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Initially, I was looking for how centripetal force is produced on the surface of the rotating earth for a mass kept at any latitude. I went through the following threads -

  1. Which force provides the centripetal acceleration that makes objects on earth's surface rotate about Earth's axis of rotation?
  2. Is the normal force equal to weight if we take the rotation of Earth into account?
  3. Question about the Normal Force exerted by Planet Earth in relation to centripetal force
  4. If the ground's normal force cancels gravity, how does a person keep rotating with the Earth?

From there, I understood that the resultant of normal force (N) and gravity (mg) is the required centripetal force. But what is bothering me now is HOW? According to the answers, the normal force is slanted such that it is not exactly opposite to gravity. Thus, they don't cancel out, resulting in a horizontal centripetal force. I'm still confused and have the following questions :

  1. Why is normal force slanted in the first place ? (Is it because of the earth's bulge, friction or centrifugal force?)

  2. I think there's also a vertically-upwards component of the resultant, why is that? (the resultant of normal force and gravity)

enter image description hereenter image description here This is what I see, what is the reason behind this?

Source of the image - https://en.wikipedia.org/wiki/Equatorial_bulge

Edit - I've gone through this question Is the normal force equal to weight if we take the rotation of Earth into account? but this doesn't clear my doubt regarding the upward component of the resultant of gravity and the normal force. I posted this question because I wanted some more insight into that poleward force and its relation to the bulge of the earth, which isn't emphasised in that question. kindly reopen my question.

Qmechanic
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Apogee Point
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2 Answers2

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First, some preliminaries. If there is no force on you, you are traveling in a straight line at a constant speed. (Perhaps $0$ speed.) If you are traveling in a circle, the total force on you is deflecting you from a straight line into a circle. That total force is centripetal force. The total acceleration is centripetal acceleration.

The total force may be the sum of multiple forces. These forces might be gravity, normal force from the surface of the Earth, friction, and anything else acting on you. So the question is how do these forces arrange themselves so they add up to the total needed to keep you moving in a circle?

Take a look at my answer to Why does a metal ball not trace back its original path if it hits a wall?. It says why the force from a rigid object like the surface of the Earth is divided into two parts. A reaction force perpendicular to the surface is present because the object is rigid. And a friction force along the surface may or may not be present.

Let's start with a simple situation. Suppose the Earth was a perfect sphere and not rotating. Gravity and the reaction force are equal and opposite. You are at rest.

Suppose the Earth was a rotating perfect sphere of ice. At a particular instant, you are on the equator moving at the same speed and direction as the surface. Again the normal force and gravity are perpendicular. The normal force is a little weaker that gravity. The total is just right for you to move in a circle.

If you are not on the equator, gravity and the reaction force don't add up to a force in the right direction to keep you moving in a circle around the axis. If you are moving in a circle, there must be another force. Perhaps you hold on to the surface. Looking at your diagram, what force do you need to make the total add up to the purple arrow?

The reaction force will adjust itself to keep you on the surface. Gravity pulls you onto the surface. The reaction keeps you from going beneath it. Gravity will be a little stronger. You need to pull yourself along the surface in the direction of the blue arrow. If you do, you will stay fixed to one spot on the Earth as it rotates. If you don't you will slide in the direction opposite the blue arrow.

That leaves the question of why the normal reaction force is not aligned with gravity. It has to do with the Earth's bulge.

Earth is not perfectly rigid. The interior is liquid. The ocean is liquid. The crust is solid rock. But over millions of years, rock bends and flows as this example from Eastern Connecticut State University shows. See Structural Geology Photos - Folds

If you have water sitting where you are, it cannot hold on. It will flow toward the equator. Over time, so will rock. It will pile up. Lower latitudes will become fatter. The shape of the surface will change like the diagram shows. When the shape has changed enough that the tilted reaction forces provides needed blue component, there is no further pull toward the equator.

enter image description here

mmesser314
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    As we know: when the solar system formed all the clumps of matter that would go on to form the planets started out as protoplanetary disks. The proto-Earth contracted from disk towards spherical shape in a process of dissipation of rotational kinetic energy. In the process of contraction gravitational potential energy is released, and converts to rotational kinetic energy. The process of contraction halts at the point where further contraction would require more input of rotational kinetic energy than is released by the contraction.. – Cleonis Jun 10 '23 at 06:53
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    I added the comment about contraction from proto-planetary disk because of the following: in any tabletop demonstration or amusement park ride: we see an increase of radial distance as a system is given rotational velocity. Explaining the Earth's equatorial bulge in terms of 'lower latitudes will become fatter' will be the first thing that comes to mind. So there's a bit of a dilemma there. The actual Earth did not get to its current equatorial bulge from a spherical starting point. But then: the OP is already struggling badly, so I can see the need to keep the explanation very accessible. – Cleonis Jun 10 '23 at 07:03
  • Thanks for the answer. There are a few things which I'm unclear about. First of all, why "The normal force is a little weaker than gravity." If both gravity and normal are perpendicular to the equator, why would normal be slightly lesser than gravity? – Apogee Point Jun 10 '23 at 09:59
  • Also, kindly elaborate on this part - "If you are moving in a circle, there must be another force. Perhaps you hold on to the surface." – Apogee Point Jun 10 '23 at 10:00
  • "Looking at your diagram, what force do you need to make the total add up to the purple arrow?" - I think we would need a force equal in magnitude and opposite in direction to the yellow vector, that way we'll only be left with purple vector i.e, centripetal force as a resultant. Is this right? – Apogee Point Jun 10 '23 at 10:17
  • @Shreya Given the sheer amount of follow-up questions: in effect you are asking mmesser314 to spoonfeed you. That is not a reasonable thing to ask. The yellow arrow that you added in the diagram makes no sense. The very fact that you placed it in there suggests you are into a pervasive misapprehension. I suggest you try the following forum: physicsforums.com The reason I suggest going there: physicsforums is a threaded forum. That means you can keep asking follow-up questions. (It may be - I don't know - that at some point people will stop answering.) – Cleonis Jun 10 '23 at 15:04
  • @Cleonis - You have a point about the formation of the Earth. But this is why it stays almost spherical, even with processes going on that form mountains and valleys. – mmesser314 Jun 10 '23 at 15:39
  • @Shreya - If you are moving in a circle, you are accelerating toward the center. The total force is toward the center. As the Earth orbits the Sun, the only force on it is gravity from the Sun. It is the same as you rotate with the surface of the Earth. If the reaction force was just as strong as gravity, you would have a total force of $0$. You would travel in a straight line. This would take you off the surface. The surface cannot push hard enough to lift you off. – mmesser314 Jun 10 '23 at 15:46
  • @Cleonis - There is a point at which I will stop answering. But so far I see reasonable conceptual confusions, and some thought going in to resolving them. I may draw the line at the yellow arrow. That sounded like a guess. Shreya needs to think about what has been said so far. It may become clearer with more thought. – mmesser314 Jun 10 '23 at 15:50
  • To Cleonis and mmesser314, Sorry for causing trouble. I did do a bit of studying on my part before asking, and didn't intend any 'spoonfeeding'. Ultimately, I figured out the answer, so thanks a lot for answering :) – Apogee Point Jun 10 '23 at 18:47
  • You did not cause trouble. You asked a good question and your thought showed. We do get a lot of questions from people who just want their homework done for them. We don't answer that kind of question. You are clearly not doing that. I am glad you thought some more and figured it out. The answer is clearer when you do. If you have more questions, be sure to ask. We also close questions that have been asked before, just to keep the number of repeats down. It is no reflection on you. – mmesser314 Jun 10 '23 at 18:57
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EDIT: Added an animation at the end that might be easier to follow.

First let's consider a simpler model of a perfect spherical planet rotating about the vertical axis with North at the top.

Diagram 1:

enter image description here

The only real force present is the force of gravity (a) acting inwards. This force is broken down into two components, (g) and (c). (g) is the centripetal force which is just a component of the gravitational force and there is no balancing force for this component, so the force accelerates the particle inwards and this keeps the particle moving in a circle as the planet rotates. This leaves component force (c) which is dissected into two more components (d) and (e) in the diagram below.

Diagram 2:

enter image description here

(d) is the inward weight force and (f) is a balancing reaction force. (e) is a sideways force and if there is insufficient friction, this unbalanced force tends to accelerate everything on the surface towards the equator. If the Earth was a sphere, this sideways force would eventually deform the Earth and any surface liquid tends to pile up at the equator. (The molten rock below the surface ensures the Earth is not a perfectly rigid object)

Eventually an equilibrium point is reached where the sphere becomes oblate as in the diagram below. The oblateness of the Earth is greatly exaggerated (as is the centripetal/centrifugal force) for clarity.

Diagram 3:

enter image description here

Again, the only real force is the force of gravity (a) acting towards the centre. (c) is the inward component normal to the surface that is felt as weight and (f) is the reaction force to (c). Component force (g) is not balanced and provides the inward centripetal acceleration. Since g is fully accounted for as providing the centripetal force, the only forces left are component (c) and reaction force (f). Since these are parallel and both normal to the surface, there is no longer any sideways force on surface particles. You can view the centripetal force simply as a component of the inward gravitational force. You could interpret the centrifugal force as the resultant of forces (a) and (f), but this is difficult to justify, as force (f) is cancelled by force (c). It also worth noting that the reaction force (f) is not really a 'tilted' normal. It is perpendicular or normal to the surface. It is only tilted in the sense that is not parallel to the inward acting gravity force (a). It is also worth noting that something like the Earth that has had millions of years to find equilibrium should be almost the perfect oblate shape required for equilibrium.

===========================================================

Food for thought: If you build a tall tower using a plumb bob to ensure it is vertical, will the tower be parallel to (f) or (a)?

P.S. While I narrated this from the point of view of starting with a sphere for simplicity, it should be obvious from the points above that if the the Earth starts as an over oblate smarty shaped planetoid, then the sideways forces will tend to contract the smarty to something closer to spherical at the equilibrium point.

Edit: Decided to add this final free body diagram to illustrate the last point. It represents a zoomed in portion of an over oblate planetoid.

Diagram 4:

enter image description here

(a) is the centre pointing gravitational force. (g) is the component of (a) that provides the centripetal force. That leaves component(c) which is subdivided into two more components, (d) which is the downward normal force, which is balanced by the reaction normal force (f) and sideways force (e) which is not balanced and drives surface material towards the pole until the equilibrium oblate shape is achieved.

Just a doubt. In the first diagram why don't you put a component of the normal force from the ground projected in the direction opposite to the centripetal force? – Mark_Bell

The force opposite to the centripetal force (in the non rotating reference frame) is called the centrifugal force. This is a 'fictitious' force and does not appear in an inertial reference frame. If there was a counter force balancing the centripetal force, there would be no net force to keep the particle moving in a circular path. In the reference frame of a non inertial observer rotating with the Earth, the centrifugal force is real.

enter image description here

In the above animation, the initial primary forces are the gravitational force (a) and the centrifugal force (b) as seen by an observer on the surface rotating with the planet. Force a is resolved to component (d) that opposes the centrifugal force and component (c). The centripetal force (d) and the centrifugal force (b) cancel each other and there is no motion parallel to the equatorial plane. This leaves component (c) that is resolved to components (f) normal to the surface and (e) tangential to the surface. There is a reaction force (g) also normal to the surface and this cancels the force (f). The only force left is component (e) tangential to the surface that acts towards the equator. This could be opposed by sufficient friction force, but in the absence of friction (or very little friction such as surface liquid) the representative particle is propelled towards the equator. In the inertial reference frame, the fictitious centrifugal force vanishes and there is no force to oppose the centrifugal force and the particle follows a circular path as seen from the North pole.

The interactive animation is here. The oblateness of the spheroid can be altered with the R slider to find the equilibrium point (around R=5) and the over oblate case can be investigated, where the residual force is towards the pole. The animation is controlled by the video play button at the bottom left or alternatively gone through step by step or in slow motion using the T slider.

KDP
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  • Just a doubt. In the first diagram why don't you put a component of the normal force from the ground projected in the direction opposite to the centripetal force? – Mark_Bell Feb 22 '24 at 08:07
  • @Mark_Bell Added a paragraph to address your doubt. – KDP Feb 22 '24 at 16:44
  • @Mark_Bell Added the promised diagram for the co-rotating reference frame in the form of an animation gif with a link to the interactive app. – KDP Feb 22 '24 at 23:59
  • I am not talking about the centrifugal force that is a pseudo force because arise just from the fact that we consider a non inertial frame. I was saying this: in an inertial reference frame you have just two real force, gravity and the normal force of the ground (plus friction if you want). N has two component if you want, one in the radial direction and another in the direction of rotation axis of the Earth – Mark_Bell Feb 23 '24 at 19:39
  • Lets define a cartesian coordinate system with z running from South to North and the equator lying on the x,y plane. Please clarify what you mean by "another in the direction of rotation axis of the Earth". Do you mean a *force* parallel to the x axis or a tangential *velocity* parallel to the x,y plane? – KDP Feb 23 '24 at 20:48
  • Two components of N the normal force, one in the z direction (this is what i called the axis of rotation) and the other lying in the xy plane, in the direction of the centripetal force – Mark_Bell Feb 23 '24 at 22:40
  • OK, let's say the force of gravity points at the centre. this decompososes into two components, one of which is the centripetal force parallel to the xy plane and pointing at the spin axis Z. The other component points downwards and is parallel to spin axis z. We can ignore the centripetal force because it is an unopposed force that provides the centripetal acceleration that keeps the particle moving in a circle when looking from above. The remaining force is not a normal force, because a normal force is perpendicular to a surface. This remaining force decomposes into two components. .. – KDP Feb 23 '24 at 23:31
  • One is a real normal force perpendicular to the surface pointing directly at the centre and there is another matching reaction force pointing directly away from the centre ensuring the particle neither sinks into the ground or rises above the surface. The remaining sub component is a force tangential to the surface and pointing in the general direction of the equator. (If we attach a spring scale that is attached to the particle and to the surface of the Earth, it will indicate a real proper force pointing to the Equator). This tangential force propels the particle to the equator. – KDP Feb 23 '24 at 23:34
  • To put it more simply, so i can explain better myself. We have $ m\vec{a} = m\vec{g} + \vec{N} + \vec{F}_f $, where $ \vec{F}_f $ is an eventual friction. This is in general in an inertial frame. I can project this equation in whatever axis i choose. So if i choose the axis as we said, z in the direction from South to North and another axis in the plane xy, The vector N, the reaction from the ground in this case is decomposed in the z and in the perpendicular direction. There is nothing special, i can rotate the reference frame, but i can always decompose a vector in two or three components – Mark_Bell Feb 24 '24 at 08:08
  • I think we are in agreement now. – KDP Feb 24 '24 at 17:18
  • (a) is not a component of the force of gravity acting on the surface object, *It is the force of gravity acting on the object.* – KDP Feb 28 '24 at 05:13
  • Yes, but i believe your answer is not correct. In particolar the first part. You say that the component of the gravity (a) that you have called (g) is the only part that contributes to the centripetal force. If this is true so $F_{c} = mgcos(\theta) $ where $ \theta $ is the angle between the gravity vector and horizontal axis. As you can see if you are in the mid you have a centripetal acceleration of g/2, but in reality the centripetal acceleration is roughly $ 0.03 \frac{m}{s^{2}} $ – Mark_Bell Feb 28 '24 at 05:18
  • What reality are you thinking of? The diagram with the g vector, is the hypothetical case of an over oblate planet such as a protoplanet before it reaches equilibrium. This has never actually been measured in reality. – KDP Feb 28 '24 at 05:24
  • "where θ is the angle between the gravity vector and horizontal axis" so $\theta$ is the angle between vector g and and vector c or do you mean horizontal is tangential to the surface? – KDP Feb 28 '24 at 05:27
  • Oh wait, the g vector is also in the first diagram which is for a perfectly spherical rotating planet (which the Earth is not) so again it a hypothetical example and has never been measured in reality. – KDP Feb 28 '24 at 05:32
  • First diagram. Angle between g and a – Mark_Bell Feb 28 '24 at 05:34
  • Starting with $cos(\theta) = adj/hyp \ $ then $adj = hyp \ \cos (\theta) = F_a \ \cos(\theta) = F_g$. For force c the equation is $F_{c} = F_a \ \sin (\theta) = mg \ \sin(\theta).$ – KDP Feb 28 '24 at 05:46
  • Where is your doubt in the 'big picture'? Do you doubt that a rotating sphere will experience surface forces that move material towards the equator until the equilibrium oblate sphere shape is arrived at and that an over oblate sphere or disk will experience surface forces that move material from near the equator towards the poles until the equilibrium oblate sphere is arrived at from the opposite direction? – KDP Feb 28 '24 at 06:01