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The Earth to the moon distance, R= 384,400 km...The front Bulge to the moon distance = R(front) = 384,400 km - 6,371 km = 378,029 KmThe far end Bulge to the moon distance = R(front) = 384,400 km + 6,371 km = 390,771 Km.F(front bulge) = G Mb m / R(front)^2 = G Mb m / (142.9 10^9)F(rear bulge) = G Mb m / R(front)^2 = G Mb m / (152.7 10^9)

Because it pulls in the direction of motion (energy increase), not tangential to it (acceleration without energy increase). Yes, it is a very minor force compared to the main component, but it is always forward, so the effect is cumulative forever. The main force is always balanced in all directions, so the cumulative effect is zero after each month.

However, I still wonder why you insist to call it "Thrust".

The meaning of Thrust by Google is : "the propulsive force of a jet or rocket engine." "Push (something or someone) suddenly or violently in the specified direction."

I couldn't find any sort of "engine" in the Bulges activity. It is just increases the gravity force.

Never the less, I think that I understand the source for your statement:Please see Figure 7.24 in the following article:

"Figure 7.24 The tidal bulge raised in Earth by the Moon does not point directly at the Moon. Instead, because of the effects of friction, the bulge points slightly "ahead" of the Moon, in the direction of Earth's rotation. (The magnitude of the effect is greatly exaggerated in this diagram.)

Because the Moon's gravitational pull on the near-side part of the bulge is greater than the pull on the far side, the overall effect is to decrease Earth's rotation rate."So, if I understand it correctly, as the bulge points slightly "ahead" of the Moon, we believe that it pulls in the direction of motion (energy increase)"Is it correct?

I will try to illustrate some of the vector work, using the guess that friction pushes the bulges 1000 km off center.So we've defined two triangles will 1000km on the short side and 378029 or 390771 on the other.F(front) was computed at GmMb * 7e-12, so the forward thrust from that bulge isF(forward) = GmMb * 1.8517 e-14 F(backward) = GmMb * 1.6762 e-14The 1.8517 figure is 700 force * 1000km / 378029 km, or essentially the F multiplied by the sin of the angle formed at the moon between the COG line and the line to the near bulge.Difference is 1.755 e-15 GmMb cumulative forward thrust forever adding energy/angular momentum to the orbit of the moon.

So, we believe that as the bulge points slightly "ahead" of the Moon, it pulls in the direction of motion (energy increase)".In order to understand the real impact of that offset is, let's verify the following conditions:1. There is no offset (0 degree) and the bulges are pointing directly to the moonDo you agree that the effective distance between the bulges to the moon is maximal?. Therefore, can we assume the diffrence between their gravity force is maximal. However, can we assume that the thrust is Zero?

2. The offset is maximal (90 degree). The bulge are located at/almost the poles.

therefore, the distance between the front bulge to the moon is actually equal to the distance of the rear bulge to the moon. Hence, can we assume that they have the same gravity force? However, what is the expected thrust? Is it zero or infinite?

3. The offset is 60 degree. Cos (60) = 0.5 Hence, the effective radius is 0.5 * r = 0.5 * 6371 = 3185.5 K.m (Let's assume 3000 Km)In this case, the effective distance to the moon is as follow:R(front) = 384,400 km - 3,000 km = 381,400 KmR(rear) = 384,400 km + 3,000 km = 387,400 Km.So, it's easy to calculate the gravity force of each one, however, I can't understand why there will be any thrust.I wonder if it is related to the idea that there are two bulges.

Hence, if there was just only one bulge, (let's assume only the front bulge), does it mean that the thrust will be zero?

I have read your following answer in thread 40 and couldn't understand the source of the thrust:

In order to understand the idea.Let's eliminate the Earth.Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).Try to put them at any offset as you wish (With regards to the Moon).How can they set any sort of thrust on the moon?

I only see gravity force.Do we base the idea of thrust on Newton gravity force? How?If no, which law proves that there is a thrust?

The 1.8517 figure is 700 force * 1000km / 378029 km, or essentially the F multiplied by the sin of the angle formed at the moon between the COG line and the line to the near bulge.Difference is 1.755 e-15 GmMb cumulative forward thrust forever adding energy/angular momentum to the orbit of the moon.

Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).Try to put them at any offset as you wish (With regards to the Moon).How can they set any sort of thrust on the moon? I only see gravity force.Do we base the idea of thrust on Newton gravity force? How?If no, which law proves that there is a thrust?

They don't. They're in orbit, so they'll go around each other every X many hours, not every month. They'd not add any cumulative thrust to the moon. The 3-body problem would probably make the whole system unstable after not much time.

It is only gravity, yes. Work out the forces in 2 dimensions, not as a scalar. Force is a vector, not a scalar.If I am on a skateboard going down a slope, I speed up. That means gravity force is providing me with thrust, increasing my speed. Is it so unimaginable that it might do this? If I was moving up a slope, that same gravity would slow me down, and not be thrust, but rather a braking action (or be a negative thrust if you will). If I am on a perfectly level lot, there is no speed change, so the same gravitational force results zero thrust to my skateboard.

However, when I have tried to focus in a general case of the offset, you have informed me that there is no thrust due to offset:Quote from: Dave Lev on 08/12/2018 06:25:48Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).Try to put them at any offset as you wish (With regards to the Moon).How can they set any sort of thrust on the moon? I only see gravity force.Do we base the idea of thrust on Newton gravity force? How?If no, which law proves that there is a thrust?Quote from: Halc on 08/12/2018 14:25:37They don't. They're in orbit, so they'll go around each other every X many hours, not every month. They'd not add any cumulative thrust to the moon. The 3-body problem would probably make the whole system unstable after not much time.Please be aware that I have specifically asked to set them at any offset as you wish. So, does it mean that in a general case of offset there is no thrust?If there is no thrust due to the offset, how can we use this idea for tidal friction?

I like the idea of skateboard.This shows that we can convert gravity force into thrust.Don't forget that Newton have used the example of falling Apple to find the whole idea of gravity.However, we can't limit the skateboard idea just for tidal friction explanation.If there is a possibility to convert Gravity into thrust due to the offset in a tidal system, than please set a formula for thrust for general case of offset.This is a breakthrough concept in gravity.Newon, kepler and Einstein didn't offer any solution for converting gravity force into thrust in orbital system.

Our scientists believe that it is feasible to get thrust due for tidal friction offset as follow:"1. There is no offset (0 degree) and the bulges are pointing directly to the moonNo thrust, right. The difference between their force is irrelevant then. 2. For offset 60 degreeEach bulge is off to the side by Sin(60) in this case (about 5500 km) so the force is not straight to Earth but has significant forward/backward components respectively. 3. The offset is maximal (No, 90° - eastward and westward, not at the poles)Still zero since they're equal and opposite."Therefore, we should set a formula that represents this understanding.So, can we set a formula (or graph) for thrust per Offset tidal friction phase?

However, If thrust works for tidal offset idea, it should also work for the example which I have offered.If it doesn't work for this example, than it should not work also for tidal offset.

How can we come with idea for a special case and close it only for that case?Do you agree that if we can't open the idea for any offset in orbital system, than we might have a problem with this idea?

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.

Thanks for the explanation about the moons.However, I have no clue why if the Moon orbits in one direction/periods it will be pushed outwards, while if it orbits in the other direction/periods, it should be pulled inwards.In tidal friction, the offset of the bulges sets the Thrust?

In this case, there is no offset.

Quote from: Halc on 11/12/2018 13:02:09Metis 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.Can you please explain why the orbital direction/periods can set a thrust?

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?