To find the Moon's average orbital velocity, we can use the given formula for orbital velocity:

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

where:

• $r$ is the average distance between the Earth and the Moon, $r = 384,400 \ \text{km}$.
• $T$ is the orbital period, $T = 27.3 \ \text{days}$.

Step 1: Convert Units

To ensure consistent units, we need to convert $r$ to meters and $T$ to seconds.

1. Convert $r$ to meters: $r = 384,400 \ \text{km} = 384,400,000 \ \text{m}$

2. Convert $T$ to seconds: $T = 27.3 \ \text{days} \times 24 \ \text{hours/day} \times 3600 \ \text{seconds/hour} = 2,358,720 \ \text{seconds}$

Step 2: Substitute Values and Calculate $v$

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

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

Calculating this gives:

$v \approx 1023.78 \ \text{m/s}$

Answer

Rounding to the nearest whole number, the Moon's average orbital velocity is approximately:

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

So, the correct answer is B. 1024 m/s.

Would you like further details on any of the steps?

Here are five related questions for deeper understanding:

1. What is the difference between average orbital speed and instantaneous orbital speed?
2. How would the Moon's velocity change if its distance from Earth increased?
3. How would we calculate the orbital speed of a satellite in a lower Earth orbit?
4. What role does gravitational force play in maintaining the Moon's orbit?
5. How can we calculate the orbital period if the speed and radius are known?

Tip: When working with orbital mechanics, always make sure units are consistent, especially when converting days to seconds or kilometers to meters.