Problem 92 - University Physics with Modern Physics

Question

A Parallel Universe. Imagine a parallel universe in which the electric force has the same properties as in our universe but there is no gravity. In this parallel universe, the sun carries charge \(Q\) , the earth carries charge \(-Q\) , and the electric attraction between them keeps the earth in orbit. The earth in the parallel universe has the same mass, the same orbital radius, and the same orbital period as in our universe. Calculate the value of \(Q\) . (Consult Appendix F as needed)

Step-by-Step Solution

Answer

Charge magnitude is approximately \( 9.4 \times 10^{17} \text{ C} \).

Step 1: Understanding the Problem

In this parallel universe, the electric force replaces gravitational force to keep the Earth in orbit around the Sun. We'll use the concept that electric force is keeping the Earth in circular motion. We need to find the magnitude of charges, \( Q \), on the Earth and Sun that result in this scenario.

Step 2: Relate Orbital Motion to Force

In our universe, the gravitational force \( F = \frac{G M_S M_E}{r^2} \) keeps the Earth in orbit. In the parallel universe, the electric force must do the same. We use \( F = \frac{k |Q|^2}{r^2} \), where \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \) is Coulomb’s constant. Here, the electrical force \( \frac{k |Q|^2}{r^2} \) replaces \( \frac{G M_S M_E}{r^2} \).

Step 3: Equate the Forces

Since gravitational force is replaced by electric force, we can equate: \[ \frac{k |Q|^2}{r^2} = \frac{G M_S M_E}{r^2} \] where \( G \) is the gravitational constant, \( M_S \) and \( M_E \) are the masses of the Sun and Earth, respectively. Cancelling \( r^2 \) from both sides and solving for \( |Q| \), we get: \[ |Q| = \sqrt{\frac{G M_S M_E}{k}} \]

Step 4: Insert Known Values

Insert the known values, \( G = 6.67 \times 10^{-11} \text{ N}\,\text{m}^2/\text{kg}^2 \), \( M_S = 1.99 \times 10^{30} \text{ kg} \), \( M_E = 5.97 \times 10^{24} \text{ kg} \), and \( k = 8.99 \times 10^9 \text{ N}\,\text{m}^2/\text{C}^2 \) into the formula: \[ |Q| = \sqrt{\frac{(6.67 \times 10^{-11}) (1.99 \times 10^{30}) (5.97 \times 10^{24})}{8.99 \times 10^9}} \]

Step 5: Calculate the Result

Perform the calculation to find \(|Q|\): \[ |Q| = \sqrt{\frac{7.95 \times 10^{44}}{8.99 \times 10^9}} \approx \sqrt{8.84 \times 10^{34}} \approx 9.4 \times 10^{17} \text{ C} \] Thus, in this parallel universe, each of the Earth and the Sun has a charge magnitude of approximately \( 9.4 \times 10^{17} \text{ C} \).

Key Concepts

Parallel Universe

Let's imagine a parallel universe, a concept popularized in fiction and science theory. This universe mirrors our own but functions under slightly different laws of physics. In our exercise, the absence of gravity is the unique aspect. Here, the forces that govern movement and attraction are primarily based on electric charge, rather than gravitational pull. This world would offer a curious alternate reality where the electric force sustains celestial motion. Everything else, like the mass of the Earth and Sun, remains unchanged. Considering such a universe helps us explore the fundamental forces at play in ours. It enriches our understanding of how electric forces can create motion that mimics the gravitational pull we're familiar with.

Orbital Motion

Orbital motion is the circular path that a smaller mass follows around a larger one, like the Earth around the Sun. In our universe, this is driven by gravity, calculated through Newton’s laws. But in the parallel universe described, electric forces take on this role.

• Orbit stability relies on a centripetal force pulling the smaller object towards the larger one.
• In a gravity-free universe, a strong electric attraction achieves this instead.

We derive the forces governing orbital motion from principles of physics. Specifically, in this scenario, we replace gravitational force (from gravity) with the electric force (from the charges of Earth and Sun). The formula equates the gravitational pull familiar in our world with electric forces, ensuring the Earth remains in its orbit.

Electric Charge

Electric charge is a fundamental property of matter, influencing how particles and objects interact. There are positive and negative charges, and objects with opposite charges attract each other. This basic principle governs the replacement of gravity in the parallel universe.

• The Earth holds a negative charge while the Sun carries a positive one in this scenario.
• The mutual attraction between these charges mirrors gravitational attraction, keeping Earth in orbit.

Electric force is determined by the strength of these charges and the distance between them. By calculating the required electric charge, as done in the problem, we can understand the magnitude necessary to replicate gravitational force, highlighting the potential of electric charge as a significant natural force.

Coulomb's Constant

Coulomb's constant, denoted as k, is crucial for calculating electric forces between two charged objects. It appears in Coulomb's law, which states that the electric force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

• Coulomb's constant is approximately equal to \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \).
• It allows us to quantify the electric force between charges in a formula comparable to Newton’s law of universal gravitation.

In the parallel universe problem, it’s vital to equate gravitational forces with electric forces, using Coulomb's constant to compute how much charge is required for the electric force to maintain Earth's orbital path around the Sun. This highlights its importance in understanding interactions between electrically charged objects.