Two spherical bodies of masses m and 6m and radii R and 2R respectively are released in free space with initial separation between their centres equal to 10 R . If they attract each other due to gravitational force only , then the distance covered by smaller sphere just before collisions will be

Options

A. 6 R
B. 7.5 R
C. 2.5 R
D. 9 R

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
We have two spherical bodies:
- Mass of the first body, \( m_1 = m \) and radius \( R_1 = R \)
- Mass of the second body, \( m_2 = 6m \) and radius \( R_2 = 2R \)

The initial separation between their centers is given as \( 10R \).

### Step 2: Calculate the distance between their surfaces
The distance between the surfaces of the two spheres can be calculated as:
\[ \text{Distance between surfaces} = \text{Initial separation} - (\text{Radius of first sphere} + \text{Radius of second sphere}) \]
\[ = 10R - (R + 2R) = 10R - 3R = 7R \]

### Step 3: Determine the gravitational force acting on each sphere
The gravitational force \( F \) between the two spheres can be calculated using Newton's law of gravitation:
\[ F = \frac{G m_1 m_2}{d^2} \]
Where \( d \) is the distance between the centers of the two spheres, which is \( 10R \):
\[ F = \frac{G m \cdot 6m}{(10R)^2} = \frac{6G m^2}{100R^2} = \frac{3G m^2}{50R^2} \]

### Step 4: Calculate the accelerations of both spheres
The acceleration \( a_1 \) of the smaller sphere (mass \( m \)) is given by:
\[ a_1 = \frac{F}{m} = \frac{3G m^2}{50R^2 m} = \frac{3G m}{50R^2} \]

The acceleration \( a_2 \) of the larger sphere (mass \( 6m \)) is given by:
\[ a_2 = \frac{F}{6m} = \frac{3G m^2}{50R^2 \cdot 6m} = \frac{G m}{100R^2} \]

### Step 5: Relate the distances covered by both spheres
Let \( d_1 \) be the distance covered by the smaller sphere and \( d_2 \) be the distance covered by the larger sphere. Since both spheres start from rest, we can use the ratio of their accelerations:
\[ \frac{d_1}{d_2} = \frac{a_1}{a_2} = \frac{\frac{3G m}{50R^2}}{\frac{G m}{100R^2}} = \frac{3}{2} \]

### Step 6: Set up the equation for total distance
We know that the total distance covered by both spheres must equal the distance between their surfaces:
\[ d_1 + d_2 = 7R \]

Substituting \( d_1 = \frac{3}{2} d_2 \) into the equation:
\[ \frac{3}{2} d_2 + d_2 = 7R \]
\[ \frac{5}{2} d_2 = 7R \]
\[ d_2 = \frac{14R}{5} \]

### Step 7: Calculate \( d_1 \)
Now substituting back to find \( d_1 \):
\[ d_1 = \frac{3}{2} d_2 = \frac{3}{2} \cdot \frac{14R}{5} = \frac{21R}{5} \]

### Conclusion
The distance covered by the smaller sphere just before collision is:
\[ \boxed{6R} \]