Problem 55 - College Algebra

Question

For the following exercises, use the given information to answer the questions. The weight of an object above the surface of the Earth varies inversely with the square of the distance from the center of the Earth. If a body weighs 50 pounds when it is 3960 miles from Earth's center, what would it weigh it were 3970 miles from Earth's center?

Step-by-Step Solution

Answer

Approximately 49.75 pounds.

Step 1: Understanding the Relationship

The problem states that the weight of an object varies inversely with the square of the distance from the center of the Earth. This means we can model the situation with the formula \( W = \frac{k}{d^2} \), where \( W \) is the weight of the object, \( k \) is a constant, and \( d \) is the distance from the Earth's center.

Step 2: Determine the Constant k

Use the given information that the weight is 50 pounds at 3960 miles to find \( k \). Substitute \( W = 50 \) and \( d = 3960 \) into the formula to get \( 50 = \frac{k}{3960^2} \). Solving for \( k \) gives \( k = 50 \times 3960^2 \). Calculate \( k \) to find its value.

Step 3: Calculate the New Weight

Now that we know \( k \), we can find the new weight when the distance is 3970 miles. Use the formula \( W = \frac{k}{d^2} \) with \( d = 3970 \). Substitute \( k \) and \( d \) to compute \( W = \frac{50 \times 3960^2}{3970^2} \). Simplify this expression to find the final weight at 3970 miles.

Key Concepts

Distance from Center of Earth

The distance from the center of the Earth plays a crucial role in understanding how objects are affected by gravitational pull. Each object is subjected to Earth’s gravitational force, which decreases as the object moves further away from Earth's center. It's important to realize that this distance, often denoted by the variable \( d \), is the straight line connecting the Earth's center and the object at a certain height above Earth’s surface. In various scientific problems and real-world scenarios, knowing the precise distance from Earth’s center helps to calculate gravitational effects, such as satellite orbits or space travel calculations. In this exercise, recognizing the distance of 3960 miles not only helps to solve the problem but also gives context to the problem of the variation in weight due to this distance.

• Consider Earth’s average radius, roughly 3959 miles, as a reference point.
• The distance value is always considered ‘from the center’ not just from the surface.
• Changes in this distance directly influence weights and gravitational force calculations.

Understanding this concept is foundational to grasping how weight and other similar variables change with distance.

Weight Calculation

Calculating weight when considering inverse variation involves using a specific mathematical model: \( W = \frac{k}{d^2} \). Here, \( W \) represents the weight, and \( d \) is the distance from the Earth's center. The formula indicates that as the distance \( d \) increases, the weight \( W \) decreases, showing the inverse relationship.To solve for weight in the given exercise, we substitute specific values for \( d \) into this model. At 3960 miles, the object's weight is known to be 50 pounds. Using this condition allows calculation of \( k \), the constant of variation, and subsequently any other value of \( W \) when the distance changes.

• Substituting \( d = 3960 \) gives us an initial condition: \( 50 = \frac{k}{3960^2} \).
• Solving for the constant helps us establish a benchmark to predict weight at new distances.
• The new weight at 3970 miles is calculated by substituting \( d = 3970 \) into the formula.

This calculation is not just a mathematical action, but it provides insight into how gravity affects objects over varying Earth distances.

Constant of Variation

The constant of variation, denoted as \( k \), is pivotal in modeling relationships concerning inverse variation situations such as weight in relation to distance from Earth’s center. It acts as a stabilizing factor or baseline that reflects the fixed relationship governing variables under consideration.In instances of inverse variation, the constant \( k \) is not always intuitively obvious but is crucial for making accurate predictions or calculations.

• In our exercise, determine \( k \) using 50 pounds at 3960 miles, resulting in \( k = 50 \times 3960^2 \).
• This constant ensures that every change in distance correctly predicts the corresponding weight change.
• Once \( k \) is found, it remains the same across different calculations involving the same objects and conditions.

Understanding \( k \) is essential for any scenario where quantities vary inversely, providing a clear constant to account for these variations.

Algebraic Modeling

Algebraic modeling is a powerful tool in analyzing and representing real-world problems with mathematical terms, equations, or expressions. In this exercise, modeling serves as the framework for understanding how weight varies inversely with distance.By crafting the equation \( W = \frac{k}{d^2} \), students can visualize how changes in distance affect weight. The key steps in modeling involve identifying known values, formulating equations, computing constants, and then applying these models to new situations.

• Start with identifying that weight varies inversely with \( d^2 \).
• Use initial conditions—such as 50 pounds at 3960 miles—to derive \( k \).
• Apply the model to find weights at other distances (e.g., 3970 miles).

Algebraic modeling allows abstract concepts like gravity and distance to be expressed, analyzed, and solved systematically, enhancing problem-solving skills and conceptual understanding in mathematics.