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 determine the minimum initial speed required for a body of mass \( m \) to reach the surface of the smaller star from the surface of the larger star. We will use the concept of gravitational potential energy and conservation of energy.

### Step-by-Step Solution:

1. **Identify the Parameters:**
- Mass of the larger star, \( M_1 = 16M \)
- Radius of the larger star, \( R_1 = 2a \)
- Mass of the smaller star, \( M_2 = M \)
- Radius of the smaller star, \( R_2 = a \)
- Distance between the centers of the two stars, \( d = 10a \)

2. **Calculate the Distance from the Surface of the Larger Star to the Smaller Star:**
- The distance from the surface of the larger star to the center of the smaller star is:
\[ d - R_1 - R_2 = 10a - 2a - a = 7a \]
- Thus, the distance from the surface of the larger star to the center of the smaller star is \( 7a \).

3. **Determine the Point Where Gravitational Forces Balance:**
- Let \( r_1 \) be the distance from the center of the larger star to the point where the gravitational forces from both stars balance, and \( r_2 \) be the distance from the center of the smaller star to that point.
- From the problem, we have:
\[ r_1 + r_2 = 10a \]
- The gravitational fields at that point due to both stars must be equal:
\[ \frac{GM_1}{r_1^2} = \frac{GM_2}{r_2^2} \]
- Substituting \( M_1 = 16M \) and \( M_2 = M \):
\[ \frac{16GM}{r_1^2} = \frac{GM}{r_2^2} \]
- This simplifies to:
\[ r_2 = 4r_1 \]

4. **Substituting into the Distance Equation:**
- From \( r_1 + r_2 = 10a \):
\[ r_1 + 4r_1 = 10a \implies 5r_1 = 10a \implies r_1 = 2a \]
- Therefore, \( r_2 = 4r_1 = 8a \).

5. **Calculate the Potential Energies:**
- The potential energy of the mass \( m \) at the surface of the larger star is:
\[ PE_1 = -\frac{GM_1 m}{R_1} = -\frac{16GmM}{2a} = -\frac{8GmM}{a} \]
- The potential energy at point P (where \( m \) is at distance \( 7a \) from the larger star and \( 2a \) from the smaller star) is:
\[ PE_P = -\frac{GM_1 m}{7a} - \frac{GM_2 m}{2a} = -\frac{16GmM}{7a} - \frac{GmM}{2a} \]
- Finding a common denominator (14a):
\[ PE_P = -\frac{32GmM}{14a} - \frac{7GmM}{14a} = -\frac{39GmM}{14a} \]

6. **Apply Conservation of Energy:**
- The total mechanical energy at the surface of the larger star (initial) equals the total mechanical energy at point P (final):
\[ \frac{1}{2} mv^2 + PE_1 = PE_P \]
- Substituting the potential energies:
\[ \frac{1}{2} mv^2 - \frac{8GmM}{a} = -\frac{39GmM}{14a} \]
- Rearranging gives:
\[ \frac{1}{2} mv^2 = -\frac{39GmM}{14a} + \frac{8GmM}{a} \]
- Finding a common denominator:
\[ \frac{1}{2} mv^2 = -\frac{39GmM}{14a} + \frac{112GmM}{14a} = \frac{73GmM}{14a} \]
- Thus:
\[ mv^2 = \frac{146GmM}{14a} \implies v^2 = \frac{146GM}{14a} \]
- Finally, taking the square root gives:
\[ v = \sqrt{\frac{73GM}{7a}} \]

### Final Expression:
The minimum initial speed \( v \) required for the body of mass \( m \) to reach the surface of the smaller star is:
\[ v = \sqrt{\frac{73GM}{7a}} \]