Math Problem Statement

The weight (in pounds) w = f(d) of an object varies inversely as the square of its distance (in thousands of miles) d from the center of Earth. An astronaut weighs 180 pounds at sea level (about 4 thousand miles from Earth's center). Find an equation of f.

Solution

The problem you're dealing with describes a situation where the weight w=f(d)w = f(d) of an object varies inversely as the square of its distance dd from the center of Earth. This is a typical example of an inverse-square law problem.

Part (a) asks to find the equation for f(d)f(d).

Since the weight varies inversely with the square of the distance dd, we can write the relationship as:

f(d)=kd2f(d) = \frac{k}{d^2}

where kk is a constant to be determined, and dd is the distance in thousands of miles from the Earth's center.

From the given information:

  • The astronaut weighs 180 pounds at sea level.
  • The distance from the Earth's center to sea level is about 4 thousand miles (so, d=4d = 4).

We can substitute these values into the equation to solve for kk:

180=k42180 = \frac{k}{4^2}

180=k16180 = \frac{k}{16}

Multiplying both sides by 16:

k=180×16=2880k = 180 \times 16 = 2880

Thus, the equation for f(d)f(d) is:

f(d)=2880d2f(d) = \frac{2880}{d^2}

Part (b) asks how much the astronaut would weigh at 4 thousand miles above Earth's surface.

If the astronaut is 4 thousand miles above the surface, the total distance from the Earth's center would be:

d=4(thousand miles to sea level)+4(thousand miles above the surface)=8thousand milesd = 4 \, (\text{thousand miles to sea level}) + 4 \, (\text{thousand miles above the surface}) = 8 \, \text{thousand miles}

Now, use the equation f(d)=2880d2f(d) = \frac{2880}{d^2} to find the weight:

f(8)=288082=288064=45poundsf(8) = \frac{2880}{8^2} = \frac{2880}{64} = 45 \, \text{pounds}

So, the astronaut would weigh 45 pounds at 4 thousand miles above Earth's surface.

Would you like more details or have any questions?

Related questions:

  1. What happens to the astronaut’s weight if they travel 6 thousand miles above Earth’s surface?
  2. How would the astronaut's weight change if the Earth’s radius was different?
  3. What other forces could affect an astronaut's weight at different distances from the Earth?
  4. How would the equation f(d)=kd2f(d) = \frac{k}{d^2} change if the astronaut were on a different planet?
  5. How could we calculate the astronaut's weight at the International Space Station's altitude?

Tip: Always double-check units and conversions when working with physical distances, especially in scientific problems involving gravitational forces.

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Math Problem Analysis

Mathematical Concepts

Inverse Variation
Square of Distance
Algebra

Formulas

w = k / d^2

Theorems

Inverse Square Law

Suitable Grade Level

Grades 9-12