Problem 80 - Elementary Algebra

Question

The weight of an object varies inversely as the square of its distance from the center of earth. If an object weighs 180 pounds on the surface of earth (approximately 4,000 miles from the center), then how much will it weigh at 2,000 miles above earth's surface?

Step-by-Step Solution

Answer

The object will weigh 80 pounds at 2,000 miles above the Earth's surface.

Step 1: Understand the Inverse Variation

The problem states that the weight of the object varies inversely with the square of its distance from the center of the Earth. We can model this relationship with the formula: \[ W = \frac{k}{d^2} \]where \(W\) is the weight, \(d\) is the distance from the center of the Earth, and \(k\) is a constant we need to determine.

Step 2: Calculate the Constant of Variation (k)

First, we need to find \(k\) using the given weight on the surface of the Earth. At the surface, the distance \(d\) is 4,000 miles, and the weight \(W\) is 180 pounds. Plug these values into the formula:\[ 180 = \frac{k}{(4000)^2} \]\[ 180 = \frac{k}{16000000} \]Solve for \(k\):\[ k = 180 \times 16000000 = 2880000000 \]

Step 3: Calculate the New Distance from the Center

The new distance from the center of the Earth is the sum of the Earth's radius and the distance above the Earth's surface. This is 4,000 miles (Earth's radius) + 2,000 miles (above the surface) = 6,000 miles.

Step 4: Determine the Weight at the New Distance

Now, use the inverse variation formula to find the weight \(W\) at the 6,000-mile distance:\[ W = \frac{2880000000}{(6000)^2} \]\[ W = \frac{2880000000}{36000000} \]\[ W = 80 \]The object weighs 80 pounds at 6,000 miles from the Earth's center.

Key Concepts

Weight Variation

The concept of weight variation in this problem is based on inverse variation. This means that as one value increases, the other decreases at a proportional rate. In our case, weight varies inversely with the square of the distance from the center of the Earth. - **Inverse Variation Formula:** In simple terms, when something varies inversely, you can express it with a formula that looks like \( W = \frac{k}{d^2} \), where: - \( W \) is the weight, - \( d \) is the distance from the Earth's center, - \( k \) is a constant. - **Calculating the Constant (k):** The constant \( k \) represents the factor that adjusts the relationship for the particular scenario. In our exercise: - At 4,000 miles from the Earth's center, the weight was 180 pounds, - Following the formula, \( k \) is calculated as \( 2880000000 \), which is crucial for determining weight at other distances. This relationship illustrates how gravitational forces decrease as objects move further from Earth.

Distance from Center of Earth

Understanding how distance affects an object's weight is essential in grasping this exercise. The Earth's center is the point from which these distances are measured. - **Initial Distance:** Our problem started with a distance of 4,000 miles, which approximates the average radius of the Earth. - **Calculating New Distances:** When moving objects away from Earth's surface: - Add the additional distance. For instance, an object located 2,000 miles above the surface means a total distance of 6,000 miles from the center. - **Effect of Distance:** This new distance is crucial because it directly affects the weight. The further away the object, the lighter it becomes due to reduced gravitational pull. In real-world terms, it explains why astronauts feel weightless in space; they are much farther from the Earth's gravitational center.

Inverse-Square Law

The heart of this exercise is the inverse-square law. This law is pivotal in understanding the variation of weight with distance. - **Inverse-Square Law Explained:** The law dictates that a given force becomes weaker with the square of the distance. So if you double the distance, the force is reduced by a factor of four (since \( 2^2 = 4 \)).- **Applying the Law in Weight Variation:** As in our scenario: - The object was initially 4,000 miles from the Earth's center, - When moved to 6,000 miles, its weight changes based on \( \frac{1}{(\text{new distance})^2 / (\text{original distance})^2} \). - **Conclusion:** At 6,000 miles, the object's weight dropped to 80 pounds, following the inverse-square law.This principle doesn’t just apply to gravity, but to all kinds of phenomena like light intensity, sound intensity, and more, proving its vital role in physics.