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The problem provided involves calculating the escape velocity from the Moon using a given formula. The formula provided is:

\[
v = 5{,}600 \left(\frac{d}{1{,}000}\right)^{-\frac{1}{2}}
\]

Where:
- \( v \) is the escape velocity in miles per hour.
- \( d \) is the distance from the center of the Moon in miles.

We are given:
- \( d = 16{,}000 \) miles (the distance from the center of the Moon).

We need to find the escape velocity \( v \).

### Step-by-Step Solution

1. **Substitute the value of \( d \) into the formula:**

   \[
   v = 5{,}600 \left(\frac{16{,}000}{1{,}000}\right)^{-\frac{1}{2}}
   \]

2. **Simplify the fraction inside the parentheses:**

   \[
   v = 5{,}600 \left(\frac{16{,}000}{1{,}000}\right)^{-\frac{1}{2}} = 5{,}600 \left(16\right)^{-\frac{1}{2}}
   \]

3. **Evaluate the exponent:**

   \[
   16^{-\frac{1}{2}} = \frac{1}{\sqrt{16}} = \frac{1}{4}
   \]

4. **Substitute back into the equation:**

   \[
   v = 5{,}600 \times \frac{1}{4} = 5{,}600 \times 0.25
   \]

5. **Calculate the final value:**

   \[
   v = 1{,}400 \text{ miles per hour}
   \]

So, the object has to be moving **1,400 miles per hour** to escape the Moon's gravity.

Would you like more details, or do you have any questions?

Here are 5 related questions for further exploration:

1. What happens to the escape velocity if the distance \( d \) is doubled?
2. How does the escape velocity formula change for a different celestial body, like Earth?
3. What is the physical significance of the exponent \(-\frac{1}{2}\) in the formula?
4. How does the escape velocity vary with respect to the mass of the celestial body?
5. What is the escape velocity from the surface of the Moon, and how is it derived?

**Tip:** Always make sure to carefully simplify and evaluate the mathematical expressions step by step to avoid errors.

Escape velocity is a measure of how fast an object must be moving to escape the gravitational pull of a planet or moon with no further thrust. The escape velocity for the moon is given approximately by the equation v = 5600 * (d/1000)^(-1/2), where v is the escape velocity in miles per hour and d is the distance from the center of the moon (in miles). If a lunar lander thrusts upwards until it reaches a distance of 16,000 miles from the center of the moon, about how fast must it be going to escape the moon's gravity?

Learn how to calculate the escape velocity for a lunar lander using the formula v = 5600 * (d/1000)^(-1/2). This problem explores the application of algebraic concepts such as rational exponents and radicals, making it suitable for high school students in Grades 9-12.

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