Two stars each of 1 solar mass `( = 2 xx 10^(30) kg )` are approaching each other for a head-on collision. When they are at a distance `10^(9) km` apart their speeds are negligible . What is the speed with which they collide? The radius of each star is `10^(4) `km. Assume the starts to remain undistorted untile they collide, `G = 6.67 xx 10^(-11) Nm^(2) kg ^(-2)`.

Explanation

To solve the problem of two stars colliding, we will use the principles of gravitational potential energy and kinetic energy conservation. Here’s a step-by-step solution:

### Step 1: Identify the Given Data
- Mass of each star, \( M = 2 \times 10^{30} \) kg (1 solar mass)
- Radius of each star, \( R = 10^{4} \) km = \( 10^{7} \) m
- Initial distance between the stars, \( d = 10^{9} \) km = \( 10^{12} \) m
- Gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)

### Step 2: Calculate the Initial Potential Energy
When the stars are at a distance \( d \) apart, the potential energy \( U_i \) is given by the formula:
\[ U_i = -\frac{G M^2}{d} \]
Substituting the values:
\[ U_i = -\frac{(6.67 \times 10^{-11}) (2 \times 10^{30})^2}{10^{12}} \]

### Step 3: Calculate the Final Potential Energy
When the stars collide, the distance between their centers will be \( 2R \):
\[ U_f = -\frac{G M^2}{2R} \]
Substituting the values:
\[ U_f = -\frac{(6.67 \times 10^{-11}) (2 \times 10^{30})^2}{2 \times 10^{7}} \]

### Step 4: Use Conservation of Energy
The total mechanical energy of the system is conserved. Therefore, the initial potential energy will equal the final potential energy plus the kinetic energy at the moment of collision:
\[ U_i = U_f + K \]
Where \( K \) is the kinetic energy given by:
\[ K = \frac{1}{2} M v^2 + \frac{1}{2} M v^2 = M v^2 \]
Thus, we can write:
\[ U_i = U_f + M v^2 \]

### Step 5: Rearranging to Find the Speed
Rearranging the equation gives:
\[ M v^2 = U_i - U_f \]
\[ v^2 = \frac{U_i - U_f}{M} \]

### Step 6: Substitute Values and Calculate
Now we will substitute the values for \( U_i \) and \( U_f \):
1. Calculate \( U_i \):
\[ U_i = -\frac{(6.67 \times 10^{-11}) (4 \times 10^{60})}{10^{12}} = -2.668 \times 10^{40} \, \text{J} \]
2. Calculate \( U_f \):
\[ U_f = -\frac{(6.67 \times 10^{-11}) (4 \times 10^{60})}{2 \times 10^{7}} = -1.334 \times 10^{40} \, \text{J} \]
3. Now substitute \( U_i \) and \( U_f \) into the equation for \( v^2 \):
\[ v^2 = \frac{-2.668 \times 10^{40} - (-1.334 \times 10^{40})}{2 \times 10^{30}} \]
\[ v^2 = \frac{-1.334 \times 10^{40}}{2 \times 10^{30}} = 6.67 \times 10^{9} \, \text{m}^2/\text{s}^2 \]

### Step 7: Calculate the Final Speed
Taking the square root gives:
\[ v = \sqrt{6.67 \times 10^{9}} \approx 2.58 \times 10^{5} \, \text{m/s} \]

### Final Answer
The speed with which the stars collide is approximately \( 2.58 \times 10^{5} \, \text{m/s} \).

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