Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Because the Earth is nearly perfectly spherical, gravitational forces act on it effectively through its center. Explain why the Earth’s axis points at the North star all year long. Also explain why the earth’s rotation speed stays the same throughout the year (one rotation per 24h). In your analysis, does it matter that the Earth is going around the sun?

In actual fact, the Earth is not perfectly spherical. It bulges out a bit at the equator, and tides tend to pile up water at one side of the ocean. As a result, there are small torque exerted on the Earth by other bodies, mainly the sun and the moon. Over many thousands of years there are changes in what portion of sky the Earth’s axis points towards (Change of direction of rotational angular momentum), and changes in the length of a day (change of magnitude of rotational angular momentum).

Short Answer

Expert verified

The rotating angular momentum of the Earth-sun system has no effect on the rotational angular momentum of the earth's center of mass. As a result, the fact that the Earth revolves around the sun is irrelevant.

Hence, it doesn’t matter that Earth is going around the sun.

Step by step solution

01

Given Data

Gravitational forces act on the Earth effectively through its core because it is nearly perfectly spherical.

The Earth is not exactly round in reality. The equator bulges a little, and tides tend to pile up water on one side of the ocean. As a result, other bodies, primarily the sun and the moon, exert minor torques on the Earth. The location of the Earth's axis in the sky has changed throughout thousands of years.

02

Concept of Angular Momentum

Angular momentum is a characteristic that describes the rotating inertia of an object or set of objects moving around an axis that may or may not pass through it.

03

Determine where the Earth's axis points to the North and the earth's rotation speed remains constant throughout the year

As the earth is nearly perfectly spherical, the gravitational force acts on it effectively through its center, which means the gravitational force and the center of mass act at the same point then the torque acting about the center of mass is zero. Thus, the corresponding rotational angular momentum stays constant in magnitude and direction (the direction of the rotational angular momentum indicates that the Earth axis points at the North Star all over the year, and the magnitude of the rotational angular momentum indicates that the Earth's rotational speed stays the same throughout the year). The rotational angular momentum relates to center of mass of the earth is not affected by the rotational angular momentum of the Earth-Sun system. Hence, it doesn't matter that Earth is going around the sun.

No torques around the center of mass, so rotational angular momentum stays constant in magnitude and direction, hence, it doesn't matter that Earth is going around the sun.

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!

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

A uniform-density wheel of mass 6 kg and radius 0.3 m rotates on a low-friction axle. Starting from rest, a string wrapped around the edge exerts a constant force of15Nfor0.6s(a) what is the final angular speed? (b) what is the average angular speed? (c) Through how big an angle did the wheel turn? (d) How much string come off the wheel?

A small rock passes a massive star, following the path shown in red on the diagram. When the rock is a distance 4.5×1013m(indicatedas in figure) from the center of the star, the magnitudeof its momentum is1.35×1017kg·m/sand the angle is126°12. At a later time, when the rock is a distance d2=1.3×1013mfrom the center of the star, it is heading in the -ydirection. There are no other massive objects nearby. What is the magnitude p2of the final momentum?

Two people of different masses sit on a seesaw (Figure 11.103). M1,the mass of personis90kg,M2is42kg,d1=0.8m,andd2=1.3m.The people are initially at rest. The mass of the board is negligible.

(a) What are the magnitude and direction of the torque about the pivot due to the gravitational force on person(b) What are the magnitude and direction of the torque about the pivot due to the gravitational force on person(c) Since at this instant the linear momentum of the system may be changing, we don’t known the magnitude of the “normal” force exerted by the pivot. Nonetheless, it is possible to calculate the torque due to this force. What are the magnitude and direction of the torque about the pivot due to the force exerted by the pivot on the board? (d) What are the magnitude and direction of the net torque on the system (board + people)? (e) Because of this net torque, what will happen? (A) The seesaw will begin to rotate clockwise. (B) The seesaw will begin to rotate counterclockwise. (C) The seesaw will not move. (f) Person 2 moves to a new position, in which the magnitude of the net torque about the pivot is now0,and the seesaw is balanced. What is the new value ofd2in this situation?

You sit on a rotating stool and hold barbells in both hands with your arms fully extended horizontally. You make one complete turn in 2s.You then pull the barbells in close to your body. (a) Estimate how long it now takes you to make one complete turn. Be clear and explicit about the principles you apply and about your assumptions and approximations. (b) About how much energy did you expand?

Make a sketch showing a situation in which the torque due to a single force about some location is \(20\,\,{\rm{N}} \cdot {\rm{m}}\) in the positive \(z\) direction, whereas about another location the torque is \(10\,\,{\rm{N}} \cdot {\rm{m}}\) in the negative \(z\) direction.

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