Two spherical bodies of mass M and 5M & radii R & 2R respectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smallar body just before collision is

Options

A. 2.5 R
B. 4.5 R
C. 7.5 R
D. 1.5 R

Explanation

To solve the problem step by step, we will analyze the motion of the two spherical bodies under the influence of their mutual gravitational attraction.

### Step 1: Understanding the System
We have two spherical bodies:
- Body 1: Mass = M, Radius = R
- Body 2: Mass = 5M, Radius = 2R

The initial separation between their centers is 12R.

### Step 2: Setting Up the Problem
Let the distance covered by the smaller body (mass M) before collision be \( x_m \) and the distance covered by the larger body (mass 5M) be \( x_{5M} \).

Since they are moving towards each other, the total distance covered by both bodies when they collide will equal the initial separation minus the sum of their radii:
\[ x_m + x_{5M} = 12R - (R + 2R) = 9R \]

### Step 3: Applying Newton's Second Law
The gravitational force acting on each body can be expressed as:
\[ F = \frac{G \cdot M \cdot 5M}{(12R)^2} = \frac{5GM^2}{144R^2} \]

Using Newton's second law, we can express the accelerations of both bodies:
- For mass M:
\[ F = M \cdot a_m \implies a_m = \frac{F}{M} = \frac{5GM}{144R^2} \]
- For mass 5M:
\[ F = 5M \cdot a_{5M} \implies a_{5M} = \frac{F}{5M} = \frac{GM}{144R^2} \]

### Step 4: Finding the Ratio of Accelerations
The ratio of the accelerations can be expressed as:
\[ \frac{a_m}{a_{5M}} = \frac{\frac{5GM}{144R^2}}{\frac{GM}{144R^2}} = 5 \]

### Step 5: Relating Distances Covered
Using the ratios of their accelerations, we can relate the distances covered:
\[ \frac{x_m}{x_{5M}} = \frac{a_{5M}}{a_m} = \frac{1}{5} \]
This implies:
\[ x_m = 5x_{5M} \]

### Step 6: Substituting into the Distance Equation
Now substituting \( x_m \) into the equation \( x_m + x_{5M} = 9R \):
\[ 5x_{5M} + x_{5M} = 9R \implies 6x_{5M} = 9R \implies x_{5M} = \frac{9R}{6} = 1.5R \]

### Step 7: Finding \( x_m \)
Now substituting back to find \( x_m \):
\[ x_m = 5x_{5M} = 5 \cdot 1.5R = 7.5R \]

### Conclusion
The distance covered by the smaller body (mass M) just before collision is:
\[ \boxed{7.5R} \]