Distance between the centres of two stars is `10a`. The masses of these stars are `M` and `16 M` and their radii `a` and `2a` respectively. A body of mass `m` is fired straight from the surface of the larger star towards the surface of the smaller star. What should be its minimum initial speed to reach the surface of the smaller star? Obtain the expression in terms of `G`, `M` and `a`.

Explanation

To solve the problem, we need to find the minimum initial speed required for a body of mass \( m \) to reach the surface of the smaller star when fired from the surface of the larger star. Let's break down the solution step by step.

### Step 1: Understand the Setup
We have two stars:
- Larger star with mass \( M \) and radius \( a \).
- Smaller star with mass \( 16M \) and radius \( 2a \).
- The distance between the centers of the two stars is \( 10a \).

### Step 2: Identify the Forces
When the body of mass \( m \) is fired from the surface of the larger star, it will be influenced by the gravitational forces of both stars. We need to find the point where the gravitational attractions from both stars are equal.

### Step 3: Determine the Distance from Each Star
Let \( R_1 \) be the distance from the center of the smaller star to the point where the forces are equal, and \( R_2 \) be the distance from the center of the larger star. We know:
\[ R_1 + R_2 = 10a \]

The gravitational force exerted by the smaller star on the body is:
\[ F_1 = \frac{G \cdot 16M \cdot m}{R_1^2} \]
And the gravitational force exerted by the larger star is:
\[ F_2 = \frac{G \cdot M \cdot m}{R_2^2} \]

Setting these forces equal gives us:
\[ \frac{G \cdot 16M \cdot m}{R_1^2} = \frac{G \cdot M \cdot m}{R_2^2} \]

### Step 4: Simplify the Equation
Canceling \( G \) and \( m \) from both sides, we have:
\[ \frac{16M}{R_1^2} = \frac{M}{R_2^2} \]
This simplifies to:
\[ 16R_2^2 = R_1^2 \]
Thus,
\[ R_1 = 4R_2 \]

### Step 5: Substitute Back into the Distance Equation
Substituting \( R_1 \) into the distance equation:
\[ 4R_2 + R_2 = 10a \]
This gives:
\[ 5R_2 = 10a \]
Thus,
\[ R_2 = 2a \]
And substituting back to find \( R_1 \):
\[ R_1 = 4R_2 = 8a \]

### Step 6: Calculate the Potential Energy
The potential energy at the surface of the larger star (initial position) and at the point \( R \) (where the forces are equal) can be calculated as follows:
- Potential energy at the surface of the larger star:
\[ PE_{initial} = -\frac{G \cdot M \cdot m}{a} \]
- Potential energy at the point \( R \):
\[ PE_{R} = -\frac{G \cdot 16M \cdot m}{2a} - \frac{G \cdot M \cdot m}{8a} \]

### Step 7: Set Up the Energy Conservation Equation
Using the conservation of energy:
\[ KE_{initial} + PE_{initial} = PE_{R} \]
Where \( KE_{initial} = \frac{1}{2} m v_{min}^2 \).

Substituting the potential energies:
\[ \frac{1}{2} m v_{min}^2 - \frac{G \cdot M \cdot m}{a} = -\frac{G \cdot 16M \cdot m}{2a} - \frac{G \cdot M \cdot m}{8a} \]

### Step 8: Solve for \( v_{min} \)
Rearranging gives:
\[ \frac{1}{2} m v_{min}^2 = \frac{G \cdot M \cdot m}{a} - \left( -\frac{G \cdot 16M \cdot m}{2a} - \frac{G \cdot M \cdot m}{8a} \right) \]

Combining terms and simplifying:
\[ \frac{1}{2} m v_{min}^2 = G \cdot m \left( \frac{M}{a} + \frac{16M}{2a} + \frac{M}{8a} \right) \]

### Step 9: Final Expression
After simplification, we find:
\[ v_{min}^2 = \frac{45GM}{4a} \]
Thus,
\[ v_{min} = \sqrt{\frac{45GM}{4a}} \]

### Final Answer
The minimum initial speed required for the body to reach the surface of the smaller star is:
\[ v_{min} = \frac{3\sqrt{5}}{2} \sqrt{\frac{GM}{a}} \]