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Earth moves a tiny bit further away, and slows down because it is now moving up hill out of the gravitational well.V goes down as R goes up. Throw a ball in the air and watch V go down as the ball rises.You seem to be neglecting the fact that things slow down when moving away from a gravitational source. Kinetic energy is being lost to acquired gravitational potential energy.

Tides just apply thrust. Thrust adds (or removes) energy, and the new energy level finds a balance between those two things. V goes down as R goes up.

Based on Newton first law: In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.[2][3]"Let's start by asking the following:What will happen if we eliminate completely and at once the gravity force from the Earth/sun orbital system?If I understand it correctly, the Earth will "continue to move at a constant velocity".Therefore, do you agree that for orbital system it is incorrect to assume that "things slow down when moving away from a gravitational source"?

Hence, do you agree that the Earth will not change its inertial velocity "unless acted upon by a force"?

Based on your explanation - the Tidal is not a force. It just pushes the Earth further away from the Sun.

As we increase R we actually decrease the gravity force.Gravity force is the power which holds the Earth in the orbital cycle around the Sun.

It must be fully synchronized with the inertial velocity of the earth in order to hold it in the orbital cycle.It is clear that as Tidal increases R it actually decreases the gravity force.

Therefore, once the Tidal push away the Earth from the Sun, there is less gravity to hold the Earth in a balance with it's current inertial velocity/force.Hence, Less gravity force to hold the Earth on the orbital track means that the Earth is moving further away from the Sun.

That increasing radius, decreases the gravity force more and more.Therefore, do you agree that from Orbital point of view, once we push away the Earth from the Sun, the Earth is starting to move down the hill?

At some point it must be totally disconnected from the sun gravity and "continues to move at a constant velocity" as explained by Newton first law.

Unfortunately, I couldn't find in your explanation how tidal sets a negative force which slow down the Earth velocity in order to compensate less gravity force due to increasing in R.

Can you please explain this issue?

Would you kindly show by mathematics how tidal sets thrust that adds energy which pushes the Earth away from the Sun, while somehow the Earth reduces its velocity without implying an external force?

So that F is the thrust, pushing in the direction of orbit, not outward. That speeds up the planet, which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path. That takes Earth further away from the sun, so it gains potential energy and loses kinetic energy.

However, would you kindly explain why the Earth is losing kinetic energy due to Tidal?It was stated that the tidal trust speeds up the orbital velocity of the Earth.That speeds up the planet (Lets assume by Δv), which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path.That takes the Earth further away from the sun, so it gains potential energy.So far so good!

However, with regards to the kinetic energy. The kinetic energy of the Earth just before speeding up the planet is: mv^2.After increasing the speed by Δv, the updated kinetic energy should be: m(v+Δv )^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.

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^2The formula for potential energy is: Ep = GMmR^2I could not understand why an increasing in potential energy (Ep) should decrease Kinetic energy (Ek).Can you please explain the rational of this idea for orbital system?Do we have any formula which shows the energy transformation?

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

The formula for potential energy is: Ep = GMmR^2

No. That is the formula for gravitational force. Potential energy would be that force integrated with respect to R, orEp = -GMm/R

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.

So, every particle at the earth attracts every other particle in the Universe due to gravity force.That is the ultimate answer for the tidal friction on any orbital system.

Therefore, there is no proof for the following statement by Newton gravity formula: Quote from: Halc on 04/12/2018 23:17:00The spin of Earth is the road, so the friction is the brakes, speeding up the moon.

If we think that the spin of the Earth can speed up the moon, than by definition we must change the Newton's law of universal gravitation so it will also include that impact.

Based on the current Newton's law of universal gravitation, "every particle attracts every other particle in the universe with a force..."

So, any particle on Earth might go/down or spin faster/slower/locked due to the local impact of gravity force (tidal activity).

If we believe that the spin of a particle in one object can affect the velocity of a particle at a far end object than:

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"?