The ISS and its crew do not feel the gravity of the earth as they are in orbit around it e.g. constant freefall. However I have read that some tides are higher and lower than others because of the combined effect of the moon and the sun's gravity, but if the earth is in orbit around the sun then the sun's gravity should have no effect upon the earth's tides like the earth's gravity has does not pull the crew on the ISS to one of the walls, so why are some high tides higher than others on the earth?
Presumably there is a tidal effect in the human body as the ISS passes between Earth & Moon. Although we see tides as quite large phenomena compared to the Earths mass the effect is small.
If there are tidal forces worth counting on, it affects everything inside the station as much as it affects the station itself. Nothing can be noticed within.
Everything in free fall feels like zero gravitation, but they rather call it microgravitation.
Tidal forces occur because the gravitational field isn't constant. The variation in the gravitational field across a space station isn't very big (it will be there, just not noticeable). But across something the size of a planet it can have a significant effect.
Originally posted by mtthw Tidal forces occur because the gravitational field isn't constant. The variation in the gravitational field across a space station isn't very big (it will be there, just not noticeable). But across something the size of a planet it can have a significant effect.
Also, those tidal forces are acting on a body of water thousands of kilometers wide so the effects accumulate to a large movement of water. If you had a swimming pool, for instance, the tidal force would act just as much per cubic meter as it does on the oceans but the size of the pool being just a few cubic meters, the pool water is constrained by the container so the total movement would be miniscule, I wonder if anyone has conducted an experiment to actually measure the tides on swimming pool sized bodies of water? Say an indoor pool with little air movement above it and watch with laser reflection or something. But the gist of that is the force per cubic meter is so small, in the ISS, it would probably not be noticeable.
the volume of water on the earth stays the same, piling up on both the high tides (well kind of), so there would be virtually effect on the swimming pool, allthough the reduction in gravity caused by the pull of the moon may make the water level very slightly higher.
Originally posted by sonhouse If you had a swimming pool, for instance, the tidal force would act just as much per cubic meter as it does on the oceans but the size of the pool being just a few cubic meters, the pool water is constrained by the container so the total movement would be miniscule, I wonder if anyone has conducted an experiment to actually measure the tides on swimming pool sized bodies of water? .
just searched on google, no experiments found but a mention of swimming pool tides on yahoo answers
Originally posted by battery123 the volume of water on the earth stays the same, piling up on both the high tides (well kind of), so there would be virtually effect on the swimming pool, allthough the reduction in gravity caused by the pull of the moon may make the water level very slightly higher.
The bigger swimming pool, the bigger effects of the tidal forces.
If you have a swimming pools big as the Pacific, you get effects as the Pacific.
Originally posted by FabianFnas The bigger swimming pool, the bigger effects of the tidal forces.
If you have a swimming pools big as the Pacific, you get effects as the Pacific.
Funny how that works🙂
So the next question would be, what is the smallest body of water could you unambiguously detect the action of tides? A drop?
Maybe with a laser reflecting on the surface tension layer, since it's round, it would reflect at a given angle depending on its shape, if it were more or less round to begin with, like a dew drop on a rose, then the laser may detect changes in the shape, you could shoot the laser at a near tangent to the top surface and get a greatly amplified view of changes to the droplet. You would have to compensate for evaporation rates because the drop would disappear eventually. Now that kind of experiment is in the realm of possible for amateurs like us. Would that win a prize at a science fair if it was shown true?
Originally posted by sonhouse Funny how that works🙂
So the next question would be, what is the smallest body of water could you unambiguously detect the action of tides? A drop?
Maybe with a laser reflecting on the surface tension layer, since it's round, it would reflect at a given angle depending on its shape, if it were more or less round to begin with, like a dew drop on a rose, ...[text shortened]... f possible for amateurs like us. Would that win a prize at a science fair if it was shown true?
You'd also have to allow for waves on the surface. For a small body of water almost any disturbance in the air will cause a disturbance in the water bigger than tidal effects, I'd have thought.
Normally one thinks of tidals as an elevation of water surfaces. The reality is that the entire rocky crust is rising with tides. Several meters in fact. But you don't see it, nor feel it, but it is detectable from satellites.
Sometimes we hear from astrologers that the moon is very significant in astrology, "Just watch the tidal forces!"
In reality, the influence of the moon is quite minuscle. The gravitational influence from the nurse at a child birth to the baby is in fact far more prominent than the gravitational force from the moon.
Originally posted by FabianFnas The bigger swimming pool, the bigger effects of the tidal forces.
If you have a swimming pools big as the Pacific, you get effects as the Pacific.
I dont think this is true.
An enclosed body of water will exhibit much, much smaller tides than an enclosed one. So even if you have your Pacific sized swimming pool the effect wont be that big.
As an example consider the Mediteranean sea compared to the English Channel. The Med has negligible tides (Straits of Gibralter is too narrow to lose/gain as much water as Moon wants to 'pull'/'push'😉. The Channel has mighty big tides.
Also remember that high tide on the English side of the channel is high tide on the French side too (give or take an hour or so due to geographical effects). The water isnt just slopping about!
Originally posted by wolfgang59 I dont think this is true.
An enclosed body of water will exhibit much, much smaller tides than an enclosed one. So even if you have your Pacific sized swimming pool the effect wont be that big.
As an example consider the Mediteranean sea compared to the English Channel. The Med has negligible tides (Straits of Gibralter is too narrow to lose/gain ...[text shortened]... give or take an hour or so due to geographical effects). The water isnt just slopping about!
The tides of the English Channel is only patrly an effect of the gravitation of the sun and the moon. The rest is an effect of the waters 'slopping around'.
I can't see that the Moons gravitational effect is particularly stronger at the area of the Channel than of my swimming pool at my back yard.
Tidal Forces
07 Oct '07 12:31
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