Find the mass of the sun from the following data. The mean orbital radius of earth around the sun is `1.5 xx 10^(8) `km.

Explanation

To find the mass of the Sun using the given data, we can apply Kepler's Third Law of planetary motion, which relates the period of orbit (T) and the mean orbital radius (r). The formula derived from this law is:

\[ T^2 = \frac{4 \pi^2 r^3}{G M} \]

Where:
- \( T \) is the orbital period of the Earth,
- \( r \) is the mean orbital radius,
- \( G \) is the gravitational constant, and
- \( M \) is the mass of the Sun.

We can rearrange this formula to solve for the mass of the Sun \( M \):

\[ M = \frac{4 \pi^2 r^3}{G T^2} \]

### Step 1: Convert the orbital radius from kilometers to meters
The mean orbital radius of the Earth around the Sun is given as \( 1.5 \times 10^8 \) km. We need to convert this to meters:

\[ r = 1.5 \times 10^8 \text{ km} \times 10^3 \text{ m/km} = 1.5 \times 10^{11} \text{ m} \]

### Step 2: Determine the orbital period \( T \)
The time period \( T \) for one complete orbit of the Earth around the Sun is 1 year. In seconds, this can be calculated as follows:

\[ T = 365.25 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} = 3.156 \times 10^7 \text{ seconds} \]

### Step 3: Use the gravitational constant \( G \)
The gravitational constant \( G \) is given as:

\[ G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \]

### Step 4: Substitute the values into the mass formula
Now we can substitute \( r \), \( T \), and \( G \) into the formula for \( M \):

\[ M = \frac{4 \pi^2 (1.5 \times 10^{11})^3}{(3.156 \times 10^7)^2 \times (6.67 \times 10^{-11})} \]

### Step 5: Calculate the mass of the Sun
Calculating the values step by step:

1. Calculate \( r^3 \):
\[ (1.5 \times 10^{11})^3 = 3.375 \times 10^{33} \text{ m}^3 \]

2. Calculate \( T^2 \):
\[ (3.156 \times 10^7)^2 = 9.973 \times 10^{14} \text{ s}^2 \]

3. Calculate \( 4 \pi^2 \):
\[ 4 \pi^2 \approx 39.478 \]

4. Now substitute these values into the mass equation:
\[ M = \frac{39.478 \times 3.375 \times 10^{33}}{9.973 \times 10^{14} \times 6.67 \times 10^{-11}} \]

5. Calculate the denominator:
\[ 9.973 \times 10^{14} \times 6.67 \times 10^{-11} \approx 6.671 \times 10^{4} \]

6. Finally, calculate \( M \):
\[ M \approx \frac{133.4 \times 10^{33}}{6.671 \times 10^{4}} \approx 2.006 \times 10^{30} \text{ kg} \]

Thus, the mass of the Sun is approximately:

\[ M \approx 2.006 \times 10^{30} \text{ kg} \]