We are given the formula for the astronaut's weight as a function of her height above the Earth's surface:

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

where:

• $w_0 = 135$ pounds (the astronaut's weight on Earth),
• $h$ is the height above the Earth's surface (in miles),
• $w(h)$ is the astronaut's weight at height $h$.

We are tasked with finding the rate of change of the astronaut's weight with respect to time when she is 40 miles above the Earth's surface and traveling upward at a speed of 11 miles per second.

Step 1: Differentiate the weight function with respect to time

To find how the astronaut's weight is changing with time, we need to differentiate $w(h)$ with respect to time $t$, using the chain rule:

$\frac{dw}{dt} = \frac{dw}{dh} \cdot \frac{dh}{dt}$

We are given that $\frac{dh}{dt} = 11 \, \text{mi/s}$, which represents the rate at which the astronaut is gaining height. Now, we need to calculate $\frac{dw}{dh}$.

Step 2: Differentiate the weight function with respect to $h$

We start by differentiating:

$w(h) = 135 \left( \frac{3960}{3960 + h} \right)^2$

Let’s apply the chain rule to differentiate this expression with respect to $h$:

$\frac{dw}{dh} = 135 \cdot 2 \cdot \left( \frac{3960}{3960 + h} \right) \cdot \left( \frac{-3960}{(3960 + h)^2} \right)$

Simplifying:

$\frac{dw}{dh} = -270 \cdot \frac{3960^2}{(3960 + h)^3}$

Step 3: Substitute $h = 40$ miles

Now, substitute $h = 40$ into the equation for $\frac{dw}{dh}$:

$\frac{dw}{dh} = -270 \cdot \frac{3960^2}{(3960 + 40)^3} = -270 \cdot \frac{3960^2}{4000^3}$

First, calculate the powers:

$3960^2 = 15,681,600 \quad \text{and} \quad 4000^3 = 64,000,000,000$

Thus,

$\frac{dw}{dh} = -270 \cdot \frac{15,681,600}{64,000,000,000} = -270 \cdot 0.000245025 = -0.06615675$

Step 4: Multiply by the rate of change of height

Now, multiply by $\frac{dh}{dt} = 11 \, \text{mi/s}$:

$\frac{dw}{dt} = -0.06615675 \times 11 = -0.7277 \, \text{lb/s}$

Final Answer

The rate at which the astronaut's weight is changing when she is 40 miles above the Earth's surface is approximately:

$\boxed{-0.7277 \, \text{lb/s}}$

This means that her weight is decreasing at a rate of 0.7277 pounds per second.

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Would you like any further details or clarification on any part of the calculation?

Here are 5 related questions:

1. How would the astronaut's weight change if the altitude increased to 100 miles?
2. What happens to the astronaut's weight as the altitude approaches infinity?
3. How does the speed of the rocket affect the rate of weight change?
4. What is the astronaut's weight at 40 miles above Earth?
5. How does the formula for weight differ at lower altitudes?

Tip: The gravitational force decreases rapidly with height, but not linearly—it’s inversely proportional to the square of the distance from the center of the Earth.