Problem 55 - College Algebra with Corequisite Support

Question

For the following exercises, use the given information to answer the questions. The weight of an object above the surface of Earth varies inversely with the square of the distance from the center of 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

The object would weigh approximately 49.75 pounds.

Step 1: Understanding Inverse Variation

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

Step 2: Finding the Constant of Variation

We know that when the distance \( d = 3960 \) miles, the weight \( W = 50 \) pounds. Plugging these values into the inverse variation formula gives \( 50 = \frac{k}{3960^2} \). To find \( k \), solve for \( k \): \( k = 50 \times 3960^2 \).

Step 3: Calculating the Constant

Compute \( k = 50 \times 3960^2 \). First, find \( 3960^2 = 15681600 \). Then multiply by 50: \( k = 784080000 \).

Step 4: Using the Constant to Find New Weight

With the constant \( k = 784080000 \), find the new weight when \( d = 3970 \) miles. Substitute into the formula: \( W = \frac{784080000}{3970^2} \).

Step 5: Calculating the New Weight

Compute \( 3970^2 = 15760900 \). Then, \( W = \frac{784080000}{15760900} \). Divide \( 784080000 \) by \( 15760900 \) to find \( W \).

Step 6: Final Weight Calculation

Perform the division to find \( W \): \( W = 49.75 \). Thus, the weight of the object at 3970 miles from Earth's center is approximately 49.75 pounds.

Key Concepts

Weight Calculation

Weight is not constant and can vary with distance when discussing celestial objects. When we calculate weight in the context of gravitational relationships like the Earth's pull, we use specific formulas. The one used in this exercise is for inverse variation. Inverse variation suggests that as one factor increases, the result decreases. In our context: - The weight of the object changes with the square of the distance from Earth's center.Given the formula: \[ W = \frac{k}{d^2} \] The weight \( W \) is calculated as the constant \( k \) divided by the square of the distance \( d \). This means as \( d \) increases, \( W \) decreases because \( d^2 \) will grow larger, reducing the value of \( \frac{k}{d^2} \). For example, an object weighed 50 pounds at 3960 miles from Earth's center, but its weight decreased to 49.75 pounds at 3970 miles, showing how a small increase in distance affects weight inversely.

Distance from Earth's Center

Distance from the center of the Earth plays a major role in determining how much an object weighs. This is because the gravitational pull weakens as you move farther from the Earth's center. The exercise uses two distances:

• 3960 miles
• 3970 miles

Notably, these differences in distances showcase how the inverse square law significantly impacts weight. By plugging these distances into the formula, we observe that even a minor change in distance—just 10 miles—alters the weight value when squared. This is a clear example of how gravity doesn’t change uniformly but rather follows an inverse square rule, making calculations essential where distances vary so slightly.

Constant of Variation

The constant of variation \( k \) is a crucial element in an inverse relationship, as seen in this exercise.To determine \( k \), use the known weight and distance:1. Start with the formula: \( W = \frac{k}{d^2} \) 2. Plug in the values when \( W = 50\) and \( d = 3960 \): \( 50 = \frac{k}{3960^2} \) 3. Solve for \( k \): \( k = 50 \times 3960^2 \)Calculating \( 3960^2 = 15681600 \), then multiplying by 50 gives us \( k = 784080000 \). This constant helps us calculate the weight at different distances. Once determined, it can be used for any distance to find the object’s weight as per the formula. So, accuracy in finding \( k \) is vital, making it the backbone of such calculations.