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Let's say an object is at rest in Earth's reference frame. We know that Earth's reference frame is non inertial. If we were to observe that object from an inertial frame, we would see three forces acting on it (ignoring resistive forces like air drag, etc). The three forces are gravity $mg$, Normal reaction $N$, and a 3rd force which imparts a centripetal acceleration to the object which makes it spin (along with the Earth). Which force provides that centripetal acceleration? Is it friction or some other force?

Edit : Ignoring Earth's revolution around the Sun in this case and just taking into account its spin about its axis of rotation

Qmechanic
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4d_
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3 Answers3

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The diagram below depicts the source of the centripetal force.

Every rotating celestial body that is large enough to pull itself into an energy minimizing shape has an equatorial bulge. (Over geologic time scale the solid rock of the Earth's tectonic plates is ductile. )

The diagram shows a celestial body with a far larger equatorial bulge than the Earth; for clarity the equatorial bulge is exaggerated.

Resultant force arising from equatorial bulge

Any buoyant object is subject to a normal force (red arrow).
When a celestial body has an equatorial bulge the gravitational force is not exactly opposite in direction to the normal force. So there is a resultant force. From here on I will refer to that resultant force as the poleward force.

In the case of the Earth: polar radius is 6357 kilometer, equatorial radius is 6378 kilometer. The difference is 21 kilometer.

This means that from the equator to the pole is a downhill slope. If that slope would not be there then the water of the Earth would flow to the Equator. The Earth's equatorial bulge prevents that.

Example: at 45 degrees latitude the slope is 0.1 degree. That slope provides the required centripetal force to remain co-rotating with the Earth. At 45 degrees, to find the amount of required centripetal force you divide by 580, that is the ratio.

Divide your own weight by 580, that is how much centripetal force is required for you (at 45 degrees latitude). If you have a weighing utensil you can push: that will give you a feel for it.


I noticed that other answers suggest that the required centripetal force is provided by friction. That is a whopping mistake.

70 percent of the Earth's surface is ocean, and it is not the case that the oceans are deeper at the equator. There are differences in ocean depth, but they are not correlated with latitude.

For the water in the oceans there is no friction available. The oceans are the same depth at the equator because the equatorially bulged shape of solid Earth provides just the right amount of poleward force.



About the process that arrived at the current equatorial bulge:

Prior to formation the material that was about to form the Earth was distibuted in the form of a proto-planetary disk. As the planet formed the blob of material became more and more spherical. Geologists have reconstructed that shortly after formation the rotation period of the Earth was about 6 hours; 4 times faster than today.

When a forming celestial body contracts to a more spherical form there is release of gravitational potential energy. That gravitational potential energy turns into rotational kinetic energy. That is: the contraction tends to speed up the rate of rotation. At some point the celestial body cannot contract further, as that would cost more energy than would be released.

Tidal interactions with the Moon have been slowing down the Earth's rotation thoughout the Earth's existence. Over geologic time scale the Earth as a whole is sufficiently ductile to act effectively as a fluid. A very, very viscous fluid, but fluid. For more on that: see the pitch drop experiment

In celestical mechanics the shape of a rotating celestial body is referred to as 'hydrostatic equilibrium'. With hundreds of millions of years available to adjust a celestial body has the same shape as a celestial body that is entirely an easy flowing fluid.

Cleonis
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  • great effort, deleted my answer in response to the point you raised. However I do have one question. What do you mean by a "downhill slope"? What is this "slope"? Clearly it is not the slope of the tangent at a point, given that you say the slope at $45°$ latitude is $0.1°$ – Ritam_Dasgupta May 15 '21 at 19:42
  • Another point: in my first year, I was taught that the effective value of $g$, considering rotation of earth, at a latitude $\lambda$ is given by $g_{eff}=g-\omega^2 R\cos^2(\lambda)$. This is only correct if we consider earth to be spherical, that is, ignore the bulge, right? – Ritam_Dasgupta May 15 '21 at 19:44
  • @Ritam_Dasgupta At the poles and the equator the normal force (red in the diagram) and the Earth gravitational force are exactly opposite. We can call that an angle of 180 degrees, we can call that 0 degrees; the context makes clear what is intended. At 45° latitude: technically the normal force and local Earth gravitational force are an an angle to each other of 179.9 degrees, but that is unwieldy. The 0.1 degree angle can mean only one thing. Note: when you measure local level (with a water level) then what you measure is the resultant of Earth gravitational force and centrifugal effect. – Cleonis May 15 '21 at 19:52
  • Alright, I get it. And my second question? – Ritam_Dasgupta May 15 '21 at 20:08
  • @Ritam_Dasgupta The formula you cite is not about whether or not the bulge is ignored; it is independent of that. At the equator some of the Earth gravitational force is expended in providing the required centripetal force for circumnavigating the Earth's axis at 6400 kilometer distance, one revolution per day. At the equator that is an acceleration of about 0.034 $m/s^2$ Providing required centripetal force and 'centrifugal effect' are two names for the same thing. At all latitudes there is some centrifugal effect. The magnitude of that can be calculated. – Cleonis May 15 '21 at 20:10
  • @Ritam_Dasgupta Just to be clear: it was not necessary for me to derive the shape of the Earth. I calculate how much centripetal acceleration is required at 45 degrees latitude. I compare the magnitude of that with the Earth average gravity. That gives me the ratio of the two, that ratio gives me how much downhill slope there must be in order to provide the required centripetal force. At the same time: gravitational mass and inertial mass are equivalent; they cannot be distinguished. So we do have that at every latitude a plumb line is perpendicular to the surface of the bulged Earth.. – Cleonis May 15 '21 at 20:36
  • No no, but for the derivation of that formula we had considered the normal reaction to be exactly the opposite direction of gravity. The crux of your argument is that, that is in fact not true. – Ritam_Dasgupta May 15 '21 at 20:49
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The details of this setup depend on where on earth you are.

On either of the geographical poles, the object would simply spin around its axis along with earth, no centripetal force occurs. On the equator, the centripetal force keeping it on the earth and making it spin along with it is purely gravity. The centrifugal (apparent) force is pointing straight up, gravity is pointing straight down. A stationary object on the surface will experience a much greater gravitational force than centrifugal force, so the object stays on the ground and spins along with the earth.

The situation is a bit more complicated anywhere in between these points. I will describe the abstract situation in which the earth is a perfect sphere. Other answers also address its actual oblate geometry.

Between the poles and the equator, gravity is pointing straight down, but the centrifugal force is pointing outwards perpendicular to the rotational axis. This means, standing on the ground, the centrifugal force is pointing diagonally up, where its horizontal component (the one parallel to the surface) is directed at the closest point on the equator (or, equivalently, the more distant pole). The vertical component is again cancelled by gravity, but the horizontal part would have to be compensated by some frictional force. If the earth was a perfect sphere with a frictionless surface, all objects left alone on it would collect at the equator. As other answers correctly point out, this effect is counteracted by the oblate shape of the actual earth.

noah
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  • Thank you. While I understood the explanation, I have a couple of doubts. 1) Why would you say that the horizontal component is directed at the equator? Earth is not flat, it's spherical right? 2) I want to specifically analyse the object from an inertial frame. No centrifugal force in that frame. Want to know which force is making it spin along with the Earth? – 4d_ May 15 '21 at 17:24
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  • The correct wording is possibly "the closest point on the equator" which is equivalent to "towards the more distant pole". Even though earth is not flat, there is still a sense of direction. So on the northern hemisphere the component parallel to the surface is pointing south, in the southern hemisphere it's pointing north.
    • – noah May 15 '21 at 17:28
  • It's the same story in an intertial frame, but it's easier to visualize in earth's frame. The body's inertia wants it to go straight ahead, which means off the surface at the angle described in the answer. To keep the object stationary on the surface, gravity is pulling it down, but the inertia still pushes the object towards the equator. This is what we need friction for.
    • – noah May 15 '21 at 17:32
  • You are proposing "some frictional force", but 70 percent of the Earth is covered in ocean. For the water in the oceans there is no friction force. As we know, at the equator the oceans are not particularly deeper than elsewhere. According to your reasoning the water of the oceans should all flow to the equator. – Cleonis May 15 '21 at 19:08
  • @Cleonis There is significant drag in water, no friction would make swimming and boating really easy. And I did not give any orders of magnitude. The effect is very tiny compared to the gravitational force. We also need to take into account that "piling up" mass at the equator results in higher potential energy, resisting this drift (especially for liquids). The point I was making was that if you had a perfectly frictionless, spinning sphere and put some objects on it, they would drift towards the equator (without stacking on top of each other). – noah May 15 '21 at 19:40
  • @noah I surmise that your answer isn't about the actual Earth but that instead you are discussing a thought experiment with a spherical celestial body that is rotating. You should say so explicitly. Please note that the question is about the actual Earth. – Cleonis May 15 '21 at 19:58
  • @Cleonis I take your point, I hope it's clearer now. – noah May 15 '21 at 20:52