Two balls, each of radius R, equal mass and density are placed in contact, then the force of gravitation between them is proportional to

Options

A. (a)`F prop 1/R^2`
B. (b)`F prop R `
C. (c)`F prop R^4`
D. (d)`F prop 1/R`

Explanation

To solve the problem of finding the gravitational force between two balls of equal mass, radius, and density when they are placed in contact, we can follow these steps:

### Step 1: Define the parameters
Let each ball have:
- Radius \( R \)
- Mass \( m \)
- Density \( \rho \)

### Step 2: Calculate the mass of each ball
The mass \( m \) of a ball can be calculated using the formula for the volume of a sphere and the density:
\[ m = \text{Volume} \times \text{Density} = \left(\frac{4}{3} \pi R^3\right) \rho \]

### Step 3: Determine the distance between the centers of the balls
When the two balls are placed in contact, the distance between their centers is equal to the sum of their radii:
\[ d = R + R = 2R \]

### Step 4: Apply Newton's Law of Gravitation
According to Newton's Law of Gravitation, the gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( d \) is given by:
\[ F = \frac{G m_1 m_2}{d^2} \]
In our case, since both balls have the same mass \( m \):
\[ F = \frac{G m^2}{(2R)^2} = \frac{G m^2}{4R^2} \]

### Step 5: Substitute the mass into the gravitational force equation
Now, substituting the expression for mass \( m \):
\[ F = \frac{G \left(\frac{4}{3} \pi R^3 \rho\right)^2}{4R^2} \]
Calculating \( m^2 \):
\[ m^2 = \left(\frac{4}{3} \pi R^3 \rho\right)^2 = \frac{16}{9} \pi^2 R^6 \rho^2 \]
Now substitute this back into the force equation:
\[ F = \frac{G \cdot \frac{16}{9} \pi^2 R^6 \rho^2}{4R^2} \]
This simplifies to:
\[ F = \frac{4G \pi^2 R^6 \rho^2}{9} \]

### Step 6: Identify the proportionality
From the final expression, we can see that the gravitational force \( F \) is proportional to \( R^6 \):
\[ F \propto R^6 \]

### Conclusion
Thus, the gravitational force between the two balls is proportional to \( R^6 \).