A satellite of mass 200 kg revolves around of mass ` 5 xx 10^(30)` kg In a circular orbit of radius `6.6 xx10^(6)` m . Binding energy of the satellite is

Options

A. `5xx10^(15)`J
B. `-5 xx 10^(15)` J
C. `10^(16) J `
D. `- 10^(16)J`

Answer

To find the binding energy of the satellite, we can use the formula for the binding energy \( U \) of a satellite in a gravitational field:

\[ U = -\frac{GMm}{2R} \]

Where:
- \( G \) is the gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M \) is the mass of the planet (or the central body), which is \( 5 \times 10^{30} \, \text{kg} \)
- \( m \) is the mass of the satellite, which is \( 200 \, \text{kg} \)
- \( R \) is the radius of the orbit, which is \( 6.6 \times 10^{6} \, \text{m} \)

### Step 1: Substitute the values into the formula

Substituting the known values into the formula for binding energy:

\[ U = -\frac{(6.67 \times 10^{-11}) \times (5 \times 10^{30}) \times (200)}{2 \times (6.6 \times 10^{6})} \]

### Step 2: Calculate the numerator

First, calculate the numerator:

\[ GMm = (6.67 \times 10^{-11}) \times (5 \times 10^{30}) \times (200) \]

Calculating this step-by-step:

1. Calculate \( 6.67 \times 5 = 33.35 \)
2. Then, \( 33.35 \times 200 = 6670 \)
3. Finally, multiply by \( 10^{-11} \times 10^{30} = 10^{19} \)

So, the numerator becomes:

\[ 6670 \times 10^{19} = 6.67 \times 10^{22} \]

### Step 3: Calculate the denominator

Now calculate the denominator:

\[ 2R = 2 \times (6.6 \times 10^{6}) = 13.2 \times 10^{6} \]

### Step 4: Combine the results

Now substitute the numerator and denominator back into the binding energy formula:

\[ U = -\frac{6.67 \times 10^{22}}{13.2 \times 10^{6}} \]

### Step 5: Perform the division

Calculating the division:

\[ \frac{6.67}{13.2} \approx 0.505 \]

And for the powers of ten:

\[ 10^{22} / 10^{6} = 10^{16} \]

So, we have:

\[ U \approx -0.505 \times 10^{16} \]

### Step 6: Final binding energy

Since binding energy is defined as a positive quantity, we take the absolute value:

\[ U \approx 5.05 \times 10^{15} \, \text{J} \]

### Conclusion

The binding energy of the satellite is approximately:

\[ U \approx 5 \times 10^{15} \, \text{J} \]