Problem 23 - Algebra 2

Question

The force due to gravity decreases with the square of the distance from the center of Earth. As an object moves farther from Earth, its weight decreases. The radius of Earth is approximately 3960 miles. The formula relating weight and distance is \(r=\sqrt{\frac{3960^{2} W_{E}}{W_{S}}}-3960,\) where \(W_{E}\) represents the weight of a body on Earth, \(W_{S}\) represents its weight a certain distance from the center of Earth, and \(r\) represents the distance above Earth's surface. An astronaut weighs 125 pounds on Earth. What is her weight in space if she is 99 miles above the surface of Earth?

Step-by-Step Solution

Answer

The astronaut weighs 118.85 pounds in space.

Step 1: Identify Given Values

The problem provides the weight of the astronaut on Earth, \(W_E = 125\) pounds, and the height above Earth, \(r = 99\) miles.

Step 2: Calculate Distance from Earth's Center

Add the radius of Earth to the height above the Earth's surface to find the total distance from Earth's center: \(d = 3960 + 99 = 4059\) miles.

Step 3: Rewrite the Weight Formula

Using the formula \(r = \sqrt{\frac{3960^{2} W_{E}}{W_{S}}} - 3960\), solve for \(W_S\). Start by rearranging it to isolate \(\sqrt{\frac{3960^{2} W_E}{W_S}} = r + 3960\).

Step 4: Square Both Sides of the Equation

Square both sides of \(\sqrt{\frac{3960^{2} W_E}{W_S}} = 4059\) to get \(\frac{3960^{2} W_E}{W_S} = 4059^2\).

Step 5: Solve for Weight in Space

Solve for \(W_S\) by rearranging the equation: \(W_S = \frac{3960^{2} \times 125}{4059^2}\). Compute this to find \(W_S\).

Step 6: Compute the Weight

Calculate \(W_S = \frac{3960^{2} \times 125}{4059^2} = 118.85\). So, the astronaut's weight in space is approximately 118.85 pounds.

Key Concepts

Weight Variation with Distance

The weight of an object is a measure of the gravitational force acting upon it. This force diminishes as the object moves away from Earth’s center. Why does this happen? Well, according to the universal law of gravitation, the gravitational force between two objects is inversely proportional to the square of the distance between them. This means as the object moves farther, the gravitational pull weakens rapidly.
In practical terms, this means that an astronaut, for instance, weighs less the higher up they are above the Earth's surface. If they're on Earth, they're subjected to a full gravitational force. However, as they ascend—for example, to a height of 99 miles above Earth—their weight decreases since they are farther away from Earth’s gravitational center.
It's important to notice this variation to understand phenomena such as the experience of apparent weightlessness in space even though the object never truly"loses weight" entirely.

Earth's Radius in Gravitational Formulas

The radius of the Earth plays a crucial role in gravitational calculations, specifically when determining how weight changes with altitude. For these calculations, Earth's radius is generally approximated to 3960 miles.
This value is crucial because the main formula used here, which is derived from gravitational principles, requires knowing how far an object is from the center of the Earth. In the given exercise, the formula involves both Earth's radius and the object's distance above Earth's surface to calculate the overall distance.
Without Earth's radius, we couldn't accurately compute how much gravitational force remains at various distances, significantly complicating our ability to forecast how an astronaut's weight changes as altitude varies.

Algebraic Manipulation in Physics Problems

Solving physics problems often involves a significant amount of algebraic manipulation to isolate the variable of interest. In our exercise, the main goal was to solve for the astronaut's weight in space using the given weight formula.
First, we had to rearrange the formula to manipulate the terms correctly. This involves isolating the square root expression on one side and, subsequently, removing the square root by squaring both sides of the equation.
This step-by-step manipulation is the key to simplifying complex formulas and making them solvable, much like peeling away layers to reach the core of the problem.
Such algebraic skills are not only important in physics but essential in a wide array of scientific calculations where understanding relationships and dependencies between variables allow for practical solutions to seemingly intricate problems.