Problem 16 - Elementary Algebra

Question

For each problem below, approximate the weight of a 120-pound object at the given height above the surface of earth. (1 mile \(=5,280\) feet). 90,000 feet

Step-by-Step Solution

Answer

The weight is approximately 119.04 pounds.

Step 1: Convert Feet to Miles

First, convert the height given in feet to miles. Since 1 mile is equivalent to 5,280 feet, divide 90,000 feet by 5,280 to find the height in miles. \[\text{Height in miles} = \frac{90,000}{5,280} \approx 17.05 \text{ miles}\]

Step 2: Use Inverse Square Law

The weight of an object decreases with the square of the distance from the center of the Earth. The formula to adjust the weight based on height is: \[W = W_0 \left( \frac{R}{R + h} \right)^2\]where \(W_0\) is the original weight (120 pounds), \(R\) is the radius of Earth in miles (approximately 3,963 miles), and \(h\) is the height in miles (17.05 miles).

Step 3: Plug Values into the Formula

Substitute the known values into the formula:\[W = 120 \left( \frac{3,963}{3,963 + 17.05} \right)^2\]Calculate the value inside the parentheses first, and then square it.

Step 4: Simplify the Expression

Calculate the expression:\[\text{Expression inside parentheses} = \frac{3,963}{3,980.05} \approx 0.996\]Then square the result:\[0.996^2 \approx 0.992\]Thus, the weight becomes:\[W = 120 \times 0.992 \approx 119.04 \text{ pounds}\]

Step 5: Conclusion

The approximate weight of the object at a height of 90,000 feet above the Earth's surface is 119.04 pounds.

Key Concepts

Weight Calculation

To calculate the weight of an object above the Earth's surface, we need to consider how gravity influences weight. The Earth's gravitational pull decreases as you move away from its center. This means that an object weighs less at higher altitudes than at sea level. The weight of an object is directly proportional to the gravitational force acting on it.

In this exercise, we use a formula based on the inverse square law to determine the object's weight at a specific height. The inverse square law is crucial because it describes how the gravitational force decreases with the square of the distance from the source, in this case, the Earth's center. The formula used is:

• \[W = W_0 \left( \frac{R}{R + h} \right)^2\]
• Where \(W_0\) is the original weight (at the Earth's surface), \(R\) is the Earth's radius, and \(h\) is the height above the Earth in miles.

Substituting the values into this equation will allow us to find out how much the weight decreases at the specified height as seen in the problem.

Earth's Radius

The Earth's radius plays a significant role in calculating the weight of an object above the planet's surface. The mean radius of the Earth is approximately 3,963 miles. This value is a crucial component of our weight calculation formula.

• It serves as a reference distance from the center of Earth to the surface, which is used to compare against any additional height above the surface.
• In the formula \( \frac{R}{R + h} \), the denominator represents the total distance from the Earth's center to the object in question, combining the radius and height.

Understanding this element is important because it helps in determining how the gravitational field strength diminishes with increased distance.

Unit Conversion

Unit conversion is essential in physics to ensure consistency and accuracy in calculations. In exercises like this one, distances may be given in units that require conversion, such as feet or miles. It's important to convert all measurements to the same unit system.

• In the problem, the height is provided as 90,000 feet. Since the Earth's radius is in miles, converting feet to miles is necessary for further calculations.
• The conversion makes it simpler to use the formula without encountering errors related to unit mismatch, ensuring that the height is correctly calculated as approximately 17.05 miles.

By adhering to consistent units, students develop better intuition for mathematical problems and solutions.

Height Above Earth

The height above Earth's surface can alter an object's weight due to changes in gravitational pull. When an object is lifted 90,000 feet above the Earth, it is at a distance where the gravitational force affects its weight measurement.

• At sea level, an object like the one in our problem has a weight of 120 pounds, due to the full effect of Earth's gravitational pull.
• As the height above Earth increases—a distance of 17.05 miles in this exercise—there's a weaker gravitational attraction, leading to a decrease in weight.
• By using the inverse square law, we calculate the weight reduction, resulting in a lighter weight of approximately 119.04 pounds.

Understanding how height influences weight provides insights into basic gravitational interactions in physics.