A `10kg` satellite circles earth once every `2hr` in an orbit having a radius of `8000km`. Assuming that Bohr's angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite.

Options

A. `5.3xx10^40`
B. `5.3xx10^45`
C. `7.8xx10^48`
D. `7.8xx10^50`

Explanation

To find the quantum number of the orbit of a satellite using Bohr's angular momentum postulate, we can follow these steps:

### Step 1: Understand the Angular Momentum Postulate
According to Bohr's postulate, the angular momentum \( L \) of an orbiting body is quantized and given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is the quantum number, and \( h \) is Planck's constant.

### Step 2: Write the Expression for Angular Momentum
For a satellite of mass \( m \) moving in a circular orbit of radius \( r \) with velocity \( v \), the angular momentum can also be expressed as:
\[ L = mvr \]
Setting the two expressions for angular momentum equal gives:
\[ mvr = n \frac{h}{2\pi} \]

### Step 3: Rearranging to Find Quantum Number \( n \)
We can rearrange the equation to solve for \( n \):
\[ n = \frac{mvr \cdot 2\pi}{h} \]

### Step 4: Substitute Known Values
Given:
- Mass of the satellite \( m = 10 \, \text{kg} \)
- Radius of the orbit \( r = 8000 \, \text{km} = 8 \times 10^6 \, \text{m} \)
- Time for one revolution \( T = 2 \, \text{hours} = 2 \times 3600 \, \text{s} = 7200 \, \text{s} \)

### Step 5: Calculate the Velocity \( v \)
The velocity \( v \) of the satellite can be calculated using the formula:
\[ v = \frac{2\pi r}{T} \]
Substituting the values:
\[ v = \frac{2\pi \times 8 \times 10^6}{7200} \]
Calculating this gives:
\[ v \approx 7 \times 10^3 \, \text{m/s} \]

### Step 6: Substitute All Values into the Quantum Number Equation
Now we can substitute \( m \), \( v \), \( r \), and \( h \) (where \( h \approx 6.626 \times 10^{-34} \, \text{Js} \)):
\[ n = \frac{10 \times (7 \times 10^3) \times (8 \times 10^6) \cdot 2\pi}{6.626 \times 10^{-34}} \]

### Step 7: Calculate \( n \)
Performing the calculations:
1. Calculate \( 10 \times 7 \times 10^3 \times 8 \times 10^6 \):
\[ = 560 \times 10^9 = 5.6 \times 10^{11} \]
2. Multiply by \( 2\pi \):
\[ = 5.6 \times 10^{11} \times 6.2832 \approx 3.51 \times 10^{12} \]
3. Finally, divide by \( 6.626 \times 10^{-34} \):
\[ n \approx \frac{3.51 \times 10^{12}}{6.626 \times 10^{-34}} \approx 5.3 \times 10^{45} \]

### Final Answer
Thus, the quantum number \( n \) for the orbit of the satellite is approximately:
\[ n \approx 5.3 \times 10^{45} \]