Problem 108 - College Algebra with Modeling and Visualization

Question

The weight of an object varies inversely as the second power of the 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: Define the Relationship

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

Step 2: Calculate the Constant

First, determine the constant \( k \) using the weight at the Earth's surface where the distance is equal to the Earth's radius (4000 miles). Substituting \( w = 160 \) pounds and \( d = 4000 \) miles into the equation: \[ 160 = \frac{k}{4000^2} \] Solve for \( k \): \[ k = 160 \times 4000^2 \] \[ k = 160 \times 16,000,000 \] \[ k = 2,560,000,000 \]

Step 3: Determine the New Distance

The person is now 8000 miles above the Earth's surface. We need to calculate the total distance from the center of the Earth. The new distance \( d \) is \[ d = 4000 + 8000 = 12000 \text{ miles} \]

Step 4: Calculate New Weight

Use the value of \( k \) obtained in Step 2 and the new distance \( d = 12000 \) in the weight formula: \[ w = \frac{2,560,000,000}{12000^2} \] Calculate \( 12000^2 \): \[ 12000^2 = 144,000,000 \] Now find the new weight: \[ w = \frac{2,560,000,000}{144,000,000} \] \[ w \approx 17.78 \text{ pounds} \]

Step 5: Record the Result

Finally, the weight of the person 8000 miles above the Earth's surface would be approximately 17.78 pounds.

Key Concepts

Distance from Earth

Understanding how distance affects object weight is crucial when learning about gravitational interactions. In our exercise, we deal with the distance from the center of the Earth. This is important because it determines gravitational pull. When measuring distance from Earth, we start from the center, as gravity pulls towards this point.
Earth's radius is approximately 4000 miles. In our problem, when a person is 8000 miles above the surface, we must add Earth's radius to this distance:

• Earth's radius: 4000 miles
• Distance above the surface: 8000 miles
• Total distance from Earth's center: 4000 + 8000 = 12000 miles

This total distance is crucial for calculating how gravity affects weight changes.

Weight Calculation

Weight changes with different distances from Earth's center. Using the known formula for inverse variation, the calculations make this clear. With inverse variation, as distance increases, weight decreases proportionally.When weight varies inversely with the square of the distance, the relation is expressed by the formula:\[ w = \frac{k}{d^2} \]where:

• is the weight,
• is a constant we need to find out, and
• is the distance from the center of the Earth.

To calculate the person's weight at a new distance, solve the formula for \( k \) using known conditions, like 160 pounds at the Earth's surface. With this constant, we can determine weight at any distance above Earth.

Inverse Square Law

The inverse square law plays a vital role in understanding gravitational forces. This law states that a physical quantity (like weight due to gravity) is inversely proportional to the square of the distance.In our exercise, weight decreases by a factor related to the increase in distance squared. So the further you go from the Earth, the less you will weigh. The relation is given by:\[ w = \frac{k}{d^2} \]Every time distance is doubled, the weight is quartered. For instance, at 12000 miles from the center, the weight is far less than at 4000 miles (Earth's surface).

• Double distance means weight decreases to one-fourth.
• Three times distance results in one-ninth the original weight.

This principle helps us predict and understand changes in forces like gravity based on distance variation.