Two neutron stars are separated by a distance of `10^(10) m`. They each have a mass of `10^(30) kg` and a radius of `10^(5) m`. They are initially at rest with respect to each other.
As measured from the rest frame, how fast are they moving when
(a) their separation has decreaed to one - half its initial value,
(b) they are about to collide .

Explanation

To solve the problem, we will use the principles of conservation of energy and conservation of momentum.

### Given Data:
- Mass of each neutron star, \( m = 10^{30} \, \text{kg} \)
- Initial separation, \( r_i = 10^{10} \, \text{m} \)
- Radius of each neutron star, \( R = 10^{5} \, \text{m} \)
- Gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)

### Part (a): Speed when separation is halved

1. **Initial Energy Calculation**:
The initial potential energy \( E_i \) when the stars are at rest is given by:
\[ E_i = -\frac{G m^2}{r_i} \]

2. **Final Energy Calculation**:
When the separation is halved, \( r_f = \frac{r_i}{2} = 5 \times 10^{9} \, \text{m} \).
The final potential energy \( E_f \) is:
\[ E_f = -\frac{G m^2}{r_f} = -\frac{G m^2}{\frac{r_i}{2}} = -\frac{2G m^2}{r_i} \]
The kinetic energy of both stars when they are moving with speed \( v \) is:
\[ K_f = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = mv^2 \]

3. **Conservation of Energy**:
Setting initial energy equal to final energy:
\[ E_i = E_f + K_f \]
\[ -\frac{G m^2}{r_i} = -\frac{2G m^2}{r_i} + mv^2 \]
Rearranging gives:
\[ mv^2 = \frac{G m^2}{r_i} \]
Dividing by \( m \):
\[ v^2 = \frac{G m}{r_i} \]
Taking the square root:
\[ v = \sqrt{\frac{G m}{r_i}} \]

4. **Substituting Values**:
\[ v = \sqrt{\frac{(6.67 \times 10^{-11}) (10^{30})}{10^{10}}} \]
\[ v = \sqrt{6.67 \times 10^{9}} \approx 8.16 \times 10^{4} \, \text{m/s} \quad \text{(or 81.6 km/s)} \]

### Part (b): Speed just before collision

1. **Final Separation**:
When the stars are about to collide, the separation is equal to twice their radius:
\[ r_f = 2R = 2 \times 10^{5} \, \text{m} = 2 \times 10^{5} \, \text{m} \]

2. **Final Energy Calculation**:
The potential energy when they are about to collide is:
\[ E_f = -\frac{G m^2}{r_f} = -\frac{G m^2}{2R} \]

3. **Conservation of Energy**:
Setting initial energy equal to final energy:
\[ -\frac{G m^2}{r_i} = mv^2 - \frac{G m^2}{2R} \]
Rearranging gives:
\[ mv^2 = -\frac{G m^2}{r_i} + \frac{G m^2}{2R} \]
Dividing by \( m \):
\[ v^2 = -\frac{G m}{r_i} + \frac{G m}{2R} \]
\[ v^2 = G m \left( \frac{1}{2R} - \frac{1}{r_i} \right) \]

4. **Substituting Values**:
\[ v = \sqrt{G m \left( \frac{1}{2 \times 10^{5}} - \frac{1}{10^{10}} \right)} \]
\[ v = \sqrt{(6.67 \times 10^{-11})(10^{30}) \left( \frac{1}{2 \times 10^{5}} - \frac{1}{10^{10}} \right)} \]
\[ v = \sqrt{(6.67 \times 10^{9}) \left( 5 \times 10^{-6} - 10^{-10} \right)} \approx \sqrt{(6.67 \times 10^{9})(5 \times 10^{-6})} \]
\[ v \approx 1.8 \times 10^{7} \, \text{m/s} \quad \text{(or 1.8 \times 10^{4} km/s)} \]

### Final Answers:
- (a) Speed when separation is halved: \( v \approx 81.6 \, \text{km/s} \)
- (b) Speed just before collision: \( v \approx 1.8 \times 10^{4} \, \text{km/s} \)