# Rotation of the Earth and Apparent Weight?

1. ### wmrunner24

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 .

2. Relevant equations
F$$_{centripetal}$$=ma$$_{centripetal}$$

F$$_{gravity}$$=mg

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.

F$$_{centripetal}$$=F$$_{gravity}$$-F$$_{normal}$$

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$$_{normal}$$=F$$_{gravity}$$-F$$_{centripetal}$$

F$$_{normal}$$=mg-ma$$_{centripetal}$$

F$$_{normal}$$=1153.3646 N

Is this line of thinking correct? Thanks in advance for any help provided.

2. ### Xerxes1986

50
Yes that is correct.