Consider two solid uniform spherical objects of the same density `rho`. One has radius `R` and the other has radius `2R`. They are in outer space where the gravitational fields from other objects are negligible. If they are arranged with their surface touching, what is the contact force between the objects due to their traditional attraction?

Options

A. `Gpi^(2)R^(4)`
B. `128/81Gpi^(2)R^(4)rho^(2)`
C. `128/81Gpi^(2)`
D. `128/87pi^(2)R^(2)G`

Explanation

To solve the problem of finding the contact force between two solid uniform spherical objects due to their gravitational attraction, we can follow these steps:

### Step 1: Determine the Mass of Each Sphere
1. **For the smaller sphere (radius R)**:
- Volume \( V_1 = \frac{4}{3} \pi R^3 \)
- Mass \( m_1 = \rho \cdot V_1 = \rho \cdot \frac{4}{3} \pi R^3 \)

2. **For the larger sphere (radius 2R)**:
- Volume \( V_2 = \frac{4}{3} \pi (2R)^3 = \frac{4}{3} \pi \cdot 8R^3 = \frac{32}{3} \pi R^3 \)
- Mass \( m_2 = \rho \cdot V_2 = \rho \cdot \frac{32}{3} \pi R^3 \)

### Step 2: Calculate the Gravitational Force Between the Two Spheres
The gravitational force \( F \) between two masses is given by Newton's law of gravitation:
\[ F = \frac{G m_1 m_2}{r^2} \]
where \( r \) is the distance between the centers of the two spheres. Since the spheres are touching, the distance \( r \) is equal to the sum of their radii:
\[ r = R + 2R = 3R \]

### Step 3: Substitute the Masses and Distance into the Gravitational Force Equation
Substituting \( m_1 \) and \( m_2 \) into the gravitational force equation:
\[ F = \frac{G \left(\rho \cdot \frac{4}{3} \pi R^3\right) \left(\rho \cdot \frac{32}{3} \pi R^3\right)}{(3R)^2} \]

### Step 4: Simplify the Expression
1. Calculate the numerator:
\[ F = \frac{G \cdot \rho^2 \cdot \frac{128}{9} \pi^2 R^6}{9R^2} \]

2. Simplify the denominator:
\[ (3R)^2 = 9R^2 \]

3. Combine the terms:
\[ F = \frac{G \cdot \rho^2 \cdot \frac{128}{9} \pi^2 R^6}{9R^2} = \frac{G \cdot \rho^2 \cdot 128 \pi^2 R^4}{81} \]

### Final Result
Thus, the contact force between the two spheres due to their gravitational attraction is:
\[ F = \frac{128 G \rho^2 \pi^2 R^4}{81} \]

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