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A body at the surface of earth experiences at least two obvious forces - gravity and its weight (normal of the surface). Both of these forces act on the radial axis that goes through the center of earth.

fbd.jpg

But for a body to rotate about the rotation axis of earth, it needs to accelerate towards the said axis. This acceleration is perpendicular to this axis and maintains an angle (equal to the latitude) with the radial axis where the forces act along.

kd.jpg

My question is, how does these forces create this acceleration and nothing else. What am I missing?

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5 Answers 5

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Gravity always acts to the center of the Earth (if you assume the Earth as a sphere). The rotation (which has nothing to do with gravity) creates an additional centrifugal force acting outwards perpendicular to the rotation axis. This means you will weigh a little bit less the more you go from the pole to the equator.

I have given the mathematical derivation of the direction of gravity on the Earth's surface in an answer to a related question. The result is that for mid-latitudes this direction misses the center of the Earth by about 0.2 deg, half of which is due to ellipsoidal shape of the Earth and the other half due to the centrifugal force because of the rotation. This is consistent with published figures.

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    $\begingroup$ Why should we bring centrifugal force at all? Clearly I am talking about the cause (FBD) and the effect (KD). Centrifugal force is neither a cause, nor an effect. It is a pseudo force that we need when considering a non-inertial reference frame. And the weight part is completely irrelevant to my question. Explain why the causes don't add up to make the mentioned effect $\endgroup$ May 1 at 8:50
  • $\begingroup$ @ImtiazKabir The centrifugal force is the effect of the rotation. It has nothing to do with gravity at all. It would be the same if you just had a rotating hollow sphere of zero mass. Only that you would fly off then. With the gravity as it is it only results in a weight reduction $\endgroup$
    – Thomas
    May 1 at 9:06
  • $\begingroup$ I think we disagree on what centrifugal force is. As I understand, this is a non-existent force so it can't push or pull you. So what is it? A rotating body can't explain the motion of any object with fundamental forces. That is because it is in a non-inertial reference frame. To explain other motion, it borrows a non-existent (fictitious) force that acts on every object. This force is parallel to the radial axis and points outward from the center. $\endgroup$ May 1 at 11:36
  • $\begingroup$ @ImtiazKabir you can't really disagree on centrifugal force here, it's a real force in the context of the frame in which it exists, it's called a pseudoforce because it only arises due to the rotation of the reference frame. But it's effects are real and measurable, just as the transverse and Coriolis forces are $\endgroup$
    – Triatticus
    May 1 at 15:00
  • $\begingroup$ @ImtiazKabir You were asking what forces an object on the surface on the Earth experiences, so this object is at rest in a rotating reference frame, which means the centrifugal force is a real force for this object and this adds to the gravitational force the object experiences. $\endgroup$
    – Thomas
    May 1 at 15:10
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Both of these forces act on the radial axis that goes through the center of earth. … My question is, how does these forces create this acceleration and nothing else. What am I missing?

You are missing the fact that neither of those forces actually go through the center of the earth.

First, the contact force between the earth and the body has both a vertical and a horizontal component. This horizontal component arises from static friction and is often larger than the centripetal acceleration. This is the force, for example, that keeps the object from sliding down a hill or sliding around from a light breeze. It also provides whatever small force is needed for the centripetal motion.

Second, the earth is not uniform. There are variations in density and local features like mountains and so forth. All of these perturb the local gravitational vector so that it does not actually point towards the center of the earth. If you took those deviations and constructed a shape such that the surface is everywhere perpendicular to the local gravitational vector, that shape is called the geoid. That shape is, in some sense, the actual shape of the earth, and on it there would be no static friction. The force for the centripetal motion would come from the sum of the normal force and the local gravitational force, neither of which would necessarily point towards the center.

Now, suppose that we were not on the earth, but rather on a perfectly uniform density planet with the same mass and angular velocity as the earth. Suppose further that the planet is formed of some fluid so that there are no mountains or hills and that there is no wind or moons so there are no waves or tides. Such a planet would have everywhere a gravitational vector that points exactly to the center of the earth, and since it is a fluid the surface is everywhere normal to the local gravitational force so there is no horizontal forces. Now, on such an idealized planet the geoid would be an ellipsoid, and although the gravitational force would be through the geometric center, the normal force would not except at the equator and the poles. The sum of the center-directed gravitational force and the off-center normal force would give the required centripetal force.

Finally, suppose that the idealized planet was not a fluid ellipsoid, but a solid sphere, and suppose that it were frictionless. In that case both the normal force and the gravitational force would point perfectly towards the center, and with no friction the net force could not cause only the desired centripetal motion. Instead, the body would begin to slide towards the equator. This is the scenario you were analyzing. The fact that we don’t observe this behavior is because the earth is not a uniform frictionless sphere.

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  • $\begingroup$ This is the only answer that makes sense. Thanks a lot @Dale. Just one more confusion. Doesn't it mean that the standard textbook proofs on affect of rotation on $g$ are incomplete and a bit misleading? $\endgroup$ May 1 at 12:54
  • $\begingroup$ @ImtiazKabir it has been so long since I looked at this material in a standard textbook that I couldn’t say if they were incomplete or misleading. But standard textbooks are full of frictionless inclined planes and massless ropes and pulleys. So they often prioritize clarity over accuracy, and I wouldn’t be surprised to find that happens on this topic too $\endgroup$
    – Dale
    May 1 at 13:16
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    $\begingroup$ @ImtiazKabir Have look at the following answer I gave in a related context. This derives the effect both of the rotation and the ellipsoidal shape on $g$ in great detail physics.stackexchange.com/questions/659032/… $\endgroup$
    – Thomas
    May 1 at 15:36
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The Earth rotates because it formed in the accretion disk of a cloud of hydrogen that collapsed down from mutual gravity and needed to conserve its angular momentum. It continues to rotate because of inertia.

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  • $\begingroup$ My question was not why the earth continues to rotate or why it started rotating in the first place. I am confused as to why the forces in the free body diagram creates such acceleration on a body on earth's surface $\endgroup$ May 1 at 8:46
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The force of gravity acts vertically downward towards the centre of the Earth. If you are neither at the poles or the equator, but at some arbitrary latitude as shown in your diagrams, then you can represent the gravitational force as having two components, one directed towards the Earth's axis of rotation and the other directed parallel to it. It is the component directed towards the Earth's axis that provides the centripetal attraction to keep you in motion in a circle.

The centrifugal effect is quite small, ranging from nothing at the poles to about a 0.5% reduction in the acceleration due to gravity at the equator, and it acts normal to the Earth's axis of rotation.

At the poles and at the equator the effect is easy to understand- at the former it doesn't exist and at the latter the centrifugal effect points directly away from the Earth's centre (more or less) and so just reduces your weight straightforwardly.

At other points on the Earth's surface the impact of the centrifugal effect is to reduce the component of your weight acting towards the Earth's axis, without changing the component that acts parallel to the Earth's axis, which causes the effective direction of your weight to change fractionally, which would tend to pull you towards the Earth's equatorial plane. However, the effect is very small compared with the force of friction that keeps you on your feet, so in practice you wouldn't notice it.

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  • $\begingroup$ Break the force of gravity in any component you wish, it would not change the fact that the resultant of the components still points to the center of the earth. Lets make this rigorous. Let $\overrightarrow{g_1},\overrightarrow{g_2},...,\overrightarrow{g_n}$ be the components of gravity. Then The forces acting upon the body will be \overrightarrow{g_1} + $\overrightarrow{g_2} + ... \overrightarrow{g_n} + \vec{N} = \vec{G} + \vec{N}$. Because both $\vec{G}$ and $\vec{N}$ acts radially, their resultant should also act radially towards the center and not towards the center of rotation $\endgroup$ May 1 at 11:43
  • $\begingroup$ Yes, the resultant points to the centre of the Earth. So what? You asked what provided the centripetal attraction- it is gravity, and specifically the component of the gravitational force that acts towards the Earth's axis of rotation. $\endgroup$ May 1 at 12:17
  • $\begingroup$ Please read the question carefully. "My question is, how does these forces create this acceleration and nothing else." If we agree that it points towards the center of earth, that means not only it helps us rotate, it also does something more. If I say this in your language, where does the other component go? What does it do? Shouldn't it move us towards the equator? $\endgroup$ May 1 at 12:50
  • $\begingroup$ You are overlooking the fact that the centrifugal effect is tiny- it will reduce your weight by maybe ten grams, depending on where you are. So your weight in the direction of the Earth's axis is hardly reduced at all. There will therefore be some overall tendency to make you move towards the equator, but it will be so small that it will have no noticeable effect. $\endgroup$ May 1 at 13:53
  • $\begingroup$ I've extended my answer to cover the point, so please let me know if that helps. $\endgroup$ May 1 at 14:43
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Friction between the body and earths surface provides a tangential force necessary to get a body rotating around earths axis in the first place.

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  • $\begingroup$ That would mean, friction is a function of latitude. Hence, given the latitude of a place, we should be able to figure out the necessary friction to keep a body in rotation. This feels counter intuitive, because it is up to us, what surface we put the body on. No matter what type of surface we use, we observe a body being at rest w.r.t the surface of earth. I can use a very slippery surface that doesn't provide enough friction to keep it in rotation, but it will still rotate! $\endgroup$ May 1 at 17:00
  • $\begingroup$ Suppose we had no atmosphere. Suppose also we had a frictionless ice track going around the earth. Now suppose you come from another planet with a ball. You use your rocketship to approach earth with the ball. As you approach earth it will be rotating underneath you. If you place the ball on the frictionless icetrack it will NOT rotate around with the earth. It will stay stationary and the earth will rotate below us. Most of use are born on the earth so we already have that tangential velocity (wrt the axis of earth) that our mother had when we were born. $\endgroup$
    – Jagerber48
    May 1 at 20:48

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