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This question already has an answer here:

Okay, so this may seem a very "simple" question.

I quote few lines from one website:

"We can’t feel Earth’s rotation or spin because we’re all moving with it, at the same constant speed."

"Why don’t we feel Earth rotating, or spinning, on its axis? It’s because Earth spins steadily – and moves at a constant rate in orbit around the sun – carrying you as a passenger right along with it."

Okay, I get the fact that the ground, the atmosphere and all the "things" on Earth are moving at the same rate. It is almost like we are travelling in a train or an airplane, where we can walk normally as if on the ground, without feeling anything unusual.

But, my "Physics" intuition seems to contradict this explanation. This is how I view this problem :

We dwell on the ground and move with the Earth, very much like in a "$2$-D Circular Motion", with the radius of the Earth as the radius of the "circle". But the velocity changes every instant as its direction (which is tangent to the circle at the given point) changes constantly. The velocity is changing --- So there must be some acceleration , as Uniform Circular Motion is always an accelerated motion.

So, I see that there is "some acceleration". According to Newton's Equation $F_{\text{net}}=m.a$, we must be experiencing some Force, which is required to provide the necessary Centripetal Acceleration$(=\dfrac{v^2}{r})$.

So, in short, I state my Problem as:

When we ride a Merry-Go-Round, we feel a "certain" force pushing on us, acting radially away from the centre. Why don't we feel the same "certain" force in case of Rotation of Earth?

EDIT : Can anyone please remove that "Duplicate" mark ? My question is not specifically about "Why don't we feel the Earth's rotation ?".

It is rather about "Why (apparently) is Earth's Rotation not congruent/similar to some normal Uniform Circular Motion (as in a Merry-Go-Round) ?". Surely, I will re-word the question to be more specific.

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marked as duplicate by Frédéric Grosshans, ZeroTheHero, Emilio Pisanty, sammy gerbil, Michael Seifert Apr 30 '17 at 18:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ No physicist here, but maybe the only way for your body to actually 'feel' a force is that particles around it experience different forces. In other words, for our body, everything is stationary, as it moves around with the particles around it, and they are all experiencing the same force. As for the difference to the Merry-Go-Round, in that case it is only one part of your body experiencing the centripetal force and all the remaining parts your body are kept in place by pushing against other parts of your body, and this is what makes you feel the force. $\endgroup$ – Bananach Apr 30 '17 at 7:55
  • $\begingroup$ Suppose you're at the equator, where the effect of the centripetal force is the greatest. Even here the centripetal force acting on you is only 0.34% of the gravitational force. So, "feel" then becomes a relative term. You'd certainly "feel" it if gravity could be "switched off". Also, if it's 0.34 % at the equator, this falls down even further as you move away from it. $\endgroup$ – RedHelmet Apr 30 '17 at 8:00
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/12487/2451 and links therein. $\endgroup$ – Qmechanic Apr 30 '17 at 8:43
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    $\begingroup$ Inertial forces, such as centrifugal force, are apparent, but not real. How much centrifugal force would you feel if the merry go round were making one circle every 24 hours? $\endgroup$ – Walter Mitty Apr 30 '17 at 10:56
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    $\begingroup$ Possible duplicate of Why can't we feel the Earth turning? $\endgroup$ – Frédéric Grosshans Apr 30 '17 at 11:32
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You're right, and this is a great question. The centripetal force is provided by gravity on Earth. If a person is $80kg$, however, it is a very small force, $\frac{80*465^2}{6.37*10^6} = 2.71N$, which is actually a discrepancy on your bathroom scale due to the Earth's rotation!

However, if you're wondering why the rotation doesn't make us feel dizzy, imagine you're spinning in place. What's your linear velocity? $0\ m/s$, as you're still. But you're still feeling dizzy. This is because of a high angular velocity $\omega$, and that's what we really feel. But $\omega$ on earth is very small - $7.2921159 × 10−5\ rad/s$, so we feel pretty stationary.

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  • $\begingroup$ Part of the variation that used have used is due o ther non-spherical shape of the Earth. $\endgroup$ – Farcher Apr 30 '17 at 11:50
  • $\begingroup$ Your first paragraph is ALL that I wanted as an explanation. Thank You very much !!!! I have re-framed the question to "fit my needs". Thanks Again ! $\endgroup$ – user139580 Apr 30 '17 at 12:31
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Your reasoning is absolutely correct. We do feel a centrifugal force owing to the fact that we are in a frame that is performing the rotational motion. This is the reason why the weight of an object will be slightly more near the poles as compared to near the equator. Because the centrifugal forces will be maximum near the equator but will be vanishing near the pole. But this effect is quite small as compared to the gravitational pull of the earth. This is why we don't feel it in any obvious or trivial way in day-to-day life.

The centrifugal accelerations produced in the merry-go-round (or giant wheels) are quite large as compared to the centrifugal accelerations produced due to the rotation of the earth. Thus, they appear quite noticeable. Also, the net direction of the centrifugal acceleration keeps changing its direction because the direction of the centrifugal acceleration due to the rotation of the merry-go-round keeps changing its direction (in the frame of the earth). In the case of the motion of the earth, the centrifugal force doesn't change its direction (in the frame of the earth).

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    $\begingroup$ Wait, what do you mean by "merry-go-round" exactly? I imagine something like this: google.com/… , with a vertical axis of rotation. In such a configuration, the centrifugal acceleration is never either "along" or "away from" the gravitational pull, but always remains perpendicular to it! $\endgroup$ – Ilia Smilga Apr 30 '17 at 11:34
  • $\begingroup$ @IliaSmilga Thanks for pointing out the mistake. I had confused the merry-go-round with the giant wheel. I have edited the answer. $\endgroup$ – Feynmans Out for Grumpy Cat Apr 30 '17 at 13:57
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You do "feel" the rotation it that the reading on a spring balance does vary between the Equator and one of the poles being larger at the poles than at the equator.
You do not notice the effect because it is so small - about $0.02 \,\rm Nkg^{-1}$ in $9.81 \,\rm Nkg^{-1}$.

Another example is the circulation of large air currents in the atmosphere the direction of which are determined by the rotation of the Earth.
These are due to the Coriolis force.

enter image description here

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  • $\begingroup$ 80*0.02 = 1.6 N, so it appears your value is different than the answer from rb612. Where is the discrepancy? $\endgroup$ – Mike Wise Apr 30 '17 at 11:26
  • $\begingroup$ The variation is approximately from $9.78$ at the Equator to $9.83$ at the poles. Only about half of that variation is due to the current rotation of the Earth the other is due to the earth being roughly an oblate spheroid (squashed sphere) and so the distance to the centre of the Earth form the surface is less at the poles.. That change of shape is also due to the rotation of the Earth but achieved when the earth was being formed. So I have only used the variation due to the rotation not the distance from the centre. $\endgroup$ – Farcher Apr 30 '17 at 11:49
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Actually, you are experiencing a centrifugal force right now. How would you know if you were feeling it? Forces are vectors. You think you're only experiencing gravity right now, but in reality you're experiencing a net force which is the sum of gravity and centrifugal force. It is because of this centrifugal force that the direction of vertical is slightly altered. People at the equator feel less weight because the centrifugal force there is maximum and is in the complete opposite direction of gravity.

Please note that it is always the net force that you experience.

Actually, gravity is far greater than the centrifugal force we experience. At equator, centrifugal force opposite to gravity, so the only contribution of the centrifugal force is decreasing the net downward force that we feel by a small amount. At other places (except poles where there is no centrifugal force), the centrifugal force acts at an angle to gravity, so in addition to decreasing effect of gravity, it alters the vertical too.

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Because centrifugal force does act upon you.

When you're standing on a plane surface (in your reference not earth , earth is oblate). This centrifugal force is known as normal reaction.

For you to be in constant state and not fly away or sink in ground, the normal reaction balances your weight.

Comparison with merry go round is not apt one as on merry go round your weight is too big while the force that keeps you with the base plate is very small and variable depending on how you sit/stand (frictional force).

In case you earth you are the size of a speck. Huge forces act upon you to balance out each other.

While you feel centrifugal force on merry go round the small speck of dust on it usually doesn't (unless you make the base plate too smooth, less friction)

The dust particle always stays on the ride of fun :)

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  • $\begingroup$ The centrifugal force is not the normal reaction force. The centripetal force is the gravitational attraction, which is not fully cancelled by the normal reaction force - the remainder is the centrifugal force. $\endgroup$ – Emilio Pisanty Apr 30 '17 at 13:29