Problem 28 - College Physics

Question

If an object's weight is \(W\) on the earth, what would be its weight (in terms of \(W\) ) if the earth had (a) twice its present mass but was the same size, (b) half its present radius but the same mass, (c) half its present radius and half its present mass, (d) twice its present radius and twice its present mass?

Step-by-Step Solution

Answer

(a) 2W; (b) 4W; (c) 2W; (d) 0.5W

Step 1: Understanding the Problem

The weight of an object on Earth, denoted as \( W \), is determined by the gravitational force: \( F = \frac{G \, M_\text{earth} \, m}{R_\text{earth}^2} \), where \( G \) is the gravitational constant, \( M_\text{earth} \) is the mass of the Earth, \( m \) is the mass of the object, and \( R_\text{earth} \) is the radius of the Earth. We need to find how the weight changes under different conditions.

Step 2: Case (a): Twice the Earth's Mass

If the Earth's mass becomes \( 2M_\text{earth} \) and its radius remains the same, the new weight \( W' \) is given by: \[ W' = \frac{G \, (2M_\text{earth}) \, m}{R_\text{earth}^2} = 2 \times \frac{G \, M_\text{earth} \, m}{R_\text{earth}^2} = 2W. \] Thus, the weight would be twice its original value, \( 2W \).

Step 3: Case (b): Half the Earth's Radius

If the Earth's radius becomes \( \frac{1}{2} R_\text{earth} \) and its mass remains the same, the new weight \( W' \) is: \[ W' = \frac{G \, M_\text{earth} \, m}{(\frac{R_\text{earth}}{2})^2} = \frac{G \, M_\text{earth} \, m}{\frac{R_\text{earth}^2}{4}} = 4 \times \frac{G \, M_\text{earth} \, m}{R_\text{earth}^2} = 4W. \] The weight would be four times the original weight, \( 4W \).

Step 4: Case (c): Half the Earth's Radius and Mass

If both the Earth's radius and mass are halved, the new weight \( W' \) is: \[ W' = \frac{G \, \frac{1}{2}M_\text{earth} \, m}{(\frac{R_\text{earth}}{2})^2} = \frac{G \, \frac{1}{2}M_\text{earth} \, m}{\frac{R_\text{earth}^2}{4}} = 2 \times \frac{G \, M_\text{earth} \, m}{R_\text{earth}^2} = 2W. \] The weight would again be twice the original weight, \( 2W \).

Step 5: Case (d): Twice the Earth's Radius and Mass

If the Earth's radius and mass are both doubled, the new weight \( W' \) is: \[ W' = \frac{G \, (2M_\text{earth}) \, m}{(2R_\text{earth})^2} = \frac{2G \, M_\text{earth} \, m}{4R_\text{earth}^2} = \frac{1}{2} \times \frac{G \, M_\text{earth} \, m}{R_\text{earth}^2} = \frac{1}{2}W. \] The weight would be half the original weight, \( \frac{1}{2}W \).

Key Concepts

Weight Variation

The weight of an object is determined by the gravitational force acting on it. On Earth, this force depends on both the mass of the Earth and the distance from the center of the Earth, which is the radius. When considering different scenarios about how weight changes, it's essential to understand that weight can vary depending on these variables.

• If the mass of Earth doubles, your weight also doubles because the gravitational pull is stronger.
• If Earth's radius decreases by half, your weight increases by four times due to a more concentrated gravitational force.
• Conversely, if both mass and radius decrease, the effect isn't as straightforward, needing more calculation.
• Finally, if both the mass and radius of Earth double, your weight is halved.

Every change in these factors results in a new experience of weight, illustrating the fascinating complexity behind gravitational interactions.

Mass and Radius of Earth

The mass and radius of Earth are fundamental physical properties that shape our understanding of gravity. These properties are crucial for calculating the gravitational force experienced by objects on Earth.
- **Mass of Earth**: The average mass of Earth provides a measure of the gravitational pull that it can exert on objects. It's measured at approximately 5.97 x 10^24 kg. A greater mass means a stronger gravitational pull.
- **Radius of Earth**: This is approximately 6,371 kilometers. It represents the distance from the Earth's center to its surface. As the radius changes, the gravity felt by objects changes according to the inverse square law. For instance, halving the radius increases gravitational force by four times.
Understanding these parameters is essential for tackling various problems involving gravitational force and for exploring conceptual changes, like those in hypothetical scenarios.

Gravitation Problems

Gravitation problems often involve calculating how gravitational forces act under changing conditions. These problems require understanding the formula for gravitational force: \[F = \frac{G \, M_{ ext{earth}} \, m}{R_{ ext{earth}}^2}\] Where:

• \(F\) is the gravitational force or weight.
• \(G\) is the gravitational constant, a fundamental value in physics.
• \(M_{\text{earth}}\) is the Earth's mass.
• \(m\) is the mass of the object.
• \(R_{\text{earth}}\) is the Earth's radius.

Solving these problems entails understanding how changes to \(M_{\text{earth}}\) and \(R_{\text{earth}}\) affect \(F\). By learning to manipulate this equation, you can predict the outcomes of different hypothetical scenarios. This skill is integral to physics education, as it builds a robust foundation for more complex gravitational concepts.

Physics Education

Physics education is crucial for developing an understanding of fundamental concepts like gravitational force. Learning physics involves more than just grasping formulas; it requires a deep understanding of the underlying principles that govern physical phenomena. When tackling problems such as weight variation under different earthly conditions, the aim is not just to find the answer but also to understand the reasoning behind each step.

• Physics aids in developing critical thinking skills by prompting students to apply theoretical knowledge to practical scenarios.
• Engaging with real-world problems helps students to appreciate the relevance of physics in everyday life.
• Interactive simulations and problem-solving exercises can make abstract concepts tangible for learners.

The ultimate goal of physics education is to develop a curious mindset that questions how the world works and seeks evidence-based explanations.