<< Chapter < Page Chapter >> Page >

Apparent weight: accounting for earth’s rotation

As we saw in Applications of Newton’s Laws , objects moving at constant speed in a circle have a centripetal acceleration directed toward the center of the circle, which means that there must be a net force directed toward the center of that circle. Since all objects on the surface of Earth move through a circle every 24 hours, there must be a net centripetal force on each object directed toward the center of that circle.

Let’s first consider an object of mass m located at the equator, suspended from a scale ( [link] ). The scale exerts an upward force F s away from Earth’s center. This is the reading on the scale, and hence it is the apparent weight    of the object. The weight ( mg ) points toward Earth’s center. If Earth were not rotating, the acceleration would be zero and, consequently, the net force would be zero, resulting in F s = m g . This would be the true reading of the weight.

An illustration of the earth, rotating on its north-south axis, with masses on spring scales shown at three locations. The radius of the earth is labeled as R E, its center is labeled as O. One spring scale is above the north pole. An upward force F S N and a downward force m g are shown acting on the mass on this spring scale. A dashed line is shown from the center of the earth to the north pole. Another spring scale is shown to the right of the equator and a dashed line connects the center of the earth to the equator on the right side of the earth. The forces on the mass on this second spring scale are shown as a force F S E to the right and m g to the left. A third spring scale is shown at an angle lambda to the horizontal. A dashed line at this angle is shown from the center to the surface of the earth. The horizontal distance from the surface of the earth at this angle lambda to the vertical dashed line connecting the center to the north pole is labeled as r. The point on the dashed vertical line where r meets it is labeled P. Three forces are shown for the third mass. One force is labeled F S and points radially outward. A second force, labeled m g points radially inward. A third force, labeled F c, points horizontally to the left.
For a person standing at the equator, the centripetal acceleration ( a c ) is in the same direction as the force of gravity. At latitude λ , the angle the between a c and the force of gravity is λ and the magnitude of a c decreases with cos λ .

With rotation, the sum of these forces must provide the centripetal acceleration, a c . Using Newton’s second law, we have

F = F s m g = m a c where a c = v 2 r .

Note that a c points in the same direction as the weight; hence, it is negative. The tangential speed v is the speed at the equator and r is R E . We can calculate the speed simply by noting that objects on the equator travel the circumference of Earth in 24 hours. Instead, let’s use the alternative expression for a c from Motion in Two and Three Dimensions . Recall that the tangential speed is related to the angular speed ( ω ) by v = r ω . Hence, we have a c = r ω 2 . By rearranging [link] and substituting r = R E , the apparent weight at the equator is

F s = m ( g R E ω 2 ) .

The angular speed of Earth everywhere is

ω = 2 π rad 24 hr × 3600 s/hr = 7.27 × 10 −5 rad/s.

Substituting for the values or R E and ω , we have R E ω 2 = 0.0337 m/s 2 . This is only 0.34% of the value of gravity, so it is clearly a small correction.

Zero apparent weight

How fast would Earth need to spin for those at the equator to have zero apparent weight? How long would the length of the day be?

Strategy

Using [link] , we can set the apparent weight ( F s ) to zero and determine the centripetal acceleration required. From that, we can find the speed at the equator. The length of day is the time required for one complete rotation.

Solution

From [link] , we have F = F s m g = m a c , so setting F s = 0 , we get g = a c . Using the expression for a c , substituting for Earth’s radius and the standard value of gravity, we get

a c = v 2 r = g v = g r = ( 9.80 m/s 2 ) ( 6.37 × 10 6 m ) = 7.91 × 10 3 m/s .

The period T is the time for one complete rotation. Therefore, the tangential speed is the circumference divided by T , so we have

v = 2 π r T T = 2 π r v = 2 π ( 6.37 × 10 6 m ) 7.91 × 10 3 m/s = 5.06 × 10 3 s .

This is about 84 minutes.

Significance

We will see later in this chapter that this speed and length of day would also be the orbital speed and period of a satellite in orbit at Earth’s surface. While such an orbit would not be possible near Earth’s surface due to air resistance, it certainly is possible only a few hundred miles above Earth.

Got questions? Get instant answers now!

Questions & Answers

I'm interested in biological psychology and cognitive psychology
Tanya Reply
what does preconceived mean
sammie Reply
physiological Psychology
Nwosu Reply
How can I develope my cognitive domain
Amanyire Reply
why is communication effective
Dakolo Reply
Communication is effective because it allows individuals to share ideas, thoughts, and information with others.
effective communication can lead to improved outcomes in various settings, including personal relationships, business environments, and educational settings. By communicating effectively, individuals can negotiate effectively, solve problems collaboratively, and work towards common goals.
it starts up serve and return practice/assessments.it helps find voice talking therapy also assessments through relaxed conversation.
miss
Every time someone flushes a toilet in the apartment building, the person begins to jumb back automatically after hearing the flush, before the water temperature changes. Identify the types of learning, if it is classical conditioning identify the NS, UCS, CS and CR. If it is operant conditioning, identify the type of consequence positive reinforcement, negative reinforcement or punishment
Wekolamo Reply
please i need answer
Wekolamo
because it helps many people around the world to understand how to interact with other people and understand them well, for example at work (job).
Manix Reply
Agreed 👍 There are many parts of our brains and behaviors, we really need to get to know. Blessings for everyone and happy Sunday!
ARC
A child is a member of community not society elucidate ?
JESSY Reply
Isn't practices worldwide, be it psychology, be it science. isn't much just a false belief of control over something the mind cannot truly comprehend?
Simon Reply
compare and contrast skinner's perspective on personality development on freud
namakula Reply
Skinner skipped the whole unconscious phenomenon and rather emphasized on classical conditioning
war
explain how nature and nurture affect the development and later the productivity of an individual.
Amesalu Reply
nature is an hereditary factor while nurture is an environmental factor which constitute an individual personality. so if an individual's parent has a deviant behavior and was also brought up in an deviant environment, observation of the behavior and the inborn trait we make the individual deviant.
Samuel
I am taking this course because I am hoping that I could somehow learn more about my chosen field of interest and due to the fact that being a PsyD really ignites my passion as an individual the more I hope to learn about developing and literally explore the complexity of my critical thinking skills
Zyryn Reply
good👍
Jonathan
and having a good philosophy of the world is like a sandwich and a peanut butter 👍
Jonathan
generally amnesi how long yrs memory loss
Kelu Reply
interpersonal relationships
Abdulfatai Reply
What would be the best educational aid(s) for gifted kids/savants?
Heidi Reply
treat them normal, if they want help then give them. that will make everyone happy
Saurabh
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 2

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 1' conversation and receive update notifications?

Ask