Problem 97 - A Graphical Approach to College Algebra

Question

Solve each problem. The weight of an object varies inversely with the square of its distance from the center of Earth. The radius of Earth is approximately 4000 miles. If a person weighs 160 pounds on Earth's surface, what would this individual weigh 8000 miles above the surface of Earth?

Step-by-Step Solution

Answer

The individual would weigh approximately 17.78 pounds 8000 miles above Earth's surface.

Step 1: Understand the Inverse Variation

The weight of an object varies inversely with the square of its distance from the center of the Earth. This can be expressed by the formula \[ W = k \times \frac{1}{d^2} \]where \(W\) is the weight, \(k\) is a constant, and \(d\) is the distance from the center of the Earth.

Step 2: Calculate the Distance from the Center of the Earth

The person is 8000 miles above the Earth's surface. Adding the Earth's radius (4000 miles), their total distance from the center of the Earth is: \[ d = 8000 + 4000 = 12000 \text{ miles} \]

Step 3: Determine the Constant of Variation

On the Earth's surface, the distance \(d\) is 4000 miles, and the person's weight \(W\) is 160 pounds. Use these values to find \(k\).\[ 160 = k \times \frac{1}{4000^2} \]Solving for \(k\), we get:\[ k = 160 \times 4000^2 \]

Step 4: Solve for Weight at the New Distance

Use the constant \(k\) to find the new weight at 12000 miles distance.\[ W = k \times \frac{1}{12000^2} \]Plugging in the value of \(k\):\[ W = (160 \times 4000^2) \times \frac{1}{12000^2} \]Simplifying, we calculate:\[ W = 160 \times \left(\frac{4000}{12000}\right)^2 \]\[ W = 160 \times \left(\frac{1}{3}\right)^2 \]\[ W = 160 \times \frac{1}{9} \]\[ W \approx 17.78 \text{ pounds} \]

Step 5: Confirm Calculation and Result

Check the mathematical operations and verify the breakdown logic to ensure the solution correctly employs inverse variation. Everything appears consistent and calculated appropriately.

Key Concepts

Distance from the Center of Earth

The concept of distance from the center of the Earth is crucial because it helps determine how forces like gravity act on an object. When calculating the distance for problems of inverse variation, you must include the radius of the Earth if the distance above the Earth's surface is given. For instance, if someone is 8000 miles above the Earth's surface, you add that to the Earth's radius (4000 miles) to find the total distance from the center of the Earth. Therefore, the total distance is 12000 miles from the center, combining the Earth's radius with the distance above it. This concept ties back to how gravity decreases with distance, according to the inverse square law.

Weight Calculation

Weight calculation in such problems involves understanding how weight is affected by the distance from the center of the Earth. The further you are from the Earth, the less you weigh, due to the inverse square law of gravity. Using the formula \[ W = k \times \frac{1}{d^2} \] we see how weight \(W\) is inversely proportional to the square of the distance \(d\). Start by determining the constant \(k\) using known values, such as the weight at the Earth's surface and then compute for the new distance. This iterative approach allows you to find the correct weight when you are at a different point above the surface.

Mathematical Formulas

Mathematical formulas are integral to solving inverse variation problems. The primary formula in this context is \( W = k \times \frac{1}{d^2} \), where weight \(W\), constant \(k\), and distance \(d\) interact to describe how weight changes with distance.

• To find \(k\), use a situation where both \(W\) and \(d\) are known, such as a person's weight at the Earth's surface.
• For example, you solve \( 160 = k \times \frac{1}{4000^2} \) to get \( k = 160 \times 4000^2 \).
• Then, apply \(k\) to calculate new weights at different distances by substituting new \(d\) values. Understanding this formula and each component's role ensures accurate calculations.

Constant of Variation

The constant of variation \(k\) is a significant part of solving inverse variation problems. It allows calculations of weight at different distances by acting as a scale factor. Given an initial situation—like the weight at Earth's surface—you can determine \(k\) with that known weight and distance. The formula\[ 160 = k \times \frac{1}{4000^2} \]helps isolate \(k\) to reflect the relationship where the weight is known. Solving gives \( k = 160 \times 4000^2 \). This constant holds for any further calculations along this line of variation and is used in subsequent steps to determine weight at another distance.