Our sun with mass `2 xx 10^(30)kg` revolves on the edge of our milky way galaxy which can be assumed to be spherical having `10^(20)m` Also assumee that many stars identical to our sun are unifromly distributed in the spherical milky way galaxy If the time period of the sun is `10^(15)` second and number of stars in the galaxy are nearly `3 xx 10^((a))` find value of 'a' (take `pi^(2)` = 10, `G (20)/(3) xx 10^(-11)`inMKS) .

Answer

To solve the problem, we need to find the value of 'a' in the expression for the number of stars in the galaxy, given that the mass of the sun is \(2 \times 10^{30} \, \text{kg}\), the radius of the Milky Way galaxy is \(10^{20} \, \text{m}\), and the time period of revolution is \(10^{15} \, \text{s}\).

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the Sun, \(M_s = 2 \times 10^{30} \, \text{kg}\)
- Radius of the Milky Way galaxy, \(R = 10^{20} \, \text{m}\)
- Time period of revolution, \(T = 10^{15} \, \text{s}\)
- Gravitational constant, \(G = \frac{20}{3} \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2\)
- \(\pi^2 = 10\)

2. **Calculate the angular velocity \(\omega\):**
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \(T\):
\[ \omega = \frac{2 \times \sqrt{10}}{10^{15}} = \frac{2\sqrt{10}}{10^{15}} \]

3. **Set up the equations for centripetal force and gravitational force:**
The centripetal force acting on the Sun is given by:
\[ F_c = M_s \omega^2 R \]
The gravitational force due to all the stars is:
\[ F_g = \frac{G \cdot n \cdot M_s^2}{R^2} \]
where \(n\) is the number of stars.

4. **Equate the two forces:**
\[ M_s \omega^2 R = \frac{G \cdot n \cdot M_s^2}{R^2} \]

5. **Simplify the equation:**
Cancel \(M_s\) from both sides:
\[ \omega^2 R = \frac{G \cdot n \cdot M_s}{R^2} \]
Rearranging gives:
\[ n = \frac{\omega^2 R^3}{G M_s} \]

6. **Substitute \(\omega\) into the equation:**
\[ n = \frac{\left(\frac{2\sqrt{10}}{10^{15}}\right)^2 \cdot (10^{20})^3}{G \cdot (2 \times 10^{30})} \]
This simplifies to:
\[ n = \frac{4 \cdot 10 \cdot 10^{60}}{G \cdot (2 \times 10^{30}) \cdot 10^{30}} \]

7. **Substitute the value of \(G\):**
\[ n = \frac{40 \cdot 10^{60}}{\frac{20}{3} \times 10^{-11} \cdot 2 \times 10^{30}} \]
Simplifying further:
\[ n = \frac{40 \cdot 10^{60} \cdot 3 \cdot 10^{11}}{20 \cdot 2 \cdot 10^{30}} = \frac{120 \cdot 10^{71}}{40 \cdot 10^{30}} = 3 \cdot 10^{41} \]

8. **Compare with the given form of \(n\):**
Given \(n \approx 3 \times 10^a\), we can equate:
\[ 3 \times 10^{41} = 3 \times 10^a \]
Thus, we find:
\[ a = 41 \]

### Final Answer:
The value of \(a\) is \(41\).