The weight $W$ of an object varies inversely with the square of the distance $d$ from the center of the Earth. This relationship can be expressed as:

$W \propto \frac{1}{d^2}$

or in equation form:

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

where $k$ is a constant.

Step 1: Find the constant $k$

We know the weight $W$ is 50 pounds when the distance $d$ is 3,960 miles.

$50 = \frac{k}{3960^2}$

Solving for $k$:

$k = 50 \times 3960^2$

$k = 50 \times 15,681,600$

$k = 784,080,000$

Step 2: Calculate the weight at 4,040 miles

Now we use the constant $k$ to find the new weight $W$ when $d = 4040$ miles:

$W = \frac{784,080,000}{4040^2}$

Calculating $4040^2$:

$4040^2 = 16,320,400$

Now substitute back:

$W = \frac{784,080,000}{16,320,400}$

$W \approx 48.04$

Therefore, the weight of the object at 4,040 miles from the Earth's center would be approximately 48.04 pounds.

Would you like further details, or do you have any questions?

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Related Questions

1. How would the weight change if the object were placed 5,000 miles from Earth’s center?
2. How does the inverse square law affect the gravitational force experienced by objects at various distances from the Earth?
3. How is the constant $k$ affected if we change the weight at the initial distance?
4. What is the weight of the object if it’s placed at twice the original distance from the Earth’s center?
5. How can the concept of inverse square law be applied to fields other than gravity, like light intensity?

Tip:

Remember that inverse square relationships are common in physics and describe how a quantity diminishes quickly as you move away from the source!