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I don't see any contradiction between this formula to your explanations.Do you agree with that?
Based on this formula there is no change in the gravity force over time.Hence, by definition each new atom has to stay at the same radius forever.
Further our discussion about new matter creation at the excretion disc of the Milky way:Please see: https://www.thenakedscientists.com/forum/index.php?topic=75261.40I wonder why any new particle/Atom/Molecular is drifting outwards from the excretion disc of the milky way.Let's look at the following formula for gravity force:F = G * M * m / R^2M = The mass of the SMBHm = The mass of the particle/Atom/Molecular.Based on this formula there is no change in the gravity force over time.Hence, by definition each new atom has to stay at the same radius forever.So how can we justify the drifting outwards mechanism of any new particle/Atom/Molecular?Could it be that something is missing in the following formula:F = G * M * m / R^2Could it be that over time there is a change in the gravity force? (Even if it is only a very minor change).Do you know that the American continent is drifting away from the European continent by only 1.5 cm per year.In the past they were fully connected.So, just based on that small change per year, we have got the vast ocean between those two continents after long time.We know that:1. The moon is drifting away from the Earth (by about 1.5 cm per year)2. The Earth is also drifting away from the Sun.3. All the planets in the solar system are drifting away from the Sun.4. All/Most of the moons are drifting away from their host planet. But objects can also drift inwards.There are plenty of orbital objects (satellites...) around the Earth that drift inwards and eventually fall down.Hence, could it be that gravity force must be changed over time?Actually, if we think about it, it is clear that nothing can stay the same forever. Although there is no friction in space, it seems to me that even gravity can't stay the same forever.There must be some small change in gravity force over time (even very small change over very long time)So, could it be that we have to use the following formula:F = G * M * m / R^2 +/- F(t)F(t) = represents the change in the gravity force over time.If the object is in orbital cycle which is too close to the host, the value of F(t) must be positive.However, if the radius is long enough, the value must be negative.So, with regards to our solar system:All the planets are located far enough from the Sun.Therefore, the value of F(t) for each one of them must be negative. Hence, over time (long enough), the gravity force on each planet is decreasing. In order to compensate that decreasing in gravity force, the planet must drift outwards.In the same token, could it be that each particle in the plasma is located long enough from the SMBH.Therefore, its F(t) is negative.If so, the total gravity force on each particle in the plasma is decreasing over time.That decreasing gravity force drifts away the particle from the center. Do you agree with that?
Quote from: mad aetherist on 26/11/2018 22:14:42If gravity travels at c then all orbits (ie of your new particle, & everything else) should spiral into the spiral galaxy.Actually, it can be shown that if gravity travels at c or some other finite speed, then orbiting objects tend to spiral out from each other, violating conservation of energy.This article shows the basic concepts of the argument: http://www.flight-light-and-spin.com/proof/instant-gravity.htm
If gravity travels at c then all orbits (ie of your new particle, & everything else) should spiral into the spiral galaxy.
Sure, I agree (F is changing, not G)
It isn't drifting. It is being pushed away.
Artificial satellites crashing to Earth are a consequence of air resistance.
So, the corrected formula is as follow:F(t) = G * M(t) * m(t) / R(t)^2 = F(0) - Δ(t)F(t) = the gravity force at time t.
Δ(t) = Lost of gravity force due to Tidal friction (or due to any other interference)....With regards to the Tidal friction:Δ(t) is the estimated lost of gravity force due to Tidal friction.
In the Earth/Moon orbital cycle, the "tidal bulges" is very clear (few cm.).
If the Moon mass will be decreased, while the Earth mass will be increased, we might not notice this - "tidal bulges".However, do you agree that the tidal effect is still valid?
Hence, do you agree that even if we will decrease the Moon mass into a single Atom while we increase the earth mass into a SMBH, the tidal friction will still be there and the Atom will be pushed away?
So, can we assume that at any orbital system without air resistance, the orbital object is losing energy over time?
Hence, do you agree that in those kind of orbital systems, as the objects are losing energy over time, they are also losing gravity force over time and therefore they are pushed away from their host?
You might as just well say that the correct formula for force F at a given time is F = G * M * m / R2 as per your OP. Over time, any of those things might change, especially R if the path of m is not a circle about mass M, so F would then not be constant over time.
Friction reduces kinetic energy. It doesn't reduce force or momentum.
A potato on a table has constant gravitational force acting on it, because the table holds the R constant, putting the potato in a circular path. Take away the table, and the R starts decreasing. The path is no longer a circle, and F goes up. Neither M nor m has changed.It needs no external force. There already is one, as computed by F.
No. There is no forward force. Any object might have a trajectory that puts it on something other than a perfect circular orbit.So for an object already in a perfect circular orbit (you don't say this, but it needs to be qualified), the only way to change the orbital radius is by changing the gravity force or by applying a secondary force to one of the objects.
The former can be changed by altering the mass of the primary mass. Altering the mass of the orbiting thing makes no difference. I can take half the moon away (halving the gravitational force acting upon it) and it will have zero effect on its orbit. But take away half of Earth, and the moon will need to orbit further out.
But the tides put a cumulative thrust on the moon that actually adds to its momentum, and so it is always moving to a higher orbit
However, what do you mean by: " applying a secondary force to one of the objects"?
Quote from: Halc But the tides put a cumulative thrust on the moon that actually adds to its momentum, and so it is always moving to a higher orbitI can't understand this idea.1. Why tides put cumulative thrust on the moon?
2. If tides put a cumulative thrust on the moon that actually adds to its momentum, why this extra momentum can't be considered as a force?
(Don't you agree that extra momentum should increase the speed and therefore it should increase the force)
Please look at the following diagram:http://www.scienceline.ucsb.edu/images/earth_sun_bodydiagram.png
Do you agree that:1. The Centrifugal force is a direct outcome of the moon speed?
2. In order to keep the moon in the same radius its Centrifugal force must be equal to its gravitational force?
3. In order to increase R (while keeping it in a perfect orbital cycle) we must decrease at the same moment and at the same amplitude the Centrifugal force and the gravitational force?
Sorry, I really can't understand how tidal changes R while we claim that it doesn't change force?
QuoteQuote from: Dave Lev on 30/11/2018 18:56:40QuotePlease look at the following diagram:http://www.scienceline.ucsb.edu/images/earth_sun_bodydiagram.pngThat diagram is really bad science. Centrifugal force does not exist in the inertial frame implied by that diagram. If it did, it would counter gravity force and the path of Earth would not curve.
QuotePlease look at the following diagram:http://www.scienceline.ucsb.edu/images/earth_sun_bodydiagram.png
In an inertial frame, no. Only the force of gravity exists, and that force continuously accelerates the moon, keeping it in a circular orbit.
QuoteQuote from: Dave Lev on 30/11/2018 18:56:40QuoteSorry, I really can't understand how tidal changes R while we claim that it doesn't change force?Tides push with the rotation (to the left in your picture). Gravity pushes tangential to the motion (down in your picture). The former adds energy and thus increases R. The latter adds no energy, so it merely bends the path from following the straight line it would otherwise follow in the absence of gravitational force.
QuoteSorry, I really can't understand how tidal changes R while we claim that it doesn't change force?
Thanks Halc1. Centrifugal forceQuote from: HalcCentrifugal force does not exist in the inertial frame implied by that diagram. If it did, it would counter gravity force and the path of Earth would not curve.I don't understand why you disqualified the centrifugal force.
Centrifugal force does not exist in the inertial frame implied by that diagram. If it did, it would counter gravity force and the path of Earth would not curve.
http://scienceline.ucsb.edu/getkey.php?key=4569
It is stated:"All massive objects in our universe are attracted to each other through a force known as gravity. If gravity were the only force acting between the sun and earth, the two bodies would indeed collapse on one another. Therefore, there must be other forces acting on this system.
What is wrong with this explanation?
Quote from: HalcIn an inertial frame, no. Only the force of gravity exists, and that force continuously accelerates the moon, keeping it in a circular orbit.How could it be that only the force of gravity exists?
In the following article: [same link]
It is stated in answer 1:"the earth has two main forces: inertia, to keep moving straight, and the gravity, to pull it to the sun. It moves in an average of the two directions, as in this image, and moves in a circle around the sun."It is stated specifically that Inertia is force.
Can we prove this explanation with mathematics?For example:Assuming that the current radius is R1ThanF1 = G M m / R1 ^2If the former adds energy and thus increases RThan, at the radius = R2 F2 = G M m / R2 ^ 2As R1 bigger than R2 then F2 is lower than F1.Hence, there is a decrease in the gravity force ΔF = F1 - F2.
So, how energy can increase R, therefore it changes the gravity force, but it doesn't consider as force.
There is another key issue (in an ideal Earth/Sun system).Let's assume that the tidal sets Energy.This energy Push the Earth further away from the Sun.However, due to inertia, the Earth keeps its orbital velocity.Hence, it this new location, the updated gravity force will be too low to hold the Earth in the orbital path.Therefore, the Earth will be disconnected from the Sun.
So, in order to keep the Earth in the orbital cycle around the Sun, two actions should be taken at the same moment and with full synchronization:1. Increases R2. Decreases V So, how Tidal can control on those two vectors with full synchronization?
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.
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.