Calculate the force of gravitation between the earth the sun, given that the mass of the earth `=6xx10^(24)`kg and mass of the sun `=2xx10^(30)`kg. The average distance between the two is `1.5xx10^(11)m`.

Explanation

To calculate the force of gravitation between the Earth and the Sun, we will use Newton's law of universal gravitation, which is given by the formula:

\[ F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} \]

Where:
- \( F_g \) is the gravitational force,
- \( G \) is the gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),
- \( m_1 \) is the mass of the first object (mass of the Sun),
- \( m_2 \) is the mass of the second object (mass of the Earth),
- \( r \) is the distance between the centers of the two objects.

### Step-by-Step Solution:

1. **Identify the values:**
- Mass of the Earth, \( m_2 = 6 \times 10^{24} \, \text{kg} \)
- Mass of the Sun, \( m_1 = 2 \times 10^{30} \, \text{kg} \)
- Distance between the Earth and the Sun, \( r = 1.5 \times 10^{11} \, \text{m} \)
- Gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)

2. **Substitute the values into the formula:**
\[ F_g = \frac{6.67 \times 10^{-11} \cdot (2 \times 10^{30}) \cdot (6 \times 10^{24})}{(1.5 \times 10^{11})^2} \]

3. **Calculate the denominator:**
\[ (1.5 \times 10^{11})^2 = 2.25 \times 10^{22} \, \text{m}^2 \]

4. **Calculate the numerator:**
\[ 6.67 \times 10^{-11} \cdot 2 \times 10^{30} \cdot 6 \times 10^{24} = 6.67 \cdot 2 \cdot 6 \times 10^{-11 + 30 + 24} \]
\[ = 80.04 \times 10^{43} \, \text{N m}^2/\text{kg}^2 \]

5. **Combine the results:**
\[ F_g = \frac{80.04 \times 10^{43}}{2.25 \times 10^{22}} = \frac{80.04}{2.25} \times 10^{43 - 22} \]
\[ = 35.56 \times 10^{21} \, \text{N} \]

6. **Final result:**
\[ F_g \approx 3.56 \times 10^{22} \, \text{N} \]

### Final Answer:
The gravitational force between the Earth and the Sun is approximately \( 3.56 \times 10^{22} \, \text{N} \).