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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

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).

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.

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 didnt. 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 moons speed, but you dont need to to account for the moon when computing the orbital speed of the ISS. Thats what the rule says.

What do you mean by: "2nd rule about only the matter"?Is it Newton's second law?

http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html"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.

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.

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.

Quote from: Halc on 15/12/2018 05:47:04QuoteWhat 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.The "second rule" is just a direct outcome from the "first rule" which is the Newton's Shell Theorem.With all the respect to ohio-state (and I have a respect...), they can't just invent new rules for gravity.So, the second statement is not a second rule, it is just a logical outcome from Newton's Shell Theorem.

Hence, by Kepler law we can calculate the total mass which is requested to meet the orbital velocity of the sun around the galaxy, while Newtons First Theorem tell us that: "M is the total mass in a sphere of radius a, centered on the galactic center.

Newtons First Theorem - If the density distribution of a ball of mass M is spherically symmetric, then the size of the force between the ball and a point mass m, that lies outside the interior of the ball, is given by the left-hand side of (1), where r is the distance between the point and the center of the ball."

This law is very relevantIt actually confirms that the shape of the object is none relevant for its central point of mass.

As long as all the masses in the object are fully connected the center of mass of this object is none relevant with its shape. (Spanner, dog, cat or even elephant).

http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html"A set of masses connected by springs will follow a path such that its center of mass moves along the same path that a point mass of the same total mass would follow under the influence of the same net force."

So, even if that spanner has an offset, it won't set any extra thrust.

In our calculation we just need to focus on its center of mass.

Conclusion -Based on the following laws (each one by itself):1. Newton's Shell Theorem 2. Newton's second law

The rules don't say that at all. They talk about net force and those net forces acting on the center of gravity of each object in question. So if the spanner puts a net force on an object that is anything but perpendicular to its motion, it will be exerting thrust to it.

You can't take laws by themselves. Let's take these two laws. Consider me on Earth with the sun directly overhead, and Mercury and Venus don't exist. In a sphere of radius R where R is the distance from the sun to me, only the mass in that radius determines my orbit. It is the Sun which is spherically symmetric. So the first law applies.Now let's apply the 2nd law. I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon), and the mass outside the radius R has no effect on my orbit, therefore I should drop in towards the sun, getting sucked into the sky. The Earth cannot hold me down because it is outside R when the sun is directly overhead, and thus does not contribute to my orbit, per the 2nd law.

Wrong!! Center of mass has no angular momentum, and this tidal thrust effect is all about net forces resulting from transfer of angular momentum. None of the laws above describe angular effects on the tumbling spanner and such.I throw a rapidly spinning pool noodle, and it is spinning far slower before it hits the ground. The net forces on the noodle do indeed determine the path of its center of gravity, but do not in any way describe the loss of spin. That rules is inadequate for the situation being described.The orbit about the galaxy is less about angular momentum and forces since there is no significant transference going on. You can treat a lot of things as point masses on that scale, but not the galaxy as a whole since it is not spherically symmetrical

Can you please prove that a spanner can put a net force on an object just by pointing to some offset?

Quote from: HalcYou can't take laws by themselves. Let's take these two laws. Consider me on Earth with the sun directly overhead, and Mercury and Venus don't exist. In a sphere of radius R where R is the distance from the sun to me, only the mass in that radius determines my orbit. It is the Sun which is spherically symmetric. So the first law applies.Now let's apply the 2nd law. I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon), and the mass outside the radius R has no effect on my orbit, therefore I should drop in towards the sun, getting sucked into the sky. The Earth cannot hold me down because it is outside R when the sun is directly overhead, and thus does not contribute to my orbit, per the 2nd law.That example is not clear to me.If the sun is just above me while I am waking on Earth, than based on Newton second law, I'm an integrated mass of Earth.

Therefore, the Sun has no impact at all about my location. If there is no Earth or moon, and it is all about me and the Sun, than my orbital velocity must be a direct outcome of gravity force based on R. In this case, the Earth and the Moon are not there to have any impact on my orbital cycle around the Sun.

With regards to the following message:"I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon)"This is a severe mistake.If I have to orbit around the Sun, there is no way to slow down due to that Earth spin velocity cancelation.

So, you don't have to maintain your orbital velocity. The Sun gravity force works for you. This is a key element.Newton didn't specify even one word about the impact of spinning velocity.This is a new idea which had not been confirmed.

If you believe that there is a possibility to slow down the earth orbital velocity

We do not discuss about rapidly spinning pool of noodle.

If the angular momentum or the revolving speed of the Moon or the Earth can slow down the orbital velocity of the Erath or the Moon - than please prove it.

I still can't understand how any sort of offset in bulges can set any sort of thrust.

Please use the spanner as an example. you are more than welcome to force the spanner in any sort of offset as you wish. Try to prove why by doing so we shall get extra trust on the orbital object.Please try to prove this idea by mathematics.

I give it a permanent stationary offset of 45°. The spanner is 40 cm long, 20 on each side of the axis. The lower side is 6372000 meters from the center of earth and has a force of GmM/40602384000000 acting on it. The upper end is a tiny bit further away from earth and has a force of around GmM/40602384382320 resulting in a 1e-6 % difference in force.

However, I really don't understand why there is a difference in forces.

Let's set the whole Earth at the shape of spanner.So, the Earth will look like an extended object with all of its mass while the length of each side is 10,000 Km. Based on Newton's Second Law for an Extended Objecthttp://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html"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:F = M a

In order to understand the calculation, we need to look at the Following "Newton's Second Law for a System of Particles"

http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html#c2"Newton's Second Law for a System of Particles:The form of Newton's second law for a system of particles will be developed with the understanding that the result will apply to any extended object where the particles are in face connected to each other.The center of mass of a system of particles can be determined from their masses and locations."

So, Our dear Newton set a complicated calculation in order to get the outcome of:M a = FNewton actually tells us that the shape of the object and its offset can't contribute any extra force as long as all the particles of the object are connected.

So, if you still think that there is an error in Newton calculation, please offer the updated calculation to prove why each side can contribute different force (while all the particles are fully connected).