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Quote from: Halc on 01/12/2018 20:28:34And in moving out of the gravity well, the kinetic energy is lost to that potential energy. It slows with a greater negative Δv than the positive Δv from the thrust, just like a roller coaster slows (exchanges kinetic energy for potential) as it moves further from Earth center.That's an interesting idea.The formula for Kinetic emery is: Ek = mv^2
And in moving out of the gravity well, the kinetic energy is lost to that potential energy. It slows with a greater negative Δv than the positive Δv from the thrust, just like a roller coaster slows (exchanges kinetic energy for potential) as it moves further from Earth center.
The formula for potential energy is: Ep = GMmR^2
The formula for potential energy is: Ep = GMmR^2No. That is the formula for gravitational force. Potential energy would be that force integrated with respect to R, orEp = -GMm/R Thus the total energy of an object in orbit is Et = mv^2/2-GMm/RThe total energy of an object in orbit is conserved assuming no additional energy is added.
Conclusion:We have actually increased the Ek by ΔEk (due to tidal) and therefore got an increased velocity by Δv.This increase Δv set an increased radius ΔR.This change in the radius increased the Ep.Now we want that this new Ep will help us not just to eliminate that ΔEk which we have added (due to tidal), but also more that that in order to achieve our goal that V3 must be lower than V1.Actually, lower velocity by itself is not good enough.Too low is also problematic.We must get a very specific velocity to meet the new gravity force.Is it a realistic goal?Only if we set full calculation we can verify if it works or not.
Et = Kinetic energy (Ek) + potential energy (Ep) = mv^2/2+ (-GMm/R) However, as "The total energy of an object in orbit is conserved assuming no additional energy is added." Et = constant for a given orbital cycle.Therefore, if we increase the potential energy (Ep) we by definition decrease EkLet's verify this idea by starting at point 1Hence:Et1 = Kinetic energy (Ek1) + potential energy (Ep1)Et1 = m(v1)^2/2+ (-GMm/(R1))In our case, the tidal set a thrust that increased the kinetic Energy by ΔEk and therefore we have got an increased velocity by (Δv).
Now we can claim that the new Total energy (Et2) due to additional energy which had been added is:Et2 = Et1 + ΔEkEt2 = Ek1 + ΔEk + Ep1This is the new starting point (just before increasing the radius)."
At that moment we can claim that:Et2 = Ek1 + ΔEk + Ep1 = constantEt2 = Ek2 + Ep1 = constantEt2 = m(v1+Δv)^2/2+ (-GMm/(R1)) = constant
However, due to velocity increase, the radius must be increased let's call it (ΔR)Hence, Due to the radius increase:Ep2 = (-GMm/(R2)) = (-GMm/(R1 + ΔR))That increase in the Ep must now decrease back the Ek So, let's call it Ek3Et2 = Ek3 + Ep2 = constantEt2 = m(v3)^2/2+ (-GMm/(R1+ ΔR )) = m(v1+Δv)^2/2+ (-GMm/(R1)) Et2 = m(v3)^2/2+ (-GMm/(R1+ ΔR )) = m(v1+Δv)^2/2+ (-GMm/(R1))
Hencem(v3)^2/2+ (-GMm/(ΔR )) = m(v1+Δv)^2/2
If so, we need to proof that this new v3 is fully correlated to the expected new orbital velocity due to the gravity force at R2 = R1 + ΔR.Do you agree with that?
Conclusion:We have actually increased the Ek by ΔEk (due to tidal)
and therefore got an increased velocity = Δv.This increased Δv set an increased radius = ΔR.This increased ΔR increased the Ep by ΔEp.Now we expect that this ΔEp will help us not just to decrease the Ek by ΔEk (in order to get back to v1), but also more than that in order to achieve our goal that v3 must be lower than v1.Even in a perfect system, we can't request to get back all the energy that had been invested.
So, it is very challenging to expect that the ΔR which was a product of Δv will help us to cancel completely the Δv.Now, we expect that this ΔR will help us to gain move than just Δv.Is it realistic?
Even if we assume that it is realistic, how do we know that we get a perfect new velocity?Too low is also problematic.
We must get a very specific velocity to meet the new gravity force (at R2 = R1 +ΔR).
So, don't you think that it is too challenging goal?In any case, only if we set full calculation we can verify if it works or not.
However, it seems to me that there is a fundamental problem with the concept of Tidal Friction.let's start by understanding the meaning of Friction by Google: "The resistance that one surface or object encounters when moving over another".
Is there any possibility to increase the velocity of car by pressing the brakes?
Therefore, if we transform some of this energy into heat, than there must be less kinetic energy. Less kinetic energy means less velocity.
So, how could it be that a brake/resistance system (like a tidal friction) can increase the orbital velocity instead of decreasing it?
It is angular velocity in this case, but yes. The Earth is always slowing its spin due to tidal friction. That's why they keep having to add leap-seconds now and then. The day used to be about 10 hours long.
The spin of Earth is the road, so the friction is the brakes, speeding up the moon.
The near bulge is ahead of the moon, and thus pulls the moon forward. The far bulge rotates to behind the moon and pulls it backwards. The near one is closer, so that forward pull is greater than the backwards pull. Net effect is gravitational thrust on the moon.
Tidal forces cause a bulge on the near and far sides of Earth.
Exactly. And the thrust on the moon can be explained if you apply the law that way. Earth is not a point-mass. Point-masses cannot be susceptible to tidal friction.
ThanksBased on Wiki the highest point of the tidal bulge is 54 Cm
The Earth moon distance is 384,400 km. The radius of the Earth is 6,371 km.We can try to calculate the impact of each bulge. In order to do so, we need to estimate the mass of each bulge with reference to the Earth and verify the total impact.However, do you agree that 6,371 X2 is neglected with regards to 384,400 km?Do you also agree that the net mass in that 54 centimeters Bulge is also neglected with regards to the total mass of the Earth?
If so, it is quite clear that the impact of the bulges is virtually zero or close to zero.
In any case, it seems to me that the key point in our discussion is the center of mass based on Newton's second law for the description of the motion of extended objects:http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
If the bulges at the near side and at the far side are similar, do you agree that there is no change in the Earth center of mass location with regards to the moon.
Hence, based on this idea, those bulges shouldn't have any impact on the Moon orbital cycle as they cancel each other.
Never the less, even if we ignor the far end bulge, what might be the impact of that near side bulge to the Earth center of mass?
Hence, let's assume that we can find that those bulges increase the total gravity force on the moon by 0.00...01.
Do you agree that it should have a similar impact as we increase the effective mass of Earth by a relative quantity - Let's say by ΔM (M=Earth mass)?
The outcome is that the moon's gravity force will be based on M+ ΔM instead of M.This effective Earth mass (M+ΔM) should increase the velocity of the Moon.
So, instead of orbiting at velocity v, it will orbit at v +Δv.However, as it is a constant gravity force, the orbital velocity should stay at the same amplitude.
Hence, I don't see any transient thrust which temporarily increases the velocity.
If there is no transient increase in the velocity, do you agree that there is also no transient increase in R?
Therefore, without this transient increase in R how can we justify the whole idea of pushing the moon away from Earth?
Neglected? By what? No real clue what you mean by that choice of words.The impact is real, so what ever 'neglected' means, I think we should not neglect those numbers.I would have run the computations in terms of torque, but you method works as well.
Quote from: Halc on 05/12/2018 21:07:50Neglected? By what? No real clue what you mean by that choice of words.O.KLet's set a calculation.
Neglected? By what? No real clue what you mean by that choice of words.
Mb = Bulge mass = V(Bulges) / V (Earth) * M = 2.51 10 ^(-8) MHowever, the core of the Earth is made of metal which should be quite heavier than water. So, the ratio in mass should be higher.
In any case, it seems to me that the bulges mass is less than the total mass in the Mountains or even in one big chain of Mountains.
F = Earth Gravity force = G M m / R^2 = G M m / (147.7 10^9) = G M m 6.99 * 10^(-3) * 10^(-9) = G M m 6.99 * 10^(-12)
Bulges distance to the moon:The bulges don't point directly to the Moon.Therefore, the effective distance is less than full Earth Radius.However, I will use the full radius just as a worst case.
Hence: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)F(Bulge Total) = F(front) - F (rear) = G Mb m * 10^(-9) * (1/142.9 -1/152.7) = G Mb m * 10^(-9) * 4.49 * 10^(-4)
= G 2.51 10 ^(-8) M m * 10^(-9) * 4.49 * 10^(-4) = G M m * 4.49 * 10 ^ (-21)
The ratio in the gravity force between the Bulge and the Earth is:F(Bulge Total)/F(Earth) = G M m * 4.49 * 10 ^ (-21) / G M m 6.99 * 10^(-12) = 0.643 * 10^(-9).Hence
F(Bulge Total) = 0.643 * 10^(-9) * F(Earth)Conclusions:I have calculated the gravity force impact of the Bulges.
It seems to me as a very minor gravity force (comparing to the Earth gravity force).It might be even weaker than one big chain of mountains.So, why do you call it "thrust"?
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?