(r )`
D. zero

Answer

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.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two masses: \( m \) and \( 4m \).
- They are separated by a distance \( r \).
- We need to find a point \( P \) on the line joining them where the gravitational field is zero.

2. **Finding the Point Where Gravitational Field is Zero**:
- Let the distance from mass \( 4m \) to point \( P \) be \( x \).
- Therefore, the distance from mass \( m \) to point \( P \) will be \( r - x \).
- The gravitational field \( E \) due to a mass \( M \) at a distance \( d \) is given by:
\[ E = \frac{GM}{d^2} \]
- For the point \( P \) to have zero gravitational field, the gravitational fields due to both masses must be equal:
\[ \frac{G \cdot 4m}{x^2} = \frac{G \cdot m}{(r - x)^2} \]
- Simplifying this equation (the \( G \) cancels out):
\[ \frac{4m}{x^2} = \frac{m}{(r - x)^2} \]
- Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{4}{x^2} = \frac{1}{(r - x)^2} \]
- Cross-multiplying gives:
\[ 4(r - x)^2 = x^2 \]

3. **Expanding and Rearranging**:
- Expanding the left side:
\[ 4(r^2 - 2rx + x^2) = x^2 \]
- This simplifies to:
\[ 4r^2 - 8rx + 4x^2 = x^2 \]
- Rearranging gives:
\[ 3x^2 - 8rx + 4r^2 = 0 \]

4. **Using the Quadratic Formula**:
- We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3, b = -8r, c = 4r^2 \):
\[ x = \frac{8r \pm \sqrt{(-8r)^2 - 4 \cdot 3 \cdot 4r^2}}{2 \cdot 3} \]
- Calculating the discriminant:
\[ 64r^2 - 48r^2 = 16r^2 \]
- Thus, we have:
\[ x = \frac{8r \pm 4r}{6} \]
- This gives two solutions:
\[ x = \frac{12r}{6} = 2r \quad \text{(not valid, as it is outside the distance)} \]
\[ x = \frac{4r}{6} = \frac{2r}{3} \]

5. **Calculating the Gravitational Potential**:
- The gravitational potential \( V \) at a distance \( d \) from a mass \( M \) is given by:
\[ V = -\frac{GM}{d} \]
- The total gravitational potential at point \( P \) is the sum of the potentials due to both masses:
\[ V = V_{4m} + V_{m} \]
- The distance from \( P \) to \( 4m \) is \( x = \frac{2r}{3} \) and from \( P \) to \( m \) is \( r - x = r - \frac{2r}{3} = \frac{r}{3} \):
\[ V = -\frac{G \cdot 4m}{\frac{2r}{3}} - \frac{G \cdot m}{\frac{r}{3}} \]
- Simplifying:
\[ V = -\frac{12Gm}{r} - \frac{3Gm}{r} = -\frac{15Gm}{r} \]

### Final Answer:
The gravitational potential at the point where the gravitational field is zero is:
\[ V = -\frac{15Gm}{r} \]