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The problem provided relates the astronaut's weight \( w \) as a function of her height \( h \) above the Earth's surface, given by the equation:

\[
w = w_0 \left( \frac{3960}{3960 + h} \right)^2
\]

Where:
- \( w_0 \) is the astronaut's weight on the Earth's surface (130 pounds in this case),
- \( h \) is the height above the Earth's surface in miles.

We are tasked with finding the rate at which her weight is changing, \( \frac{dw}{dt} \), when she is 40 miles above the Earth's surface, and she is moving upward at a speed of 12 miles per second. This can be approached using the chain rule.

### Step-by-step approach:

1. **Differentiate the weight function** with respect to time \( t \). By the chain rule:

\[
\frac{dw}{dt} = \frac{dw}{dh} \cdot \frac{dh}{dt}
\]

We know that \( \frac{dh}{dt} = 12 \, \text{mi/s} \) (the speed at which she is moving upward).

2. **Differentiate \( w \) with respect to \( h \)**:

\[
w = w_0 \left( \frac{3960}{3960 + h} \right)^2
\]

To differentiate this, use the power rule and chain rule:

\[
\frac{dw}{dh} = w_0 \cdot 2 \left( \frac{3960}{3960 + h} \right) \cdot \left( \frac{-3960}{(3960 + h)^2} \right)
\]

Simplifying:

\[
\frac{dw}{dh} = w_0 \cdot 2 \cdot \left( \frac{3960}{3960 + h} \right)^3 \cdot (-1)
\]

3. **Substitute values**:
   - \( w_0 = 130 \) pounds,
   - \( h = 40 \) miles,
   - \( \frac{dh}{dt} = 12 \, \text{mi/s} \).

   At \( h = 40 \):

\[
\frac{dw}{dh} = 130 \cdot 2 \cdot \left( \frac{3960}{3960 + 40} \right)^3 \cdot (-1)
= 130 \cdot 2 \cdot \left( \frac{3960}{4000} \right)^3 \cdot (-1)
\]

Simplifying the fraction:

\[
\frac{3960}{4000} = 0.99
\]

\[
\left( 0.99 \right)^3 = 0.970299
\]

Thus:

\[
\frac{dw}{dh} = 130 \cdot 2 \cdot (-0.970299) = -252.277
\]

4. **Calculate \( \frac{dw}{dt} \)**:

\[
\frac{dw}{dt} = -252.277 \cdot 12 = -3027.33 \, \text{lb/s}
\]

So, the astronaut's weight is decreasing at a rate of approximately **3027.33 pounds per second** when she is 40 miles above the Earth's surface, moving upward at 12 miles per second.

Would you like more details or have any questions?

### Related Questions:
1. How does the astronaut's weight change as she moves further away from Earth?
2. How would the result differ if the rocket speed increased to 15 miles per second?
3. What is the astronaut's weight at 100 miles above the Earth's surface?
4. How does the gravitational force formula relate to this weight equation?
5. What happens to the rate of weight loss as height \( h \) approaches infinity?

**Tip:** Always double-check units (miles, pounds, seconds) when applying calculus to physics problems to avoid errors!

The weight w of an astronaut (in pounds) is related to her height h above the surface of the earth (in miles) by a given formula. If the astronaut weighs 130 pounds on Earth and is in a rocket moving upward at 12 miles per second, find the rate at which her weight is changing in pounds per second when she is 40 miles above the Earth's surface.

Explore how to calculate the rate of change of an astronaut's weight as they move away from Earth's surface at 12 miles per second. This problem involves related rates, derivatives, and the physics of gravity, with a detailed step-by-step solution.

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