To solve these problems, we'll use the principle that the weight of an object varies inversely as the square of its distance from the center of the Earth. This can be represented by the equation:

$W = \frac{k}{d^2}$

where:

• $W$ is the weight,
• $k$ is a constant,
• $d$ is the distance from the center of the Earth.

Part (a): Astronaut's Weight Above the Surface of Earth

1. Find the constant $k$:
   On the surface of the Earth, the astronaut weighs 115 pounds, and the distance from the center of the Earth is equal to the Earth's radius (3960 miles). So:

   $115 = \frac{k}{3960^2}$

   Solving for $k$:

   $k = 115 \times 3960^2$

2. Calculate the new distance from the center of the Earth:
   The astronaut is 5753 miles above the surface, so the total distance from the center is:

   $3960 + 5753 = 9713 \text{ miles}$

3. Calculate the astronaut's weight at the new distance:

   $W = \frac{k}{9713^2}$

Now, let's calculate this value.

Calculation:

Let's find $k$ first:

$k = 115 \times 3960^2$

Then use it to find the new weight:

$W = \frac{k}{9713^2}$

Let's compute the weight.The astronaut weighs approximately 19.12 pounds when she is 5,753 miles above the surface of the Earth.

Part (b): Miner's Weight Below the Surface of Earth

1. Find the constant $k$:
   On the surface of the Earth, the miner weighs 209 pounds, and the distance from the center of the Earth is 3960 miles. So:

   $209 = \frac{k}{3960^2}$

   Solving for $k$:

   $k = 209 \times 3960^2$

2. Calculate the new distance from the center of the Earth:
   The miner is 14 miles below the surface, so the total distance from the center is:

   $3960 - 14 = 3946 \text{ miles}$

3. Calculate the miner's weight at the new distance:

   $W = \frac{k}{3946^2}$

Let's compute this value.The miner weighs approximately 210.49 pounds when he is 14 miles below the surface of the Earth.

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Would you like more details on these calculations, or do you have any further questions?

Here are some related questions you might be interested in:

1. How does the weight of an object change as it moves further away from the Earth's center?
2. What is the effect of altitude on the gravitational force experienced by an object?
3. How would the weight change if the altitude or depth changed by a different amount?
4. What factors affect the weight of an object besides distance from Earth's center?
5. How do these calculations change if we consider other celestial bodies like the Moon or Mars?

Tip: When dealing with problems involving inverse variations, always remember that if one variable increases, the other decreases proportionally to the square or cube, depending on the relationship defined.