Let's break down and solve the problem shown in the image.

Given Information:

• The weight $w$ (in pounds) of an object varies inversely as the square of its distance $d$ (in thousands of miles) from the center of the Earth.

• Inverse variation formula: $w = \frac{k}{d^2}$ where $k$ is a constant.

• The astronaut weighs 180 pounds at sea level (4,000 miles from Earth's center).

Part (a): Find the equation for $f(d)$.

We use the formula $w = \frac{k}{d^2}$ and the fact that the astronaut weighs 180 pounds at $d = 4$ (since 4,000 miles is 4 in thousands of miles).

1. Plug in the known values: $180 = \frac{k}{4^2}$
2. Solve for $k$: $180 = \frac{k}{16} \implies k = 180 \times 16 = 2880$

Thus, the equation for $f(d)$ is: $f(d) = \frac{2880}{d^2}$

Part (b): Weight of the astronaut at 4,000 miles above Earth's surface.

• Total distance from Earth's center: $d = 4 + 4 = 8$ (since the astronaut is 4,000 miles above the surface).

Now, use the equation $f(d) = \frac{2880}{d^2}$: $w = \frac{2880}{8^2} = \frac{2880}{64} = 45 \, \text{pounds}$

Part (c): At what distance from Earth's center would the astronaut weigh 1 pound?

• Set $w = 1$ and solve for $d$: $1 = \frac{2880}{d^2} \implies d^2 = 2880 \implies d = \sqrt{2880} \approx 53.66$

So, the astronaut would weigh 1 pound at a distance of approximately 54,000 miles from Earth's center.

Part (d): Weight of the astronaut on the surface of the Moon (239,000 miles from Earth).

• Use the equation again with $d = 239$: $w = \frac{2880}{239^2} = \frac{2880}{57121} \approx 0.0504 \, \text{pounds}$

Thus, the astronaut would weigh approximately 0.050 pounds on the surface of the Moon.

Would you like further details on any step or clarification?

Here are five related questions you can explore:

1. How is inverse variation used in real-world problems involving gravity?
2. What is the effect on weight if the distance is doubled?
3. How would this equation change for other celestial bodies?
4. What other factors, besides distance, can affect an astronaut’s weight in space?
5. How is the constant $k$ derived in different planetary systems?

Tip: In inverse variation problems, as the distance increases, the weight decreases drastically due to the square of the distance.