In the solar system, the Sun is in the focus of the system for Sun-earth binding system. Then the binding energy for the system will be [given that radius of the earth's orbit round the Sun is `1.5 xx 10^(11) m` and mass of the earth `= 6 xx 10^(24) kg`]

Options

A. `2.7xx10^(33)J`
B. `5.4xx10^(33)J`
C. `2.7xx10^(30)J`
D. `5.4xx10^(30)J`

Explanation

To find the binding energy of the Sun-Earth system, we can use the formula for gravitational binding energy, which is given by:

\[ E = -\frac{G \cdot M \cdot m}{r} \]

Where:
- \(E\) is the binding energy,
- \(G\) is the gravitational constant (\(6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\)),
- \(M\) is the mass of the Sun (\(1 \times 10^{30} \, \text{kg}\)),
- \(m\) is the mass of the Earth (\(6 \times 10^{24} \, \text{kg}\)),
- \(r\) is the distance between the Sun and the Earth (\(1.5 \times 10^{11} \, \text{m}\)).

### Step-by-Step Solution:

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

2. **Substitute the values into the formula**:
\[ E = -\frac{(6.67 \times 10^{-11}) \cdot (1 \times 10^{30}) \cdot (6 \times 10^{24})}{1.5 \times 10^{11}} \]

3. **Calculate the numerator**:
\[ \text{Numerator} = (6.67 \times 10^{-11}) \cdot (1 \times 10^{30}) \cdot (6 \times 10^{24}) = 6.67 \times 6 \times 10^{30 + 24 - 11} = 40.02 \times 10^{43} \]

4. **Calculate the denominator**:
\[ \text{Denominator} = 1.5 \times 10^{11} \]

5. **Now divide the numerator by the denominator**:
\[ E = -\frac{40.02 \times 10^{43}}{1.5 \times 10^{11}} = -26.68 \times 10^{32} \, \text{J} \]

6. **Convert to scientific notation**:
\[ E \approx -2.67 \times 10^{33} \, \text{J} \]

7. **Final answer**:
The binding energy for the Sun-Earth system is approximately \(2.67 \times 10^{33} \, \text{J}\).

### Conclusion:
The binding energy for the Sun-Earth system is \(2.67 \times 10^{33} \, \text{J}\).