Two identical solid copper spheres of radius `R` placed in contact with each other. The gravitational attracton between them is proportional to

Options

A. `R`
B. `R-`
C. `R`
D. `R`

Answer

To solve the problem of finding the gravitational attraction between two identical solid copper spheres of radius \( R \) placed in contact with each other, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Mass of Each Sphere:**
The mass \( M \) of a solid sphere can be calculated using the formula:
\[ M = \text{Density} \times \text{Volume} \]
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass of each sphere is:
\[ M = \rho \times \frac{4}{3} \pi R^3 \]
where \( \rho \) is the density of copper.

2. **Determine the Gravitational Force:**
The gravitational force \( F \) between two masses \( M_1 \) and \( M_2 \) separated by a distance \( r \) is given by Newton's law of gravitation:
\[ F = G \frac{M_1 M_2}{r^2} \]
Since both spheres are identical, we have \( M_1 = M_2 = M \). The distance \( r \) between the centers of the two spheres when they are in contact is \( 2R \). Thus, we can rewrite the gravitational force as:
\[ F = G \frac{M^2}{(2R)^2} \]

3. **Substituting the Mass:**
Substitute \( M \) into the gravitational force equation:
\[ F = G \frac{\left(\rho \frac{4}{3} \pi R^3\right)^2}{(2R)^2} \]

4. **Simplifying the Expression:**
Now simplify the expression:
\[ F = G \frac{\left(\rho^2 \left(\frac{4}{3}\right)^2 \pi^2 R^6\right)}{4R^2} \]
\[ F = G \frac{\rho^2 \left(\frac{16}{9}\right) \pi^2 R^6}{4R^2} \]
\[ F = G \frac{4 \rho^2 \pi^2 R^6}{9R^2} \]
\[ F = G \frac{4 \rho^2 \pi^2 R^4}{9} \]

5. **Conclude the Proportionality:**
From the final expression, we can see that the gravitational force \( F \) is proportional to \( R^4 \):
\[ F \propto R^4 \]

### Final Answer:
The gravitational attraction between the two identical solid copper spheres is proportional to \( R^4 \).