Calculat the binding energy of the - sum system . Mass of the earth `=6xx 10 ^(24) ` kg , mass of the sun =` 2 xx 10^(30)` kg, distance between the earth and the sun =` 1.5 xx 10^(11)` and gravitational constant = ` 6.6 xx 10 ^(-11)` `N m^(2) kg^(2)`

Options

A. `8.8 xx 10^(10)J`
B. `8.8 xx 10^3 J`
C. `5.2 xx 10^33 J`
D. `2.6 xx 10^33 J`

Answer

To calculate the binding energy of the Earth-Sun system, we can use the formula for gravitational potential energy, which is also the binding energy in this context:

\[ U = -\frac{G \cdot M \cdot m}{R} \]

where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Sun,
- \( m \) is the mass of the Earth,
- \( R \) is the distance between the Earth and the Sun.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the Earth, \( m = 6 \times 10^{24} \) kg
- Mass of the Sun, \( M = 2 \times 10^{30} \) kg
- Distance between Earth and Sun, \( R = 1.5 \times 10^{11} \) m
- Gravitational constant, \( G = 6.6 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)

2. **Substitute the Values into the Formula:**
\[ U = -\frac{(6.6 \times 10^{-11}) \cdot (2 \times 10^{30}) \cdot (6 \times 10^{24})}{1.5 \times 10^{11}} \]

3. **Calculate the Numerator:**
- First, calculate \( G \cdot M \cdot m \):
\[ G \cdot M \cdot m = 6.6 \times 10^{-11} \cdot 2 \times 10^{30} \cdot 6 \times 10^{24} \]
- Calculate \( 2 \cdot 6 = 12 \):
\[ 6.6 \times 10^{-11} \cdot 12 \times 10^{54} = 79.2 \times 10^{54} = 7.92 \times 10^{55} \]

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

5. **Final Calculation of Binding Energy:**
\[ U = -\frac{7.92 \times 10^{55}}{1.5 \times 10^{11}} = -5.28 \times 10^{44} \, \text{J} \]

6. **Convert to Positive Binding Energy:**
- The binding energy is taken as a positive value, so:
\[ \text{Binding Energy} = 5.28 \times 10^{44} \, \text{J} \]

### Final Answer:
The binding energy of the Earth-Sun system is approximately \( 5.28 \times 10^{44} \) Joules.