What is the escape velocity of a body from the solar system? Calculate using the following data. Mass of the sun `= 2xx 10^(30) kg`. The distance between earth and the sun `= 1.5 xx 10^(11) m`.

Explanation

To calculate the escape velocity of a body from the solar system, we can use the formula for escape velocity given by:

\[ v_e = \sqrt{\frac{2GM}{R}} \]

Where:
- \( v_e \) is the escape velocity,
- \( G \) is the universal gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),
- \( M \) is the mass of the celestial body (in this case, the Sun),
- \( R \) is the distance from the center of the celestial body to the point of escape (in this case, the distance from the Earth to the Sun).

### Step-by-Step Solution:

1. **Identify the values:**
- Mass of the Sun, \( M = 2 \times 10^{30} \, \text{kg} \)
- Distance from the Earth to 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 escape velocity formula:**
\[ v_e = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (2 \times 10^{30})}{1.5 \times 10^{11}}} \]

3. **Calculate the numerator:**
- First, calculate \( 2 \times G \):
\[ 2 \times G = 2 \times (6.67 \times 10^{-11}) = 1.334 \times 10^{-10} \, \text{N m}^2/\text{kg}^2 \]
- Now, calculate \( 2 \times G \times M \):
\[ 1.334 \times 10^{-10} \times (2 \times 10^{30}) = 2.668 \times 10^{20} \, \text{N m}^2 \]

4. **Calculate the full expression under the square root:**
\[ \frac{2.668 \times 10^{20}}{1.5 \times 10^{11}} = 1.77867 \times 10^{9} \]

5. **Take the square root to find the escape velocity:**
\[ v_e = \sqrt{1.77867 \times 10^{9}} \approx 42.17 \times 10^{3} \, \text{m/s} = 42.17 \, \text{km/s} \]

### Final Answer:
The escape velocity of a body from the solar system is approximately \( 42.17 \, \text{km/s} \).