qOriginally posted by twhitehead I could be mistaken, but a mass being raised or lowered should experience a sideways force.
On an equatorial laid out track? I thought about that frictionless scenario and came to the conclusion there would be no motion of the car on the track when it contacts because it would already be going at the same rate of rotation of the Earth when it is held in place by the crane or pulleys or whatever holds it above Earth before setting on the track.
Originally posted by sonhouse In my thought experiment, the car is carefully placed on the track so it starts out with no motion. The question is would the movement of Earth be enough to impart a motion to the car in the opposite direction of Earth's rotation. In an evacuated tunnel, good vacuum.
I am assuming the track to be going east to west not north to south, which if there was ...[text shortened]... ould lead to a situation where the car tries to tilt off the track. Assuming there such a force.
No motion relative to what? If you mean the surface of the earth then it is already moving relative to the earth's centre. In your set up there are no forces on the train that are not radial.
Originally posted by sonhouse On an equatorial laid out track?
No, just on a stationary tower.
If you take a spacecraft and raise its orbit, it orbits slower, so it would seem to me that raising an object on the surface of the earth at the equator would make it want to orbit slower and thus it would experience a force opposite to the direction of the earths rotation.
It seems space elevators agree and categorise it as the Coriolis force:
https://en.wikipedia.org/wiki/Space_elevator
Originally posted by twhitehead https://en.wikipedia.org/wiki/E%C3%B6tv%C3%B6s_effect
I wonder if a very sensitive pendulum with built in scale could detect the mass difference as it swings with the earths rotation or against it.
Very interesting. Also interesting is they could measure this over 100 years ago with the primitive equipment of that day. I wonder if more modern measurements with more sensitive equipment would add enough decimal points to show the cosign formula is correct?
So suppose you had two pendulums a few meters apart, also on the equator, one swinging east west the other north south.
Assuming identical masses and string lengths, would there be a measureable difference in the pendulum rates? I would assume they would be far enough apart but center line still on the equator so there would be no gravitational coupling, also they would have to be in a vacuum chamber so air movements would not effect results either.
Originally posted by sonhouse Assuming identical masses and string lengths, would there be a measureable difference in the pendulum rates?
I don't think so. The east-west pendulum weighs different one direction than the other, but its overall period should not be significantly affected. I believe period is a function of length not mass.
Originally posted by sonhouse Very interesting. Also interesting is they could measure this over 100 years ago with the primitive equipment of that day. I wonder if more modern measurements with more sensitive equipment would add enough decimal points to show the cosign formula is correct?
So suppose you had two pendulums a few meters apart, also on the equator, one swinging east we ...[text shortened]... also they would have to be in a vacuum chamber so air movements would not effect results either.
The force between two objects of mass 1kg at a separation of 1 metre is 6.674E-11 Newtons. The Earth's gravity provides 10 Newtons per weight and we are expecting an effect of about 0.1% or 1.0E-2 Newtons. So we need the weights to be less than of the order of 10 tonnes, I'm thinking that the rail operator is going to want that in any case. More realistic problems are associated with deviations from simple harmonic motion and air resistance. You can repeat the experiment on the return journey and count the number of oscillations of the pendulum. Measuring a difference like that will tend to correct for all these extraneous effects.
It does not matter which way the pendulum is swinging, we aren't expecting the pendulum weight to be moving fast relative to it's fulcrum. The period is independent of the amplitude provided the amplitude is small so we can make the speed of the pendulum weight arbitrarily small. What matters is how fast the train is going relative to the Earth's surface. At 125 miles per hour you should get about 7 more swings on a 25 cm pendulum going West then you would going East.
Edit: And if you are thinking you can do this with the pendulums stationary, that is to say the fulcrums stationary with respect to the Earth's surface then you'll need very high frequency pendulums because the effect size is less than 0.1% for an object moving at 125 miles per hour (about 55 m/s). So for a pendulum with a period of the order of a second and an amplitude of the order of a metre the effect will be of the order of 0.002% and controlling for errors is going to be really hard.
Originally posted by DeepThought The force between two objects of mass 1kg at a separation of 1 metre is 6.674E-11 Newtons. The Earth's gravity provides 10 Newtons per weight and we are expecting an effect of about 0.1% or 1.0E-2 Newtons. So we need the weights to be less than of the order of 10 tonnes, I'm thinking that the rail operator is going to want that in any case. More reali ...[text shortened]... the effect will be of the order of 0.002% and controlling for errors is going to be really hard.
Well that throws some stuff thinking about doing the experiment on a train but I was thinking of stationary pendulums.
Would there be more of a measurable effect if the string length was say 100 meters or so for each one? On the equator, in a giant air evacuated bell jar, and they are swinging 90 apart, one swinging due north/south the other east/west.
Would there be a difference with smaller masses, say 1 kg at the end V 100 Kg or so.
Originally posted by Andrew Kern If the earth didnt spin, then space stations would be anywhere instead of closer to the equater, because of the 10,000 mph boost of the spin.
Please clarify. The International Space Station is not particularly close to an equatorial orbit, and many other satellites are not. Although it is normal to take advantage or the spin when launching, satellites can be injected into just about any orbit, and space stations are no exception. Considering the regular visits to the space station there may be some cost savings to having it close to equatorial orbit, but do you have any actual references to confirm this?
Originally posted by twhitehead Please clarify. The International Space Station is not particularly close to an equatorial orbit, and many other satellites are not. Although it is normal to take advantage or the spin when launching, satellites can be injected into just about any orbit, and space stations are no exception. Considering the regular visits to the space station there may be ...[text shortened]... s to having it close to equatorial orbit, but do you have any actual references to confirm this?
I don't know about the exact numbers of mph you gain at the equator but Cape Canaveral in Florida sits at about 28 degrees north but where Suyuz rockets are launched at Baikunor is almost halfway between the equator and the pole at 45.6 degrees north.
So if the number he posted, 10,000 mph is the boost at the equator, at 45 degrees you would get only half that, 5K mph. And at the Cape, something like 7.5K mph boost. So rockets launched from Baikunor have to expend more fuel to get to a given orbit than those launched at the cape. A permenant disadvantage in terms of payload V fuel expended.
Test to prove Earths spins:
14 Oct '16 00:00
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