Problem 97 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem. Gravity 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 person would weigh approximately 17.78 pounds.

Step 1: Understanding the Problem

The weight of an object varies inversely with the square of its distance from the center of Earth. This means that the product of the weight and the square of the distance remains constant. We need to find the weight of a person at a certain distance above Earth's surface.

Step 2: Define the Variables

Let the person's weight on the surface be \( W_1 = 160 \) pounds. Denote the radius of Earth as \( r = 4000 \) miles, so the initial distance from Earth's center is also \( 4000 \) miles. The new distance from the center of Earth is \( D_2 = 8000 + 4000 = 12000 \) miles. Let \( W_2 \) be the person's weight at this new distance.

Step 3: Using the Inverse Variation Formula

The relationship between weight and distance states that \( W_1 \times D_1^2 = W_2 \times D_2^2 \). We need to calculate \( D_1 \) and \( D_2 \) first. Here, \( D_1 = 4000 \) miles and \( D_2 = 12000 \) miles.

Step 4: Solve for the Unknown Weight

Plug in the known values into the formula: \( 160 \times 4000^2 = W_2 \times 12000^2 \). Rearrange this to solve for \( W_2 \): \[ W_2 = \frac{160 \times 4000^2}{12000^2} \]

Step 5: Calculate the Result

Perform the calculations:1. Compute \( 4000^2 = 16000000 \).2. Compute \( 12000^2 = 144000000 \).3. Calculate \( W_2 = \frac{160 \times 16000000}{144000000} = \frac{2560000000}{144000000} \approx 17.78 \).

Key Concepts

Gravity

Gravity is a force that attracts objects toward each other. This force is what keeps us grounded on Earth. Interestingly, in the problem of **weight and distance from Earth**, we use the concept of inverse variation. This involves how gravity weakens the farther you are from the center of the Earth.

The principle of inverse variation here states that the weight of an object (like a person) varies inversely with the square of the distance from Earth's center. This means that as the distance from Earth's center increases, the weight decreases, and vice versa. The key equation involved can be written as follows:

• Initial Weight x (Initial Distance)^2 = New Weight x (New Distance)^2

This relationship helps us understand the change in force experienced by objects due to gravity when their distance from Earth's center changes. By grasping this concept, we can calculate an object's weight anywhere in space, given the appropriate data.

Distance from Earth's center

"Distance from Earth's center" is a vital part of understanding weight calculations under the influence of gravity. The center of the Earth acts as a point source of gravitational force, making the distance critical in determining how strongly gravity acts on an object.

In the exercise, the radius of the Earth is given as approximately **4000 miles**. When a person stands on the surface, this is the initial distance from the Earth's center. If they move further away, for instance, **8000 miles above the surface**, we add the radius of the Earth to their altitude to find the total distance from the center.

• **Earth's surface distance** = 4000 miles
• **Above Earth's surface** = 8000 miles
• **Total distance** = 4000 + 8000 = 12000 miles

Understanding these distances allows us to apply formulas correctly in weight calculations. Accurate measurements are crucial since they directly affect how inverse variation impacts the final weight calculation.

Weight calculation

Weight calculation in the context of gravity and distance involves applying a specific relationship, **inverse variation**, to find the weight at a particular distance from Earth's center. This idea simply asserts that the gravitational force on an object diminishes as it moves away from Earth, using the principle of inverse squares.

Let's revisit the formula:

• **Initial Weight x (Initial Distance)^2 = New Weight x (New Distance)^2**

Plugging the known values:- Initial Weight (W_1) = 160 pounds- Initial Distance (D_1) = 4000 miles- New Distance (D_2) = 12000 miles The equation becomes: \[ 160 imes 4000^2 = W_2 imes 12000^2 \] To solve for the new weight (W_2), calculate the squared distances first, then rearrange for W_2:- **Calculate Squares**: - 4000² = 16,000,000 - 12000² = 144,000,000- **Solve for** W_2: \[ W_2 = \frac{160 imes 16,000,000}{144,000,000} \approx 17.78 \]This calculated weight at 12000 miles from Earth's center illustrates how distance affects gravitational force, as the weight decreases significantly from its original 160 pounds.