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.