Centrifugal Force Experienced at the Earth's North Pole

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Discussion Overview

The discussion revolves around the concept of centrifugal force experienced at the Earth's North Pole, particularly in relation to gravity and the effects of Earth's rotation. Participants explore theoretical implications, calculations, and the nature of forces acting on objects at the poles versus the equator.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that at the equator, centrifugal force is balanced by gravity, but question how this balance changes at high altitudes or at the poles.
  • There is a discussion about the nature of gravity being perpendicular to the ground and whether this holds true at the poles.
  • One participant suggests that centrifugal force has a component parallel to the ground at the poles, leading to a net force that does not cancel out with gravity.
  • Others argue that at the poles, centrifugal forces act equally in all directions and thus cancel out to zero, challenging the idea of experiencing centrifugal force there.
  • Participants engage in calculations of acceleration due to centrifugal force and compare it to gravitational acceleration, with some noting that the values are very small.
  • There is confusion about the terminology used, with references made to the Coriolis force and whether it is relevant to the discussion of centrifugal force.
  • Some participants emphasize that even small accelerations can affect balance, while others question the significance of these forces at the poles.

Areas of Agreement / Disagreement

Participants express differing views on the existence and effects of centrifugal force at the poles, with no consensus reached. Some believe it has a significant effect, while others argue it cancels out entirely.

Contextual Notes

Discussions include assumptions about the definitions of forces and the conditions under which they apply. Participants also note the importance of considering the geometry of the Earth when discussing these forces.

Mohammad Fajar
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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 don't 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?
 
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Mohammad Fajar said:
If we standing in the equator, then centrifugal force caused by Earth rotation directly balanced by gravity force.
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.)
 
Mohammad Fajar said:
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?

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.
 
Mohammad Fajar said:
because gravity perpendicular to the ground
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:
@Mohammad Fajar ,

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.
 
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Doc Al said:
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.)
If we ride a Merry Go Round, even if it have a normal force that balanced our weight, we still experienced centrifugal force.
 
anorlunda said:
@Mohammad Fajar ,

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.

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

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.
 
gmax137 said:
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.
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
ZapperZ said:
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.
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
Mohammad Fajar said:
We will lost our balance.

really? What is the magnitude of the acceleration? Show me the number xx m/sec^2.
 
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  • #12
gmax137 said:
really? What is the magnitude of the acceleration? Show me the number xx m/sec^2.
it is around 0.06 m/s^2. Dont compare that with Earth gravity acceleration.
 
  • #13
Mohammad Fajar said:
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.

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
ZapperZ said:
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.

Sory, centrifugal force always act outward the axis of rotation.
 
  • #15
Mohammad Fajar said:
Sory, centrifugal force always act outward the axis of rotation.
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
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
Mohammad Fajar said:
Sory, centrifugal force always act outward the axis of rotation.

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
Doc Al said:
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.
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
Mohammad Fajar said:
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.
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
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.
 
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  • #21
anorlunda said:
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.
He rotated around the Earth axis.
 
  • #22
Mohammad Fajar said:
He rotated around the Earth axis.

The Earth axis and the axis in his body are co-linear when he is at the pole.

So you standing at the pole holding out a book (as @Doc Al suggested) will experience the same centrifugal force as you will by standing up right now, holding out a book, then using your feet to rotate your body one revolution per 24 hours.
 
  • #23
gmax137 said:
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.
can you post your calculation?
even if that have a small acceleation it still created a force. Just multiply that with an hour, 3600 second, and you will get that your velocity away from the Earth axis.
 
  • #24
anorlunda said:
The Earth axis and the axis in his body are co-linear when he is at the pole.

So you standing at the pole holding out a book (as @Doc Al suggested) will experience the same centrifugal force as you will by standing up right now, holding out a book, then using your feet to rotate your body one revolution per 24 hours.
Not like that. If the Earth rotated, then if we place a perfect ball at the pole it will move away from the pole during course of the day.
 
  • #25
Thread temporarily closed for moderation.

@Mohammad Fajar , if you want to continue this thread, send me a message. To do that, click on my name and then "start a conversation"
 

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