# I Centrifugal Force Experienced at the Earth's North Pole

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1. Apr 11, 2018

If we standing in the equator, then centrifugal force caused by earth rotation directly balanced by gravity force. But what if we standing in the high altitude or in the pole? In the pole at the distance like 40 km from north pole (so the north pole inside the horizon plane), the gravity have no component parallel to the ground, because gravity perpendicular to the ground, so we dont have experienced force by gravity to the left, right, back, or front. But the earth rotation makes us experienced centrifugal force that have component (projection) parallel to the ground. So at this distance we must travel with linear speed like 2 m/s. But why we never experienced this centrifugal force at the pole?

2. Apr 11, 2018

### Staff: Mentor

Don't forget the normal force of the earth on your feet. The centrifugal force at the equator is quite small. (And much less near the poles, of course.)

3. Apr 11, 2018

### gmax137

What acceleration do you calculate this causes? Compared to the acceleration due to gravity (9.8 m/sec^2)? Try the calculation. I get a very tiny number.

4. Apr 11, 2018

### jbriggs444

But is gravity, in fact, perpendicular to the ground? The ground, on average, follows a "level" that is perpendicular to the combined effect of gravity and centrifugal force. In general, it will not be perpendicular to gravity.

The radius of the Earth on a line from center to pole is less by about 32 km than the radius of the earth on a line from center to equator. On average, the ground follows an equipotential surface that slopes downward toward the pole at an angle that is always just enough to compensate for centrifugal force at the local latitude.

The force of apparent gravity is perpendicular [on average] to the local surface.

Last edited: Apr 11, 2018
5. Apr 11, 2018

### Staff: Mentor

Is it possible that you are using the wrong word in your question. You said centrifugal force, but are you really asking about the Coriolis Force? https://en.wikipedia.org/wiki/Coriolis_force The Coriolis force makes hurricanes, cyclones and typhoons spin around in spirals.

If not, then I don't understand your question. The only component of gravity is from you to the center of the earth. It is perpendicular to the surface in most places. That is true at the poles and at the equator.

6. Apr 16, 2018

If we ride a Merry Go Round, even if it have a normal force that balanced our weight, we still experienced centrifugal force.

7. Apr 16, 2018

Gravity perpendicular to the ground. But there is still centrifugal force if we travel at circle path. And in the pole our rotation plane not concentric with gravitation plane.

8. Apr 16, 2018

### ZapperZ

Staff Emeritus
You are now turning things into what it is not.

On the merry-go-round, the normal force (and gravity) is perpendicular to the "centrifugal force".

This is different than the case of us standing at the equator. Here, the normal force (and gravity) is co-linear with the "centrifugal force", not perpendicular!

So you are comparing apples to oranges here.

BTW, why are we dealing in "centrifugal force" here? Are we trying to sharpen our skills in dealing with fictitious forces?

Zz.

9. Apr 16, 2018

But it is still effected our motion. We will lost our balance. Gravity with acceleration 9.8 m/s^2 will cause a very strong force, like what we experienced when we hanging in the tree. But a small acceleration still make a significant force that can be our experienced.

10. Apr 16, 2018

Not at the equator, but at the pole. The centrifugal force have a projection (component) that parallel to the ground. And this component of force perpendicular to the normal/gravity force. So there is a net force that perpendicular to the gravity force. Say in y axis there is gravity and normal force so both cancel out. But in x axis the centrifugal not cancel out by any other force.

11. Apr 16, 2018

### gmax137

really? What is the magnitude of the acceleration? Show me the number xx m/sec^2.

12. Apr 16, 2018

it is around 0.06 m/s^2. Dont compare that with earth gravity acceleration.

13. Apr 16, 2018

### ZapperZ

Staff Emeritus
Actually, technically, NO. Right at the pole, you have "centrifugal forces" acting equally in all directions, meaning that they cancel out to be zero! This force is largest at the equator and goes to zero at the poles!

You have a very severe misunderstanding of this physics.

Zz.

14. Apr 16, 2018

Sory, centrifugal force always act outward the axis of rotation.

15. Apr 16, 2018

### Staff: Mentor

Let's get specific. You're standing at the north pole. (Silly, I know, but I'm trying to understand your issue.) You hold a book in your outstretched hand. Please estimate the centrifugal force you'd feel.

16. Apr 16, 2018

### Staff: Mentor

If a man stands at the pole, he rotates around his own vertical axis once every 24 hours. If he extends his arms, there will be a centrifugal force on his hands. Is that what you are asking about?

The magnitude of that force would be very small, almost too small to measure.

Edit: I see @Doc Al beat me to the answer.

17. Apr 16, 2018

### ZapperZ

Staff Emeritus
Yeah, so? The axis of rotation points vertically UP (or down) at the poles. If you think this is the direction of the "centrifugal force", then you are very mistaken, and re-enforces my claim that you have a severe misunderstanding of this concept.

Centrifugal force is ZERO at the poles.

Zz.

18. Apr 16, 2018

But if the earth rotating if we place perfect ball in a flat surface, then the acceleration around 0.06 m/s^2 is enough to move that ball away from axis of rotation. If it have initial velocity 0 m/s then during an hour it will have velocity 216 m/s.

19. Apr 16, 2018

### Staff: Mentor

Are you talking about at the pole? Then please estimate the centrifugal force that the ball would experience. (Make some assumptions, similar to extending your arms to the side.)

20. Apr 16, 2018

### gmax137

I calculate a smaller number (0.0002 m/sec^2) for the 40 km case you described in the OP. Either way, what makes you think you can feel such a small acceleration? You might have to "lean in" a fraction of a degree to compensate.