# The tides are the result of the rotation of the Earth and the whirlpools

Discussion in 'Pseudoscience' started by Fermer05, Apr 3, 2018.

1. ### DaveC426913Valued Senior Member

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For starters, satellites are not in zero gravity; they are in Earth's gravity, EXACTLY like the oceans and continents are.

Satellites are not floating weightless in space; a satellite 200 miles up is pulled toward the Earth with almost the same force (within a few percent) just as if it were on the ground.
All satellites are perpetually falling toward the earth; their lateral speed allows them to miss it when they fall. That's why it has to move at 25,000 mph, just to not crash into the Earth.
Geostationary sats are the same thing, just less pronounced.

And Fermer, the fact that you don't know this should ring an alarm bell in your head about how much you don't know about the relevant physics.
I won't, but it should.

3. ### Michael 345Bali 1 week here 2 to goValued Senior Member

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Veeerrry technical point - they are - just like the Earth is floating weightless in space

For something (anything) to be weighed
• It needs to be in contact with a weighing device
• Weighing device much be in contact with another body (larger mass) such that its measuring components are in-between the larger mass and the item being weighed
• The weight of any object varies depending on the mass of the object on the other side of the measuring device
Akin to this
Astronauts are NOT weightless in space because they are out of the Earth's gravity. They are in a micro gravity environment where they are attracted to the mass of the station (to which they are in contact with via the air inside the station)

The station provides the micro gravity

Since the station is not in contact with the Earth it is also weightless floating in space (disregarding the minute attraction it would have towards its occupants)

5. ### RainbowSingularityValued Senior Member

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should not be confused with a zero gravitational field

7. ### Michael 345Bali 1 week here 2 to goValued Senior Member

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Where would there be a zero gravitational field?

:}

8. ### RainbowSingularityValued Senior Member

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Deep space between 2 Galaxys ... per-say... 2 extremely distant galaxys
i am postulating the difference between competing gravitational forces Vs no gravitational force to illistrate the massive difference in scientific basics that might have been over looked by the author.
though...
RE: back ground radiation & the recent discoveries not withstanding for my example(which is purely for a mathamatical model example)
http://www.sciencemag.org/news/2017...s-physics-nobel-discovery-gravitational-waves

soo technically, black holes influence our tides, not only the moon

Last edited: Apr 9, 2018
9. ### Michael 345Bali 1 week here 2 to goValued Senior Member

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I would contend very very weak gravity but I seriously doubt zero

But from me this is just speculation

10. ### Gawdzilla SamaValued Senior Member

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An observer between galaxies would have a small gravity well, yes?

11. ### Q-reeusValued Senior Member

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I won't embarrass by quoting one or two earlier posts, but will point out that by definition, anything following a geodesic is in a state of weightlessness. That fapp includes orbiting Earth satellites - with only minuscule tidal forces (averaging to zero over the satellite etc. physical extent) spoiling the picture. Well apart from generally even more minute residual atmospheric drag, Solar radiation pressure and sporadic Solar wind. Can't think of anything else nitpicky to add there.
https://en.wikipedia.org/wiki/Weightlessness

12. ### Gawdzilla SamaValued Senior Member

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If the satellites don't have any weight why do they move in a circular orbit, constantly falling toward the Earth?

13. ### Michael 345Bali 1 week here 2 to goValued Senior Member

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Mass and momentum

Not sure if directed to me but suspect to a person on my iggy list

14. ### RainbowSingularityValued Senior Member

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i dont know the math but my laymens understanding is yes, they are in a state of constant free-fall ?
they are loosing energy that they gain constantly ?

15. ### Q-reeusValued Senior Member

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They have mass whether or not they are in a weightless state. Gravity acts on mass - and so does inertia. In Newtonian language, F + (-ma) = 0.

16. ### Michael 345Bali 1 week here 2 to goValued Senior Member

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Second paragraph not make sense

They have a forward momentum which counters the downward tug of gravity

17. ### RainbowSingularityValued Senior Member

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this ?
https://en.wikipedia.org/wiki/Standard_gravitational_parameter
and this
https://en.wikipedia.org/wiki/Circular_orbit#Equation_of_motion

18. ### Michael 345Bali 1 week here 2 to goValued Senior Member

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Did not go to the links

QUOTE

The escape velocity from any distance is √2 times the speed in a circular orbit at that distance: the kinetic energy is twice as much, hence the total energy is zero

appears to me to imply (or state) - to escape from a circular orbit will require a speed which is - the square root of the speed multiplied by that speed (the particular speed being that which is relevant to that orbit)

In other words extra energy will be required to increase the speed to move out of the circular orbit (makes sense)

If extra energy is NOT applied (to reach escape velocity) the kinetic energy will retain the orbit

Again the above does not make sense

19. ### billvonValued Senior Member

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The moon's gravity can be (and has been) directly measured in a lab - so it certainly reaches the ground. It is about 10e-7 G's.

Indeed, there is no known physical way to keep it from reaching the ground.

It does. In fact, the planets affect the tides as well. Since they are so far, their effect is measured in tenths of a millimeter instead of meters.
Not only have they been measured, they must be subtracted from accurate measurements of the Earth's gravity, since they would otherwise make those measurements inaccurate.
Correct. That is because the Earth spins.
Here is a paper that talks about the fluctuations and the problems they cause for accurate gravity measurements:

http://homepage.oma.be/mvc/pdf/ECGS_22_Efficiency_Tides.pdf

Contains a lot of data if you are interested.

From the introduction:

============
The Sun and the Moon exert tidal accelerations with maximal peak to peak amplitudes of 250 µGal. This is the most important signal affecting gravity measurements, if we except transient seismic perturbations. The calculation of tidal phenomena requires a representation of the tidal potential. Nowadays very accurate tidal potentials based on the relative position of the Moon, the Sun and the planets (Wenzel, 1996a) allow us to compute the gravimetric tides for a rigid Earth at the nanoGal level.
============

(Note - a Gal is an acceleration of 1 cm/sec^2.)

Last edited: Apr 9, 2018
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20. ### billvonValued Senior Member

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It, of course, is exposed to it. A geosync satellite is affected by both the Earth's and the Moon's gravity - but since the Moon's mass is much less than the Earth's, and is ten time farther away, its effect is eight orders of magnitude less than the Earth's.
Correct. It does indeed affect the orbit of satellites - just in a very, very small way.

Perhaps a book on basic physics would help you understand this topic?

21. ### RainbowSingularityValued Senior Member

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it was more soo a question about the nature of inert states in relation to universal gravitational effects.
i was pondering the nature of the relationship in the solar system.
at no point i would imagine the satallite is NOT effected by a gravitational field.
excuse my backward writing

22. ### DaveC426913Valued Senior Member

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If the satellite stopped moving laterally, it would plummet to the Earth at 9.8m/s^2, just like the continents and oceans would if they could.

They are attracted to the Earth. They are falling toward it. Their lateral motion takes Earth out from under them before they collide.

Last edited: Apr 10, 2018
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23. ### Q-reeusValued Senior Member

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Nitpicking. See my #108, #112.