# Homework Help: Why don't we feel the Earth moving?

1. Jun 19, 2013

### alingy1

Okay, so, this is not really a homework question but it is still related to coursework.

I wonder why we don't feel that we are moving. Our velocity is constantly changing since we are going around the Earth's center. My research tells me that we can compare this situation to a car. But, in the car, the vector of velocity doesn't change orientation... Can you help me understand this?

2. Jun 19, 2013

### voko

We do not feel velocities. We feel accelerations. What is the acceleration due to the rotation of the Earth? Compare it with the acceleration due to gravity.

3. Jun 20, 2013

### Basic_Physics

In the car example you should compare it to a car going around a bend with a very large diameter so that the direction of the velocity is changing. Voko suggests that you compare the centripetal acceleration we experience due to the earth's rotation with gravitational acceleration. What fraction is it of g?

4. Jun 20, 2013

### alingy1

Hihi, I don't have a car and I haven't taken a ride for a long time. But, I got some new insight. If the acceleration is pulling us towards the center of the Earth, that sums up the effect of gravity. I don't understand what you mean by the fraction of g. Since our velocity vector and our acceleration vector (of a normal human being on earth) are perpendicular, the magnitude of the velocity vector doesn't change, only its orientation does. But, how come our body doesn't feel that change in orientation?

EDIT: I understood that the acceleration caused by centripetal force is very small compared to g since
A=v^2/r

Last edited: Jun 20, 2013
5. Jun 20, 2013

### voko

As already indicated, our bodies (as opposed to our visual systems) feel accelerations, not velocities or "changes in orientation". And if you are standing "upright", for example, the body feels constant acceleration "downward", which it recognizes as "normal" gravity and does not signal to us in any particular way.

6. Jun 20, 2013

### Basic_Physics

Yes, we need a centripetal force to let us go around on the spinning earth. Gravity supplies this force as you rightly remarked. But it is larger than is needed for this type of motion. The required centripetal force can be calculated via

FC = mv2/R

where v is the speed with which we are going around - large at the equator and zero at the poles of the earth and R is the radius of the circle in which we are going around, which also depends on the latitude where we are located. The speed can be calculated with

v = 2∏R/T

where T is about 24 hours so that the above formula becomes

FC = mR(2∏/T)2

7. Jun 20, 2013

### D H

Staff Emeritus
We feel a constant acceleration upward, not downward. We can't feel gravity. Nothing can.

What we can feel is everything but gravitation. We feel the centripetal force that keeps us moving with a car as the car turns a corner. We feel the upward normal force from the ground whilst gravity pulls us downward.

Gravitational acceleration onboard the International Space Station is about 90% of earth surface gravity. The reason astronauts in the ISS are called "weightless" is because that's exactly what they feel. They can't feel gravity, and since gravity is the only force acting on them, they feel weightless. What the astronauts feel is the same queasy queasy sensation many of us get when a roller coaster reaches the top of a climb and then starts dropping. Gravitational force hasn't changed one iota, and yet we feel very different. What's different is that those other forces that we can feel has changed. We are very accustomed to feeling that constant one-g upward acceleration throughout our bodies. It's the sudden lack of that ever-present feeling that makes us feel a bit queasy.

Getting back to the original question, you *do* feel the consequences of the Earth's rotation. It is part of what makes you a bit lighter (scale weight) at the equator than at the poles.

8. Jun 20, 2013

### voko

I did not say we feel gravity.

I agree that the acceleration we feel is physically upward, yet our sensory organs report the sensation as "downward", even though this might also be debated. I hope we can all agree it is "vertical".

The crux of the matter is that this variation in acceleration is only in magnitude, not in direction, and is very tiny. It is probably below the noise floor of our built-in sensors.

9. Jun 20, 2013

### D H

Staff Emeritus
It's tiny, but not "very tiny". You feel a tiny bit heavier when you toss on a light jacket or when you drink two cups of water. That's about the magnitude of the effect were one to instantaneously teleport from the equator to the North Pole. The reason we don't feel it is because, sans the ability to teleport from the equator to the North Pole, it's both small and constant.

It does vary in direction with latitude. Down, as measured by the direction a plumb line points, points to the center of the Earth only at the poles and at the equator. Everywhere else, toward the center of the Earth and down are slightly different directions.

10. Jun 20, 2013

### Andrew Mason

This is an interesting question. Even if the earth was spinning 16 times faster than it is so that the centripetal acceleration to keep us on the surface was 256 x as strong as it is (about 9 m/sec^2) we would not feel the earth moving. We would just feel a lot lighter. But we would feel no centrifugal effect as we would if we were tethered to a rotating body in space. Of course if the earth rotated much faster, gravity would not be sufficient to keep us on the surface. We would then feel the earth because we would be bouncing up and down a lot.

AM

11. Jun 20, 2013

### voko

I still think that our natural chaotic vibrations and oscillations, even in a state that we would consider "stationary", could result in a greater variation of perceived acceleration. Nor do I think our sensors are so sensitive to begin with. I did a back-of-an-envelope calculation yesterday, and the variation of acceleration with latitude was, I think, about 0.003 g. Perhaps not very, but still quite tiny.

12. Jun 20, 2013

### alingy1

Andrew, if the earth spinned much faster, wouldn't we feel heavier? The centrifugal force would be pulling us towards the center?

13. Jun 20, 2013

### alingy1

Man, I love physics. Seeing the formula F=mv^2/2 popped a new question in my head. a=v^2/r

How could we calculate the maximum speed that an asteroid should have to still be in orbit around earth?
Let's consider a=9,8m/s^2 and r=7000000m. Do we simply isolate the v in the formula with the radius of the orbit?

14. Jun 20, 2013

### voko

That would not be the maximum speed. The would be the minimum speed. It is known as the first cosmic velocity. Slower than that at the given altitude, the asteroid crashes.

15. Jun 20, 2013

### alingy1

True. However, I don't understand how a smaller radius has a smaller minimum velocity at the same acceleration. Shouldn't that asteroid have more velocity to not crash on the planet?

16. Jun 20, 2013

### voko

Smaller radius generally means the orbit is more "curved". Greater curvature generally means greater acceleration. But you have the acceleration fixed. That can only be satisfied by slowing down.

This phenomenon is very important in driving vehicles and especially in motor racing involving turns. A vehicle has a finite amount of sideways friction. It is this sideways friction that causes a vehicle to turn. Which means there is only some max acceleration it may have in a turn, otherwise it will just skid away. Which pretty much forces any sensible driver to go slower as turns get tighter. In racing, drivers try very hard to stay just at this max acceleration, and frequently fail very spectacularly.

17. Jun 20, 2013

### alingy1

Awesome insight for the motor racing industry. The acceleration doesn't change much if you don't get too far from Earth no? Thank you so much all of you!

18. Jun 20, 2013

### Andrew Mason

There would be no difference in the force pulling us toward the earth - that is always GMm/r^2. The other force is the normal force. Those two forces must sum to your mass x your acceleration. The only acceleration is the centripetal acceleration: v^2/r toward the centre of the earth. So, letting the direction of gravity be $-\hat{r}$ (ie. opposite to the direction of the displacement vector from the centre of earth to your position):

$$-\frac{mGM}{r^2}\hat{r} + \vec{N} = -\frac{mv^2}{r}\hat r$$

$$\vec{N} = \frac{mGM}{r^2}\hat{r} - \frac{mv^2}{r}\hat r$$

From that you can see that the normal force N is reduced as centripetal force increases.

AM

19. Jun 20, 2013

### alingy1

Hmm, why would v^2/r be the only acceleration? Isnt there the gravitational pull?
Sorry for my incompetency in understanding your equations. I'm in grade 11. But, in my head, the centrifugal force is pulling us towards earth. The gravitational pull too. So, if you combine those two, you get a bigger force. The normal force will consequently be bigger too. I did my research on yahoo and ask a scientist, and they say you are right.
Could you please explain it to me again?

20. Jun 20, 2013

### voko

You may want to memorize that "-petal" and "-fugal" are derived from Latin words which mean "to seek" and "to flee" ("-fugal" is cognate with "fugitive"), so "centripetal" is "toward the center", and "centrifugal" is "from the center".

Knowing that, you would never have said that the centrifugal force is pulling us towards Earth.

21. Jun 20, 2013

### alingy1

Centripetal!!! Sorry :)

22. Jun 20, 2013

### D H

Staff Emeritus
There is only one acceleration. It's the derivative of velocity. Newton's second law describes the relation between the total force acting on some object and the acceleration of that object. The total force acting on some object is the superposition (vector sum) of all of the individual forces that act on that object.

Two corrections: Newton's law of gravitation, GMm/r^2, applies only to point masses and non-point masses with a spherical mass distribution. It's an OK representation of Earth's gravity field for low fidelity purposes. It would not apply to your hyper-rotating Earth (16x current rotation rate, one day = 90 minutes). That Earth would flatten out to such an extent that GMm/r^2 would only be valid at very, very large distances from the Earth.

The other correction is that the v^2/r centripetal acceleration points toward the Earth's rotation axis, not the center of the Earth (and the r is the distance to the rotation axis rather than to the center of the Earth).

23. Jun 21, 2013

### Andrew Mason

Ok. I was assuming that the earth's shape would not change which is obviously not true. Lets pretend earth is a solid metal ball held together by intermolecular bonds that are much stronger than the gravitational forces.

Quite right. I was thinking of a person was standing on the equator of a spherical earth.

AM

24. Jun 21, 2013

### Basic_Physics

There are two forces acting on us standing up straight. The gravitational force, F, acting towards the center of the earth and the supporting force, S, from the surface of the earth. We will accelerate according to the resultant of these two forces according to Newton's 1st law. Since we are spinning around the axis of the earth we know that we are should be accelerating towards the centre of the of the circle in which we are spinning (which is actually not the centre of the earth). So that means that the resultant of these two forces supplies the centripetal force giving us a centripetal acceleration towards the centre of the circle in which we are spinning.

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