# Test to prove Earths spins:

sonhouse
Science 09 Oct '16 12:44
1. sonhouse
Fast and Curious
09 Oct '16 12:44
Obviously Earth spins but there are those who deny that. I wonder if you put a bell jar with vacuum inside on the equator, and it is say 100 meters high and inside was a long thin wire with a bob at the bottom, wouldn't the bob be not vertical because of the spinning of Earth? If Earth didn't spin, wouldn't the bob point straight down?
2. 09 Oct '16 14:021 edit
Originally posted by sonhouse
Obviously Earth spins but there are those who deny that. I wonder if you put a bell jar with vacuum inside on the equator, and it is say 100 meters high and inside was a long thin wire with a bob at the bottom, wouldn't the bob be not vertical because of the spinning of Earth? If Earth didn't spin, wouldn't the bob point straight down?
But which way would it 'tilt'? And how would you know what is 'straight down'?

A much simpler experiment which you can do anywhere other than the equator is a large pendulum (a few meters is sufficient, but the longer the better). It rotates over time. You can actually measure the rotation on each swing and calculate exactly how fast the earth is rotating.
3. 09 Oct '16 16:246 edits
providing you accept c, i.e. the known speed limit of light, and accept nothing can go over c, and providing you accept the stars in the sky are very far away and certainly much more than a light-day away;
If the Earth isn't spinning then a calculation can show that the stars in the sky spin around the Earth each night over the speed of light which is causally impossible; proof by contradiction (of natural law) that the Earth must be spinning.
But I guess if someone believes the nonsense (given modern science) of the Earth not spinning then they may well believe the nonsense of c being much larger than the scientists think it is or the stars in the sky being just a few miles up or perhaps relativity is just nonsense and stars can go over c etc. In which case, no amount of scientific proof will ever be enough for them and they will always demand more (does this sound all-too familiar? ) .
4. 09 Oct '16 16:41
Some references:
https://en.wikipedia.org/wiki/Foucault_pendulum
https://www.wired.com/2014/05/wuwt-foucaults-pendulum/

Interestingly the pendulum method can also tell you what hemisphere you are in and what latitude. The guy in the video calculates his latitude correctly to within a couple of degrees.

I knew about this method because I recall seeing a pendulum that had been set up for this purpose by my grandfather who was a Physics professor at Salisbury University.
5. DeepThought
10 Oct '16 02:38
Originally posted by sonhouse
Obviously Earth spins but there are those who deny that. I wonder if you put a bell jar with vacuum inside on the equator, and it is say 100 meters high and inside was a long thin wire with a bob at the bottom, wouldn't the bob be not vertical because of the spinning of Earth? If Earth didn't spin, wouldn't the bob point straight down?
Why would the bob deviate from being vertical?
6. joe shmo
Strange Egg
10 Oct '16 03:071 edit
Originally posted by DeepThought
Why would the bob deviate from being vertical?
I think what your getting at is at the equator it won't deviate from vertical. However, at some other latitude it will,...slightly ( centrifugal force in non inertial frame). That is if we assume vertical to be in the direction of the gravitational field alone. The Foucault the pendulum that Twhitehead is describing is the result of the Coriolis force. Which won't manifest itself in a resting pendulum.
7. sonhouse
Fast and Curious
10 Oct '16 11:08
Originally posted by joe shmo
I think what your getting at is at the equator it won't deviate from vertical. However, at some other latitude it will,...slightly ( centrifugal force in non inertial frame). That is if we assume vertical to be in the direction of the gravitational field alone. The Foucault the pendulum that Twhitehead is describing is the result of the Coriolis force. Which won't manifest itself in a resting pendulum.
I was thinking BECAUSE Earth spins there would be an acceleration to a non-moving pendulum that would cause it to be slightly off from vertical. I am assuming the big bell jar would be placed on ground level pointing exactly 90 degrees from horizontal so a pendulum inside the bell jar would have been made in such a way as to be fully centered thus equidistant from all sides of the container. So if there was say some kind of bulls eye affair marking off dead center I was expecting the bob to not be on dead center because of the circular motion of Earth would deflect the bob to some extent and point in the opposite direction of spin.

That was just my thought experiment at work, obviously I never did that in real life.

But it seems there should be some kind of deflection regardless of whether it is in oscillatory motion or dead still. I would think dead still would reveal such a deflection due to the fact Earth is in fact spinning.
8. 10 Oct '16 11:58
Originally posted by sonhouse
So if there was say some kind of bulls eye affair marking off dead center I was expecting the bob to not be on dead center because of the circular motion of Earth would deflect the bob to some extent and point in the opposite direction of spin.
I somehow doubt that that would be the case although I cannot back that up with any reasoning.
Having said that, the wind patterns of earth are due to rotation.
9. joe shmo
Strange Egg
10 Oct '16 11:58
Originally posted by sonhouse
I was thinking BECAUSE Earth spins there would be an acceleration to a non-moving pendulum that would cause it to be slightly off from vertical. I am assuming the big bell jar would be placed on ground level pointing exactly 90 degrees from horizontal so a pendulum inside the bell jar would have been made in such a way as to be fully centered thus equidista ...[text shortened]... would think dead still would reveal such a deflection due to the fact Earth is in fact spinning.
It won't deflect at the equator, because the centrifugal force is aligned ( in opposing direction) with the gravitational force, radially. However, It will weigh less at the equator than at the poles. That shows rotation.
10. DeepThought
10 Oct '16 14:05
Originally posted by joe shmo
It won't deflect at the equator, because the centrifugal force is aligned ( in opposing direction) with the gravitational force, radially. However, It will weigh less at the equator than at the poles. That shows rotation.
Except that the Earth is flattened and the difference in radius is of the order of 0.2%, while the centrifugal force is of the order of 0.33 m/s^2 or 0.3%. These two effects tend to cancel, so you'll probably find you can't demonstrate spin. The problem with the method is that you cannot rule out the difference in weight being due to the shape of the Earth. Foucault's pendulum directly demonstrates the rotation of the Earth (provided one is away from the equator), there are no other effects it can be ascribed to.
11. joe shmo
Strange Egg
10 Oct '16 14:32
Originally posted by DeepThought
Except that the Earth is flattened and the difference in radius is of the order of 0.2%, while the centrifugal force is of the order of 0.33 m/s^2 or 0.3%. These two effects tend to cancel, so you'll probably find you can't demonstrate spin. The problem with the method is that you cannot rule out the difference in weight being due to the shape of the E ...[text shortened]... Earth (provided one is away from the equator), there are no other effects it can be ascribed to.
Ok, I see. I was modeling to perfect spherical, perfectly homigeneous earth. The phenomenon is on the level of precision where reality will make observation near impossibe. The Coriolis force demonstrates on the pendulum, but to be fair he wanted to use a giant plum bob. Not a swinging pendulum. No relative motion, no Coriolis force. Correct?
12. sonhouse
Fast and Curious
10 Oct '16 16:07
Originally posted by joe shmo
Ok, I see. I was modeling to perfect spherical, perfectly homigeneous earth. The phenomenon is on the level of precision where reality will make observation near impossibe. The Coriolis force demonstrates on the pendulum, but to be fair he wanted to use a giant plum bob. Not a swinging pendulum. No relative motion, no Coriolis force. Correct?
But there would still be a force on the bob by the simple rotation of Earth I would think. There is a rotation in X direction around the equator and my bell jar is on the equator. I don't see how the bob on a 100 meter string could still point directly to the center of Earth, it seems there should be some deflection. I think the circular motion represents circular acceleration so why wouldn't the bob not deflect?
13. joe shmo
Strange Egg
10 Oct '16 16:09
Originally posted by DeepThought
Except that the Earth is flattened and the difference in radius is of the order of 0.2%, while the centrifugal force is of the order of 0.33 m/s^2 or 0.3%. These two effects tend to cancel, so you'll probably find you can't demonstrate spin. The problem with the method is that you cannot rule out the difference in weight being due to the shape of the E ...[text shortened]... Earth (provided one is away from the equator), there are no other effects it can be ascribed to.
I'm missing something...why must we be away from the equator for the Coriolis Force to be present?
14. 10 Oct '16 16:47
Originally posted by joe shmo
I'm missing something...why must we be away from the equator for the Coriolis Force to be present?
https://en.wikipedia.org/wiki/Coriolis_force
The Coriolis Force is when an object is rotating relative to another object. That isn't the case at the equator. It is everywhere else.
15. joe shmo
Strange Egg
11 Oct '16 00:13
https://en.wikipedia.org/wiki/Coriolis_force
The Coriolis Force is when an object is rotating relative to another object. That isn't the case at the equator. It is everywhere else.
Its when the frame of reference is rotating ( specifically with a constant angular velocity I believe). I don't think it is when an object is rotating relative to another object. Even still...Is the Corialis force always zero at the equator for some reason that I'm missing? The mathematics of the ficticious force seem to indicate three possible scenarios.

F_cor = -2mΩxv

That leaves 4 possibilities for F_cor = 0

1) Ω=0 ...The Earth is not rotating

2) v = 0 ... The Body is not moving

3) m= 0 ...The body has no mass

or

3) Ωxv = 0 = Ωv ... The bodys motion is parallel to the axis of rotation.

None of those seem to be unaminously satisfied specially at the equator... So, since Deep Thought implied it, I must be wrong. I just still cant see it.