How far is the center of mass of the Earth-Moon system from the center of the Earth? (From Appendix C, obtain the masses of the Earth and Moon and the distance between the centers of the Earth and Moon. It is interesting to compare the answer to the Earth's radius.)

Short Answer

Expert verified
The center of mass of the Earth-Moon system is approximately \(4.73 x 10^{6}\) meters (or 4730 kilometers) from the center of the Earth.

Step by step solution

01

- Identify known quantities

From the appendix, we know that the mass of the Earth (\(m1\)) is \(5.972 x 10^{24} kg\), the mass of the Moon (\(m2\)) is \(7.348 x 10^{22} kg\), and the distance between them (\(d\)) is \(3.844 x 10^{8} m\).
02

- Substitute the known values into the formula

Substitute the known values into the formula for the center of mass. \(x_{cm} = \frac{(5.972 x 10^{24} kg * 0) + (7.348 x 10^{22} kg * 3.844 x 10^{8} m)}{5.972 x 10^{24} kg + 7.348 x 10^{22} kg}\)
03

- Simplify the complex product

Simplify the product \(7.348 x 10^{22} kg * 3.844 x 10^{8} m\) first, which equals \(2.827 x 10^{31} kg*m\).
04

- Simplify the total mass on the denominator

Simplify the addition \(5.972 x 10^{24} kg + 7.348 x 10^{22} kg\), which equals \(5.979 x 10^{24} kg\).
05

- Final computation

Finally, divide the two results to find the center of mass. \(x_{cm} = \frac{2.827 x 10^{31} kg*m}{5.979 x 10^{24} kg}\) which equals approximately \(4.73 x 10^{6} m\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Earth-Moon System
Our Earth is part of a complex cosmic dance with its sole natural satellite, the Moon. Understanding the Earth-Moon system involves knowledge of their masses, the distance between their centers, and how they interact with each other gravitationally.

Imagine the Earth and the Moon as two objects in space connected by an invisible rod. The point where this rod would balance is known as the center of mass of the system. It is not always at the center of the larger body, as one might assume, but rather is dependent on the masses of both bodies and the distance between them. In the case of the Earth-Moon system, the center of mass is actually located beneath the Earth's surface because the Earth is significantly more massive than the Moon.

The position of this center of mass is crucial for understanding how the two bodies move in tandem through space, including how they orbit each other and the effects this system has on phenomena such as tides. Determining this point is a fundamental astrophysical calculation that provides insights into the stability and dynamics of the system.
Gravitational Forces at Play
Mastering the concept of gravitational forces is key to unlocking the mysteries of celestial movements. Every object in the universe that has mass exerts a gravitational force on every other mass. This fundamental force is what keeps our feet on the ground and dictates the motion of planets, moons, and stars.

The strength of this force is directly related to the masses of the two objects involved and inversely related to the square of the distance between their centers, a concept enshrined in Newton's law of universal gravitation. Gravitational attraction brings the Earth and Moon toward each other, and consequently, their mutual forces create a balance point - the center of mass.

In the context of the Earth-Moon system, it's intriguing to consider that the Moon's smaller mass doesn't prevent it from having a substantial effect. The gravitational pull between Earth and Moon results in ocean tides and affects Earth's rotation. These forces act at a distance, unseen but deeply felt, governing motion and creating harmony in the celestial dance.
Astrophysics Calculations
To engage with astrophysics calculations is to decode the language of the universe. These calculations are vital for understanding the movement and behavior of celestial bodies. In the specific case of finding the center of mass for the Earth-Moon system, the process begins with identifying the known quantities: the masses of both the Earth and the Moon and their separation distance.

Utilizing formulas specific to celestial mechanics, these values are substituted, and calculations are performed to simplify and solve the problem. It is akin to solving a puzzle where each piece must fit perfectly to see the full picture. The equation used to calculate the center of mass involves multiplying each mass by its distance from a chosen reference point (in this case, the Earth's center is chosen as zero) and then summing these products.

These calculations are not just theoretical exercises; they have practical implications for space navigation and the planning of space missions. Whether plotting the trajectory of a spacecraft or predicting lunar eclipses, the astrophysics calculations that pinpoint the Earth-Moon center of mass are indispensable for both understanding our universe and exploring it.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Richard, mass \(78.4 \mathrm{~kg}\), and Judy, who is less massive, are enjoying Lake George at dusk in a \(31.6\) -kg canoe. When the canoe is at rest in the placid water, they change seats, which are \(2.93 \mathrm{~m}\) apart and symmetrically located with respect to the canoe's center. Richard notices that the canoe moved \(41.2 \mathrm{~cm}\) relative to a submerged log and calculates Judy's mass. What is it?

Two skaters, one with mass \(65 \mathrm{~kg}\) and the other with mass \(42 \mathrm{~kg}\), stand on an ice rink holding a pole with a length of \(9.7 \mathrm{~m}\) and a mass that is negligible. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far will the 42 -kg skater move?

In the ammonia \(\left(\mathrm{NH}_{3}\right)\) molecule, the three hydrogen (H) atoms form an equilateral triangle, the distance between centers of the atoms being \(16.28 \times 10^{-11} \mathrm{~m}\), so that the center of the triangle is \(9.40 \times 10^{-11} \mathrm{~m}\) from each hydrogen atom. The nitrogen \((\mathrm{N})\) atom is at the apex of a pyramid, the three hydrogens constituting the base (see Fig. 7-27). The nitrogen/hydrogen distance is \(10.14 \times 10^{-11} \mathrm{~m}\) and the nitrogen/hydrogen atomic mass ratio is \(13.9 .\) Locate the center of mass relative to the nitrogen atom.

Show that the ratio of the distances \(x_{1}\) and \(x_{2}\) of two particles from their center of mass is the inverse ratio of their masses; that is, \(x_{1} / x_{2}=m_{2} / m_{1}\).

A rocket of total mass \(1.11 \times 10^{5} \mathrm{~kg}\), of which \(8.70 \times 10^{4} \mathrm{~kg}\) is fuel, is to be launched vertically. The fuel will be burned at the constant rate of \(820 \mathrm{~kg} / \mathrm{s}\). Relative to the rocket, what is the minimum exhaust speed that allows liftoff at launch?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free