Two bodies of mass M and 4M kept at a distance y apart. Where should a small particle of mass m be placed from M so that the net gravitational force on it is zero

Options

A. `(y)/(5)`
B. `(y)/(2)`
C. `(y)/(4)`
D. `(y)/(3)`

Answer

To solve the problem of where to place a small particle of mass \( m \) such that the net gravitational force acting on it is zero, we can follow these steps:

### Step 1: Understand the Setup
We have two masses:
- Mass \( M \) located at point A.
- Mass \( 4M \) located at point B.
The distance between these two masses is \( y \). We need to find a position \( x \) from mass \( M \) where the small mass \( m \) can be placed such that the gravitational forces from both masses on \( m \) cancel each other out.

### Step 2: Define the Forces
The gravitational force \( F \) between two masses is given by Newton's law of gravitation:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
Where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between them.

1. The force \( F_1 \) exerted by mass \( M \) on mass \( m \) when placed at a distance \( x \) is:
\[ F_1 = \frac{G \cdot M \cdot m}{x^2} \]

2. The force \( F_2 \) exerted by mass \( 4M \) on mass \( m \) when placed at a distance \( y - x \) is:
\[ F_2 = \frac{G \cdot 4M \cdot m}{(y - x)^2} \]

### Step 3: Set the Forces Equal
For the net gravitational force on mass \( m \) to be zero, the forces \( F_1 \) and \( F_2 \) must be equal:
\[ F_1 = F_2 \]
This gives us the equation:
\[ \frac{G \cdot M \cdot m}{x^2} = \frac{G \cdot 4M \cdot m}{(y - x)^2} \]

### Step 4: Simplify the Equation
We can cancel \( G \), \( M \), and \( m \) from both sides since they are common factors:
\[ \frac{1}{x^2} = \frac{4}{(y - x)^2} \]

### Step 5: Cross-Multiply
Cross-multiplying gives:
\[ (y - x)^2 = 4x^2 \]

### Step 6: Expand and Rearrange
Expanding the left side:
\[ y^2 - 2yx + x^2 = 4x^2 \]
Rearranging this gives:
\[ y^2 - 2yx - 3x^2 = 0 \]

### Step 7: Solve the Quadratic Equation
This is a quadratic equation in \( x \). We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = -3 \), \( b = -2y \), and \( c = y^2 \):
\[ x = \frac{2y \pm \sqrt{(-2y)^2 - 4(-3)(y^2)}}{2(-3)} \]
\[ x = \frac{2y \pm \sqrt{4y^2 + 12y^2}}{-6} \]
\[ x = \frac{2y \pm \sqrt{16y^2}}{-6} \]
\[ x = \frac{2y \pm 4y}{-6} \]

### Step 8: Calculate the Values
1. For the positive root:
\[ x = \frac{6y}{-6} = -y \quad \text{(not a valid solution)} \]
2. For the negative root:
\[ x = \frac{-2y}{-6} = \frac{y}{3} \]

### Conclusion
Thus, the small particle of mass \( m \) should be placed at a distance \( \frac{y}{3} \) from mass \( M \) to ensure that the net gravitational force acting on it is zero.