# Weight at the North Pole vs. Equator Problem

1. Dec 7, 2011

### Legaldose

1. Suppose Earth is a perfect sphere of radius 6,370km. If a person weighs exactly 600N at the north pole, how much will the person weigh at the equator?

Givens:

r = 6,370km = 6,370,000m
Weight at north pole = mg = 600N

2. Relevant equations

ω$_{av}$ = θ/t

ƩF$_{net}$ = ma

a$_{c}$ = v$^{2}$/r

3. The attempt at a solution

Okay, so the first thing I did was to draw a picture. Then I noticed that ω$_{av}$ would be 0 at the North Pole because you don't have any rotational velocity! Then I set up a Free Body Diagram for a location at the equator and see that the normal force (N) is acting up perpendicular to the ground, and gravity is acting downward towards the center of the Earths gravity.

I get

ƩF$_{y}$ = ma$_{y}$

where a$_{y}$ is the centripetal acceleration a$_{c}$

so

ƩF$_{y}$ = mv$^{2}$/r = mrω$^{2}$

and substituting in N and mg for ƩF$_{y}$

N-mg = mrω$^{2}$

since we are looking for N

N = mrω$^{2}$+mg

and so when I substitute the values of m, r, ω, and g I get

N = 61.16kg(6.37 x 10$^{6}$m)(7.27 x 10$^{-5}$rad/s)$^{2}$ + 61.16kg(9.81m/s$^{2}$)

solving for N I get 602.06N

But I thought that you weighed less at the equator? I believe I missed something fundamental, can someone please help me out here? Thank you for your time.

2. Dec 7, 2011

### Matterwave

It should be mg-N=mrω^2, since the centripetal acceleration points inwards towards the center, and so does mg, but not N.

3. Dec 8, 2011

### Legaldose

Oh! So at that point I should take downward as positive and solve for N = mg-mrω^2? That makes more sense I don't know why I didn't remember that the centripetal acceleration would be pointed towards the center of gravity as well, thank you.