Problem 22 - Algebra A Combined Function

Question

The weight of an object on or above the surface of Earth varies inversely as the square of the distance between the object and Earth's center. If a person weighs 160 pounds on Earth's surface, find the individual's weight if he moves 200 miles above Earth. Round to the nearest whole pound. (Assume that Earth's radius is 4000 miles.)

Step-by-Step Solution

Answer

The individual's weight 200 miles above Earth is 137 pounds.

Step 1: Understand the Inverse Variation Relationship

The weight of an object varies inversely with the square of the distance from the center of Earth. This means that if the distance doubles, the weight becomes a quarter of its original value. The relationship can be written as \( W = \frac{k}{d^2} \), where \( W \) is the weight, \( d \) is the distance from the center of Earth, and \( k \) is a constant.

Step 2: Calculate Distance from the Center of Earth

On Earth's surface, the distance from the center is the radius of Earth, which is given as 4000 miles. If the person moves 200 miles above the surface, total distance \( d \) from the center becomes \( 4000 + 200 = 4200 \) miles.

Step 3: Find the Constant using Initial Conditions

Initially, the person weighs 160 pounds at 4000 miles distance. Substitute \( W = 160 \) and \( d = 4000 \) into the inverse variation equation: \( 160 = \frac{k}{4000^2} \). From this, solve for \( k \): \( k = 160 \times 4000^2 \).

Step 4: Substitute New Distance to Find New Weight

Use the same equation with the new distance \( d = 4200 \) miles: \( W = \frac{k}{4200^2} \). Substitute \( k = 160 \times 4000^2 \) to find the new \( W \): \( W = \frac{160 \times 4000^2}{4200^2} \).

Step 5: Perform Calculation and Round Result

Calculate \( W = \frac{160 \times 16000000}{17640000} = 137.139 \) pounds. Round the result to the nearest whole number: 137 pounds.

Key Concepts

Weight and Distance Relationship

When dealing with physical concepts like gravity, understanding how weight changes with distance is crucial. The exercise above demonstrates the relationship between the weight of an object and its distance from Earth's center. This connection is described as an "inverse variation," meaning as one value increases, the other decreases proportionally.
For instance, if you move away from Earth's center, your weight decreases. This happens in a specific mathematical way. The formula that represents this relationship is given by:

• The weight of an object (\( W \)) inversely varies with the square of the distance (\( d \)) from the Earth's center: \( W = \frac{k}{d^2} \).
• The constant \( k \) helps maintain proportionality in real-world units.

Understanding this relationship is foundational in grasping how gravity affects objects at different altitudes.

Mathematical Modeling

Mathematical modeling is a powerful tool to translate real-world situations into mathematical language. In this exercise, we are modeling how weight changes with altitude using an inverse square relationship.

Through mathematical models, abstract mathematical formulas can represent real-life situations accurately. In our example:

• We use the formula \( W = \frac{k}{d^2} \) to model the weight of a person at different distances from Earth's center.
• This allows predictions of weight under different offshore positions, using simple algebraic expressions.

Modeling helps us make calculated guesses, allowing adjustments to variables while predicting outcomes securely and effectively.

Inverse Square Law

The inverse square law is a fundamental principle in physics and mathematics. It describes how a quantity's intensity or strength weakens as you move further from the source.

In this exercise, the inverse square law explains how quickly a person's weight decreases as they ascend above Earth's surface. Here's how it applies:

• Weight varies inversely with the square of the distance from Earth's center.
• This means if you double the distance, only one-fourth of the original weight remains.
• Mathematically, \( W = \frac{k}{d^2} \) outlines this rapid decline.

This law is essential to understand because it appears in various scientific fields, such as electromagnetism, light, sound, and gravity.

Problem-Solving in Algebra

Algebra is key to translating mathematical relationships into computed results. In this exercise, we solve the algebraic problem of determining weight at a new altitude using specific problem-solving strategies:1. **Identifying Relationships**: We first need to understand how weight and distance interact, symbolized by \( W = \frac{k}{d^2} \).
2. **Substituting Known Values**: By knowing the initial weight and distance, you can calculate the constant \( k \).
3. **Using Algebraic Techniques**: With \( k \) determined, substitute the new distance to find the new weight.
4. **Calculation and Estimation**: Finally, solve the equation, simplifying and rounding as needed to a practical weight.By applying these steps, algebra helps unravel complex relationships, making calculations straightforward and the results reliable.