To calculate the Moon's average orbital velocity, we can use the formula:

$v = \frac{2 \pi r}{T}$

where:

• $r = 384,400 \, \text{km} = 384,400,000 \, \text{m}$
• $T = 27.3 \, \text{days} = 27.3 \times 24 \times 3600 \, \text{seconds}$

First, let’s convert $T$ into seconds:

$T = 27.3 \times 24 \times 3600 = 2,358,720 \, \text{seconds}$

Now, we can substitute $r$ and $T$ into the formula:

$v = \frac{2 \pi \times 384,400,000}{2,358,720}$

Calculating this:

1. $2 \pi \times 384,400,000 \approx 2,414,220,000$
2. $\frac{2,414,220,000}{2,358,720} \approx 1023.78 \, \text{m/s}$

So, the Moon's average orbital velocity is approximately:

$\boxed{1024 \, \text{m/s}}$

Answer:

B. 1024 m/s

Would you like more details or have any questions?

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Related Questions:

1. How does the Moon's orbital velocity compare to the Earth's orbital velocity around the Sun?
2. How would the Moon's orbital velocity change if it were farther from the Earth?
3. Why is it important to convert days into seconds in calculations like these?
4. How does the gravitational force between Earth and the Moon relate to the Moon’s orbital velocity?
5. What effect does the Moon’s orbital velocity have on tidal forces on Earth?

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Tip: Always ensure units are consistent, especially when working with large distances and time periods in orbital mechanics.