To give you a better understanding of what is happening, first we need to understand what 'centrifugal force' is, and how it is 'generated'.

First of all, there is no such force as centrifugal force! It is an

**apparent**force that acts outward on a body that is being accelerated, but it actually arises from the inertia of mass. It is EXACTLY the same apparent force that 'throws you forward' when you slam on the breaks in your car - but when the body in question is rotating we call it centrifugal force.Newton's first law of motion (the Law of Inertia): An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

The real force is called gravity (which provides the

**centripetal**component**)**and this is the force that is causing the actual acceleration. If you put a rock on the end of a string and spin it around it is the string that is applying the centripetal force to the rock which keeps it from flying away. On the Earth is it the force of Gravity that pulls you towards the center. On Earth the acceleration due to Gravity is 9.807 m/s² - this can be observed by dropping an object from various heights and observing the different amounts of time it takes to hit the ground (pick something heavy and round to minimize the effects of air resistance).
But the inertia inherent in the mass of the rock wants the rock to keep going along the path which it is currently going. The 'tug' of the string is like the breaks of your car, it pulls it around the center of rotation but anything that might be inside the rock would be pressed to the outside of the curve by inertia.

So the centrifugal acceleration is really just due to inertia.

**What is a Force?**

Force is the mass multiplied by the acceleration - for the remainder of this discussion we will focus on the acceleration component rather than the force. If you want to convert any of our accelerations into a force just multiply it by your mass in kilograms (kg). Or to make it super easy, assume a 1 kg object - which gives you kg*m/s² which is called a Newton of force.

So an acceleration of 10 m/s² acting on a 1 kg object would just be 10 Newtons of force. And on a 100kg object it would be 1000 Newtons. Which is how much force your body experiences from gravity (well, a 100kg person would experience 980.7 Newtons because the acceleration of Gravity is 9.807 m/s²).

**Ok then, What is an Acceleration?**

An acceleration is simply a change in the speed (which is also called velocity) of some object - if you are in a car going 1 m/s and you aren't speeding up or slowing down then you aren't accelerating.

Gravity is a field which applies an acceleration to mass upon which it acts mutually. So my mass 'tugs' on the Earth and the Earth 'tugs' on me mutually. It is from this simple observation that all of the equations of motion were first derived, it is this action that gives the planets an elliptical orbit as well. There is a great series of lectures by Richard Feynman which explain all the observations about gravity that give us our current understanding of how it works, I encourage anyone who honestly wants to understand their world better to watch them.

**So what is the effective/apparent centrifugal acceleration at Earth's equator?**

THIS is the question that we need to understand so we can determine if something would fly off.

If the

**centrifugal acceleration**is greater than the**acceleration due to gravity**then the thing would fly off.
And it turns out we can figure out the

**centrifugal acceleration**very easily if we just know the radius around which we are rotating and something called our**angular velocity**.
If we were standing on the equator then our distance from the center of Earth's rotation would be 6,378,137 meters (or 3,959 miles - NOTE: the distance to the north pole is a mere 9.1 miles difference), and we know our rotational rate is once every 24 hours.

So let's define our values:

T = 86164.09054 seconds (time, 1 sidereal day)

ω = 2π/T (angular velocity - simply the angle we turn divided by the time, 2π radians = 360°)

r = 6378137 meters (equatorial radius of Earth)

And finally

**centrifugal acceleration**simply equals the radius times the angular velocity squared.
So it turns out our

**centrifugal acceleration**is a mere ~0.0339 m/s² which is working against the acceleration of Gravity which was 9.807 m/s²
Gravity is 289 times STRONGER than the

**centrifugal acceleration**.
So we can safely say that you would NOT fly off the Earth because of this rate of rotation.

**What Is Going On Here!**

Ok -- the math is a little complex and it doesn't help us understand WHAT is going on just yet. Let's look at things from a couple of different perspectives.

We remember being on the playground merry-go-round and being thrown off that and it wasn't going anywhere NEAR 1040 miles per hour!

So why is our intuition about this so wrong? The answer again lies in understanding inertia.

Do you remember standing in the middle? It was easy to sit in the middle, you didn't get thrown off, but the closer to the edge you got the greater the force you felt.

Let's use the same math from above to work out the acceleration for our little merry-go-round where we are hanging on to the outer edge (about 5 feet or 1.524 meters from our center of rotation as that is a 10 foot merry-go-round) and assume we're a 50 kg kid.

What if we SLOWLY walked the merry-go-around taking, oh say, 42.1282 seconds to go all the way around once (just to pick a random number)?

42.1282s = 0.0339 m/s² [Wolfram|alpha]

Hey! That's the SAME acceleration as Earth spinning at 1040 miles per hour! Let's see - radius of 1.524 meters = circumference 9.576 meters around... that's only 1/2 mile per hour -- that's a SLOW walk around.

**What is going on here?**

Both the merry-go-round and the Earth are TRYING to give us the slip -- but the Earth has a HUGE radius so even at 1040 miles per hour FORWARD (or 0.288 miles per second) and we know from the drop rate equation (8 inches times miles squared) that the drop over 1 second is about 2 inches per second or just ~0.05 m/s! Meanwhile, Gravity is sucking you down at 9.807 m/s² and since both accelerations are constant, the

**centrifugal acceleration**really just cancels out that TINY portion of the acceleration due to Gravity, and that is at the equator -- as you move away from the equator the**centrifugal acceleration**drops (advanced version:*to get the correct acceleration for your latitude just multiply the radius by**cos**²(latitude) to get your true radius of rotation, it is 1 at the equator and 0 at the pole - make sure you use latitude in the correct units; degrees or radians for your software/calculator*).

*So it really has to do with how fast the 'thing' is moving 'out from under you' NOT merely how fast around you are going -- the distance from the center of rotation has as much to do with it as the raw speed.*You can also try thinking about your experience in a car. Even when traveling fairly slowly, if the car takes a sharp turn you are thrown to the side with a moderate amount of force. But in a very fast moving car, moving around a larger curve of the freeway you feel much less force than the slower car turning abruptly - despite going even 10 times faster. So the size of this curve matters a lot and the curvature of the Earth is enormous, 1000 times flatter than the curve in the road.

**Spin It Faster!**

What happens if we spin our kids ten times faster, or once around every 4.2128 seconds?

4.2128 = 3.39 m/s² [Wolfram|alpha] - no surprise, we just went 10 x faster and got 10 x the acceleration.

And if we convert that to the force felt by a 50 kg kid we get 169.5 Newtons or about 38 pounds of force pushing our kid sideways.

That's pretty accurate.

**I SAID FASTER!**

Now, what if we used a motorcycle to spin our merry-go-round at say... a mere 16.5 miles per hour... You would create the Spinning Merry-Go-Round Of Death (or at least injury - don't do that kids).

Ouch

Ouch

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ReplyDeleteAnother Great article and there is a video showing that Centrifugal Force actually reduces the apparent G in an aircraft by almost 1% when flying East. You can see it in the accelerometer.

ReplyDeleteMore evidence the aircraft is following the Curvature of the Earth.

https://www.youtube.com/watch?v=n4khR31bvpI