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However, the water bulges in oceans are very unique for the Earth/Moon system.In one hand, this is the only planet in the solar system with so much water and on the other hand, the moon is relatively big enough to form the bulges.
We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?
They don't have water. They don't have relatively big moon around them. They don't form those water bulges.So, if tidal friction can't be the answer for their drifting outwards activity, what could be the answer for that?
Because they go around slower than the spin of the primary, and in the same direction. If they orbited lower (at or below geo-sync), or if they went around the opposite way, they'd get pushed down. Such objects tend to fall into their primaries before too long.
So, the tidal friction idea between Earth/Moon is actually none relevant to all the other orbital systems in the solar system.
Now there is new idea.All the planets and moons are pushed away as they all "go around slower than the spin of the primary, and in the same direction".
1. Would you kindly explain this idea?
2. Can we prove it by mathematical calculations/formula?
3. Why our moon isn't pushed outwards due to this idea?
Quote from: Dave Lev on 10/12/2018 16:00:34We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?Because they go around slower than the spin of the primary, and in the same direction. If they orbited lower (at or below geo-sync), or if they went around the opposite way, they'd get pushed down. Such objects tend to fall into their primaries before too long.
Metis and Adrastea have orbital periods shorter than the primary, and hence have negative thrust. They will eventually fall in turn, giving Jupiter some nice rings.
Yes, the offsets set the thrust. If the offset is positive, so it the thrust.
Quote from: HalcYes, the offsets set the thrust. If the offset is positive, so it the thrust.Can you please explain how the orbital direction sets the offset?
Why orbital periods that are shorter than the primary can set a negative offset?
Do we have any drawing for that idea (as we have for the water bulges on Earth)?
So, how do we get any sort of offsets due to orbital direction or orbital periods?
Because the friction of the slower primary drags the bulge to the rear, a negative offset.
There are nine planets in the solar system.We all know that there are bulges in our planet.However, do we see any sort of bulge at any other planet in the solar system?If we don't see any bulge at any other planet, how can we think that a bulge which doesn't exist can set an offset that we wish for?
So, do you agree that without confirmed bulges per planet, there is a problem with this hypothetical idea?
Water isn't necessary. The tides are quite significant in amplitude on planets/stars that are not rocks. Venus has a thick atmosphere to drag around. Pluto and Mercury are the only planets with nothing but solids to work with, and both those have become tide locked with the most significant gravity source nearby.
Why water isn't necessary?Please remember that the tidal on Earth is based on water.
If I remember correctly, at land the tidal is just few cm (2-3 Cm?).
The mass ratio between Earth/moon is significantly higher than any other Planet/moon system.
Did we try to measure the Tidal impact on other planets?
If I understand it correctly, the offset is also due to water.
Therefore, without verifying minimal bulge amplitude and offset, I really can't understand why we are so sure that there is a minimal thrust that can push or pull the moon.
"Ganymede's mean radius is 1,635 miles (2,631.2 km). Although Ganymede is larger than Mercury it only has half its mass, classifying it as low density.Therefore, it is clear that the ratio between Jupiter and this biggest moon is very low.
I would assume that at this ratio, the tidal impact on land in Earth might be less than one millimeter
Jupiter has 79 moons. Therefore, there is good chance that those moons cancel the tidal impact of each other.
There is one more issue.For some moons we believe that there is a negative thrust.
However, did we measure if they are pulled inwards?
What is the chance that they will not be so cooperative with our theory?
Why we are so sure with our theory while we only have real measurements of only Earth/Moon and Sun/Earth system (Both drifts/Pushed outwards?
For some moons we believe that there is a negative thrust. This negative thrust should pull those moons inwards. This is the theory. However, did we measure if they are pulled inwards?
Like Phobos and most of Jupiter's moons, yesJupiters outer moons are so far out that the tides might not have measurable impact. The two inner ones very much do have measurable orbit degradation. Phobos has massive degradation, and has only some tens of millions of years left in its life.Making up your facts I see. We've plenty of measurements of the others, at least the things near their primaries, which have significant forces acting on their orbits.
Most planets don't have free flowing water. Europa does, even it there's a crust that has some inhibiting effect to its tides. I think Europa is tide locked, so no matter.
Why do we ignore the distance between the moon/rings to the planet?
As an example:The Earth Radius is 6370 Km. If Phobos would orbit the Earth at the same ratio, its orbital distance from the Surface should be about 11,000 Km.A commercial airplane is normally fly at about 10,000 Km.If it is expected that airplane should come/fall down, why is it so big surprise that Phobos is also coming/falling down?
With regards to Jupiter moons/Ring:https://en.wikipedia.org/wiki/Rings_of_Jupiter"Main ring - The narrow and relatively thin main ring is the brightest part of Jupiter's ring system. Its outer edge is located at a radius of about 129000 km (1.806 RJ;RJ = equatorial radius of Jupiter or 71398 km) and coincides with the orbit of Jupiter's smallest inner satellite, Adrastea.[2][5] Its inner edge is not marked by any satellite and is located at about 122500 km (1.72 RJ).[2]So, Jupiter radius is 71398 Km.Therefore, If this ring would orbit the Earth at the same ratio (1.806), its orbital distance from the Surface of earth should be about 11,500 Km. (Same distance at commercial airplane on Earth).
So, don't you see a similarity between airplane orbit around the Earth, to Phobos around Mars and Main ring around Jupiter?
Don't you agree that with tidal or without it, all of them must fall down?
Therefore, how can we use an object which it's orbital cycle is so close to the host to prove a negative thrust due to tidal???Can you please find one moon (only one) in the whole solar system that is located long enough from its planet which is pulled inwards due to negative thrust (But please - real prove for that)?
With Regards to Europa:Quote from: Halc on 13/12/2018 00:58:08Most planets don't have free flowing water. Europa does, even it there's a crust that has some inhibiting effect to its tides. I think Europa is tide locked, so no matter.Europa is tidally locked, so the same side faces Jupiter at all times.The surface of Europa is frozen, covered with a layer of ice, but scientists think there is an ocean beneath the surface. "So, It is located far enough from Jupiter. Ratio of about 1:10.It is covered with a layer of ice, therefore, the chance to set any significant bulges due to tidal is quite minimal (even if it has ocean beneath the surface).
Does it have tidal bulges?
Do those bulges set the positive/negative offset?
Do we know if it is pushed outwards or pulled inwards?
I assume that the answer is - No, we don't know as it is too far away to measure.
So, if all the moons are too far away from us, how do we know that all of them must obey to tidal friction idea?
Airplanes fly at 10 km, not at 10000
The laws of physics work everywhere, not just where humans confirm them in court
Quote from: Dave Lev on 14/12/2018 16:28:53Based on Newton's Shell Theorem:https://www.math.ksu.edu/~dbski/writings/shell.pdfThe gravitational field outside a spherical shell having total mass M is the same as if the entire mass M is concentrated at its center (Center of mass).So, if that is correct, than the gravitational field of all the mass of Earth (including the bulges), must be concentrated at the center (center of mass).Right, but the theorem works only for a very regular object, a sphere in particular. All kinds of funny things can be done if the objects are irregular. I can take two objects and put the centers of gravity very close to each other and actually get them to repel each other. I just can't do it with spheres.
Based on Newton's Shell Theorem:https://www.math.ksu.edu/~dbski/writings/shell.pdfThe gravitational field outside a spherical shell having total mass M is the same as if the entire mass M is concentrated at its center (Center of mass).So, if that is correct, than the gravitational field of all the mass of Earth (including the bulges), must be concentrated at the center (center of mass).
How could it be that we can't use Newton's Shell Theorem for the Earth, while the orbital velocity of stars around the galaxy is based on Newton's Shell Theorem.http://www.astronomy.ohio-state.edu/~ryden/ast162_7/notes30.html"In the above equation, M is the total mass in a sphere of radius a, centered on the galactic center. (The mass outside the sphere doesn't have any net effect on the star's orbit).
So, based on that theory we took all the matter (Stars/dust/SMBH/Dark Matter...) in the orbital radius and set the calculation as all the mass is located at the very center of the galaxy.
Hence, from the galaxy point of view, we have used all that variety of matter in the Sun' orbital radius as a very regular sphere.However, when we come look at the Earth, suddenly it is not a regular object.
Is it real?How could it be that the none regular matter in the sun' orbital radius is more regular than the Earth itself.
Somehow, it seems to me that we are using the idea of "regular" to prove an object which is by definition none regular (as the sphere inside the orbital radius of the Sun)
How can we use the Newton's Shell Theorem to prove the orbital velocity of the Sun,
while we reject this idea just because it contradicts our theory about tidal friction?
QuoteHow can we use the Newton's Shell Theorem to prove the orbital velocity of the Sun,We didn’t. We used that 2nd rule about only the matter within the orbit contributing to the velocity. You can use that rule on the Earth/moon system as well. It works just great. It means you need to take into account all the little stuff in low orbit around Earth to account for the moon’s speed, but you don’t need to to account for the moon when computing the orbital speed of the ISS. That’s what the rule says.
QuoteWhat do you mean by: "2nd rule about only the matter"?Is it Newton's second law?The first rule you quoted was Newton's Shell Theorem concerning only uniform spherical objects.The 2nd rule you quoted was the one from the ohio-state site concerning the mass that has net effect on an orbit and the mass that doesn't.I don't know the name of that rule or who came up with it, but you quoted it, and then you misrepresented what it means.
What do you mean by: "2nd rule about only the matter"?Is it Newton's second law?
Quote"The utility of Newton's second law for the description of the motion of extended objects is the key to its general practical usefulness. The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:"Please advice.I don't see the relevancy of that law to this portion of the discussion. It says that the center of gravity of something like Earth follows a smooth curve even if the various parts (like your mailbox) don't.
"The utility of Newton's second law for the description of the motion of extended objects is the key to its general practical usefulness. The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:"Please advice.