A 1500 kg satellite orbits a planet in a circular orbit of radius `6.2 xx 10^(6)` m. What is the angular momentum of the satellite in its orbit around the planet if the satellite completes one orbit every `1.5 xx 10^(4)` s?

Options

A. `3.9 xx 10^(6) kg. m//s `
B. `2.4 xx 10^(13) kg. m//s `
C. `1.4 xx 10^(14) kg. m//s `
D. `8.1 xx 10^(11) kg. m//s `

Answer

To find the angular momentum of the satellite in its orbit around the planet, we can follow these steps:

### Step 1: Understand the formula for angular momentum
The angular momentum \( L \) of an object moving in a circular path can be expressed as:
\[ L = mvr \]
where:
- \( m \) = mass of the satellite
- \( v \) = linear velocity of the satellite
- \( r \) = radius of the circular orbit

### Step 2: Find the linear velocity \( v \)
The linear velocity \( v \) of the satellite can be calculated using the formula:
\[ v = \frac{2\pi r}{T} \]
where:
- \( T \) = time period of one complete orbit

### Step 3: Substitute the values
Given:
- Mass \( m = 1500 \, \text{kg} \)
- Radius \( r = 6.2 \times 10^6 \, \text{m} \)
- Time period \( T = 1.5 \times 10^4 \, \text{s} \)

First, calculate \( v \):
\[ v = \frac{2\pi (6.2 \times 10^6)}{1.5 \times 10^4} \]

### Step 4: Calculate \( v \)
Calculating \( v \):
\[ v = \frac{2 \times 3.14 \times 6.2 \times 10^6}{1.5 \times 10^4} \approx \frac{3.884 \times 10^7}{1.5 \times 10^4} \approx 2585.33 \, \text{m/s} \]

### Step 5: Calculate the angular momentum \( L \)
Now substitute \( v \) back into the angular momentum formula:
\[ L = mvr = 1500 \times 2585.33 \times (6.2 \times 10^6) \]

### Step 6: Compute \( L \)
Calculating \( L \):
\[ L = 1500 \times 2585.33 \times 6.2 \times 10^6 \]
\[ L \approx 1500 \times 2585.33 \times 6.2 \times 10^6 \approx 2.4 \times 10^{13} \, \text{kg m}^2/\text{s} \]

### Final Answer
The angular momentum of the satellite in its orbit around the planet is approximately:
\[ L \approx 2.4 \times 10^{13} \, \text{kg m}^2/\text{s} \]

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