In a particular double star system two stars of mass `3.22xx10^(30)` kg each revolve about their common centre of mass `1.12xx10^(11)` on away.
(i). Calculate their common period of revolution, in years
(ii). Suppose that a meteoroid (small solid particle in space) passes through this centre of mass moving at right angles to the orbital plane of the stars. What must its speed be if it is to escape from the gravitational field of the double star?

Explanation

To solve the problem step by step, we will address both parts of the question.

### Part (i): Calculate the common period of revolution in years.

1. **Identify the given values:**
- Mass of each star, \( M = 3.22 \times 10^{30} \) kg
- Distance from the center of mass to each star, \( r = 1.12 \times 10^{11} \) km = \( 1.12 \times 10^{14} \) m (since 1 km = 1000 m)

2. **Use the formula for the period of revolution:**
The period \( T \) for two stars revolving around their common center of mass can be derived from the gravitational force and centripetal force balance. The formula is:
\[ T = 2\pi \sqrt{\frac{4r^3}{G(M + M)}} \]
where \( G \) is the gravitational constant, \( G \approx 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).

3. **Substituting the values:**
\[ T = 2\pi \sqrt{\frac{4(1.12 \times 10^{14})^3}{6.67 \times 10^{-11}(3.22 \times 10^{30} + 3.22 \times 10^{30})}} \]
\[ = 2\pi \sqrt{\frac{4(1.12 \times 10^{14})^3}{6.67 \times 10^{-11} \times 6.44 \times 10^{30}}} \]

4. **Calculate \( T \):**
- First, calculate \( (1.12 \times 10^{14})^3 \):
\[ (1.12 \times 10^{14})^3 = 1.41 \times 10^{42} \, \text{m}^3 \]
- Then calculate the denominator:
\[ 6.67 \times 10^{-11} \times 6.44 \times 10^{30} = 4.29 \times 10^{20} \, \text{N m}^2/\text{kg} \]
- Now plug these values back into the equation for \( T \):
\[ T = 2\pi \sqrt{\frac{4 \times 1.41 \times 10^{42}}{4.29 \times 10^{20}}} \]
\[ = 2\pi \sqrt{1.31 \times 10^{22}} \approx 2\pi \times 1.14 \times 10^{11} \]
\[ \approx 7.16 \times 10^{11} \, \text{s} \]

5. **Convert seconds to years:**
\[ T \approx \frac{7.16 \times 10^{11}}{3.154 \times 10^7} \approx 22700 \, \text{years} \]

### Part (ii): Calculate the speed of the meteoroid to escape the gravitational field.

1. **Use the escape velocity formula:**
The escape velocity \( v \) from the gravitational field of the double star system is given by:
\[ v = \sqrt{\frac{2GM}{r}} \]
where \( M \) is the total mass of the system (which is \( 2M \)) and \( r \) is the distance from the center of mass.

2. **Substituting the values:**
\[ v = \sqrt{\frac{2G(2M)}{r}} = \sqrt{\frac{4GM}{r}} \]

3. **Calculate \( v \):**
\[ v = \sqrt{\frac{4 \times 6.67 \times 10^{-11} \times 3.22 \times 10^{30}}{1.12 \times 10^{14}}} \]
\[ = \sqrt{\frac{8.56 \times 10^{20}}{1.12 \times 10^{14}}} = \sqrt{7.65 \times 10^{6}} \approx 2765 \, \text{m/s} \]

### Final Answers:
(i) The common period of revolution is approximately **22700 years**.
(ii) The speed of the meteoroid must be approximately **2765 m/s** to escape the gravitational field of the double star system.