Register to reply

Rotation of the Earth and Apparent Weight?

by wmrunner24
Tags: centripetal, force, rotation, weight
Share this thread:
Oct22-10, 11:43 AM
P: 57
1. The problem statement, all variables and given/known data

Because of Earth’s rotation about its axis, a point on the Equator experiences a centripetal acceleration of 0.034 m/s2, while a point at the poles experiences no centripetal acceleration.
What is the apparent weight at the equator of a person having a mass of 118.1 kg? The
acceleration of gravity is 9.8 m/s^2 .

Answer in units of N.

2. Relevant equations


3. The attempt at a solution

So, at first this problem greatly confused me. Now I think I have an idea of how to approach it.


This equation describes the net force toward the center of the Earth. Gravity acts toward the center of the Earth, while the normal force will resist it. If a person were to be standing on a scale, it would be the normal force that would produce the "apparent" reading. So then:



F[tex]_{normal}[/tex]=1153.3646 N

Is this line of thinking correct? Thanks in advance for any help provided.
Phys.Org News Partner Science news on
Physical constant is constant even in strong gravitational fields
Montreal VR headset team turns to crowdfunding for Totem
Researchers study vital 'on/off switches' that control when bacteria turn deadly
Oct22-10, 11:55 AM
P: 50
Yes that is correct.

Register to reply

Related Discussions
Rotation of earth and weight of an object on the earth Classical Physics 8
Apparent Weight help! Introductory Physics Homework 13
Rotation of Earth effects the weight of star. Advanced Physics Homework 2
Apparent Weight Introductory Physics Homework 6
Apparent weight Introductory Physics Homework 2