[Dave Lev (OP)]

That diagram depicts Mercury with a circular orbit, which is very wrong.  I said that in my post above.

Current orbital kinetic energy of Mercury is something like 3.1e32 joules due to a velocity of around 47 km/sec.  Potential energy is about -6.86e32 .
By mid-February, Mercury's orbital kinetic energy will rise to about 4.5e32 joules with potential energy falling to about -8.26e32 joules.  It does this by converting potential energy to kinetic energy as it's distance from the sun falls from ~70 million km to about 46 million and its speed increases to over 56 km/sec. Your incorrect (#5) statement above says this transformation cannot happen, and thus contradicts empirical observations.
Total orbital energy of Mercury is fairly stable at about -3.76e32 joules, consistent with energy not be created or destroyed.

Dear Halc
You miss the whole point.
You discuss on none circular orbit.
In that kind of orbit, every full cycle, the object gets to the same point.
So, in total, the average radius is fixed, the average potential energy is fixed and even the average orbital velocity is fixed.
This isn't the case in our discussion.
I have already informed evan_au about it:

You discuss on the "eccentricity" of the orbit.
This is an important issue by itself. However, this is not the main issue in the current discussion.
We currently discuss on the real meaning of Total orbital kinetic energy.

So, please answer the following:
1. Do you agree that the potential energy is a direct outcome of Gravity?
2. If so, what Is Gravity?
https://spaceplace.nasa.gov/what-is-gravity/en/
So do you agree that "Gravity is the force by which a planet or other body draws objects toward its center?"
Therefore in that article it is stated:
"Mercury is in a constant state of falling into the Sun"
" If Mercury were stationary, it would fall straight towards the sun (just like what happens when you drop an apple on Earth). "
Therefore, gravity by itself represents a direct falling in velocity vector.
By itself, it can't generate any sort of orbital velocity.
Unless - we discuss on airplane (an object with a wings) that is flying in an atmosphere.
So, if gravity force represents a direct falling in velocity vector, (the blue line in the diagram), than that must be the real meaning of potential energy.
Therefore, the potential energy can ONLY be converted to direct falling in velocity vector, or falling in kinetic energy as it changes the average radius of the object.
Hence, the potential energy can't contribute even 0.000...1 Joule to the Orbital kinetic energy (again - assuming that it isn't an airplane) even if the average radius is falling from R to 1/10...0 R.
Is it clear?

[Dave Lev (OP)]

Dear Halc
As you might know, we discuss on orbital energy.
1. What do you mean by "other kinds of potential energy":

No. Gravitational potential energy is related to gravity, but there are other kinds of potential energy.

Lets look at total orbital energy:
http://bogan.ca/orbits/kepler.html
Don't you agree that our scientists assume that:
"The Total energy of an object in orbit is the sum of kinetic energy (KE) and gravitational potential energy (PE).
KE = 1/2 mv2
PE = - GMm/r
r = the distance of the orbiting body from the central object and
v = the velocity of the orbiting body
E = 1/2 mv2 - GMm/r"

2. kinetic energy (KE).
KE = 1/2 mv2
Do you agree that the following formula represents the ORBITAL kinetic energy?
KE = 1/2 mv2
If so, Let's call it
KE(orbital) = 1/2 mv2
Do you also agree that v (velocity vector) is represented as a red arrow in the diagram that we have discussed?

3. Gravitational potential energy:
Do you agree that the following formula represents the gravitational potential energy (PE):
PE = - GMm/r
Based on your explanation the meaning of gravitational potential energy (PE) is:

Gravitational potential energy is arguably a direct outcome of the gravity of a system. The value is the energy yielded by bringing all matter in the system to zero potential (out of each other's gravity wells) from a stationary state (so not including kinetic energy).

PE = - GMm/r
Don't you agree that by your own explanation the direct outcome of the PE is the blue arrow in the diagram.
You can call it acceleration, velocity, force, momentum or "impact"  - as you wish.

Quote
Therefore, gravity by itself represents a direct falling in velocity vector.

11: "Falling in vector" is undefined.  Use accepted terms such as acceleration, velocity, force, or momentum, all of which are vector quantities. Your undefined nonstandard terms serve only to obfuscate (which I realize is a goal of yours). If you mean force, then say force.

Quote
So, if gravity force represents a direct falling in velocity vector, (the blue line in the diagram),

13: The blue vector is not labeled as a kind of velocity. The red line is. Your reading skills are lacking. 14: You seem to be equivocating force and velocity here, making this statement utterly wrong.

So, do you agree by now that the Blue arrow represents the "impact" of the gravitational potential energy?

4. Energy transformation
The key issue in this discussion is the energy transformation.
Especially the transformation from Gravitational Potential energy PE to ORBITAL kinetic energy KE(orbit)
Do you think that by changing the average radius of the orbital object (From r to r2), it can transform the change in the PE to added Orbital kinetic energy?
If so, would you kindly explain how that Potential energy can be transformed into added orbital kinetic energy?
Let's look at the following example:
Starting point:
KE = 1/2 mv2
PE = - GMm/r
If the average radius is changed to r2 (while there is no change in v)
The updated Potential energy will be
PE2 = -GMm/r2
So we get a change in the PE
Let's call that change PE'
As the Gravity works only on the Blue vector, would you kindly advice how that change in gravitational potential energy PE' can set any effect on orbital velocity vector -v?

[Kryptid]

I realize that my earlier explanation may have been difficult to follow without some kind of visual aid. Note: this is a simplification that uses circular orbits. The real orbits would be eccentric and thus would pass through regions of different gravitational field strengths. In particular, a decrease in orbital radius caused by a collision would probably result in an orbit even more eccentric than the original. In the case of tidal deceleration, the opposite is true: the orbit becomes less eccentric than the original:


Orbit.jpg (107.76 kB . 390x1252 - viewed 3632 times)

(1) The satellite in orbit remains in a region around the planet of constant gravitational field strength. Its velocity is constant.

(2) The satellite collides with something in space, such as an asteroid. This collision causes its orbital velocity to decrease somewhat. Since the satellite is now going slower than it was before, it can't follow its original path around the planet and instead begins to fall closer to the planet. Since it is traveling closer to the planet, it is passing through regions where the gravitational field strength is higher than it was in the original orbital radius. A stronger field means more force means more acceleration. This is what causes the satellite to accelerate to an average velocity in excess of its original orbital velocity. This is how gravitational potential energy gets converted into orbital kinetic energy when some event causes the satellite's original velocity to slow down (whether it is a collision, tidal deceleration, gravitational wave emission, or whatever else).

(3) The new orbit is closer to the planet than before, with the satellite's velocity having increased because gravity is stronger at this radius than it was at the original orbital radius.

[Kryptid]

The new orbit would still come back to the collision point every time unless a 2nd application of force (another collision or millions of years of tidal correction) circularizes the orbit like that.  The new orbit would remain like the 2nd picture, not the third.

I figured someone would bring that up (hence why I said I was simplifying it by using only circular orbits). However, for the sake of making it more correct, I'll add in this image (which would go between (2) and (3):


Orbit2.jpg (30.47 kB . 390x390 - viewed 4093 times)

Image (3) comes, as you say, after millions of years of tidal effects.

[Kryptid]

Ow ow No!
Elliptical orbits put the planet at one of the foci, not in the middle like that.

You're right. I went back and edited it.

[Dave Lev (OP)]

(2) The satellite collides with something in space, such as an asteroid. This collision causes its orbital velocity to decrease somewhat. Since the satellite is now going slower than it was before, it can't follow its original path around the planet and instead begins to fall closer to the planet. Since it is traveling closer to the planet, it is passing through regions where the gravitational field strength is higher than it was in the original orbital radius. A stronger field means more force means more acceleration

Yes, I agree with all of that.
Let's assume that due to the collision, the orbital velocity had been dropped from v to 3/4v (while it was at radius r).
At that moment, it had been lost its requested "magic velocity"
Remember:
v^2 = G M /r
Due to the collision, the velocity had dropped to 3/4v.
The outcome due to Newton is very clear
Let's look at Newton cannon
https://www.sciencelearn.org.nz/images/269-newton-s-orbital-cannon
"Newton reasoned that, if the cannon ball was fired with exactly the right velocity, the ball would travel completely around the Earth, always falling in the gravitational field but never reaching the Earth, which is curving away at the same rate that the projectile falls. It would be placed in orbit around the Earth."
As the requested velocity had been dropped from v to 3/4v that satellite MUST collide with the main object.
There is NO WAY for it to increase its orbital velocity due to increased gravity force!!!
There is another explanation:
"A stronger GRAVITATIONAL field means more force means more acceleration"
That force is working between the satellite to the center of the main object.
Therefore, the acceleration that we get represents a direct falling in acceleration.
As an example:
"https://sciencebasedlife.wordpress.com/2011/08/15/free-fall-on-the-moon/
Galileo conducted several experiments and concluded that the effect of gravity on earthly objects is the same, regardless of the mass of those objects."
So, if it is a satellite, elephant or just 1 Kg of mass, all of them must fall at the same acceleration, while they keep the same orbital velocity - 3/4v.
Therefore, it is quite clear that there is no way to increase the orbital velocity due to this falling in acceleration.Hence, the following explanation is a total MISTAKE!!!

This is what causes the satellite to accelerate to an average velocity in excess of its original orbital velocity. This is how gravitational potential energy gets converted into orbital kinetic energy when some event causes the satellite's original velocity to slow down (whether it is a collision, tidal deceleration, gravitational wave emission, or whatever else).

A falling in acceleration can't increase any current Orbital velocity.
It must be clear to all of us.

[Kryptid]

As the requested velocity had been dropped from v to 3/4v that satellite MUST collide with the main object.

The velocity doesn't stay that way, though. It accelerates again once it passes into the stronger regions of the gravitational field. This is exactly the same thing that happens in an eccentric orbit: an orbiting object speeds up when it approaches perihelion because the force of gravity is stronger closer to the main object.

There is NO WAY for it to increase its orbital velocity due to increased gravity force!!!

And yet that's exactly what happens in an eccentric orbit. Or do you deny the empirical data that shows planets accelerate when they move closer to the Sun in their orbit? If not gravity, then what force is it that increases the planet's orbital velocity as it moves towards perihelion?

A falling in acceleration can't increase any current Orbital velocity.

It does in an eccentric orbit.

[Dave Lev (OP)]

Quote from: Dave Lev on Yesterday at 18:31:43
As the requested velocity had been dropped from v to 3/4v that satellite MUST collide with the main object.

The velocity doesn't stay that way, though. It accelerates again once it passes into the stronger regions of the gravitational field. This is exactly the same thing that happens in an eccentric orbit: an orbiting object speeds up when it approaches perihelion because the force of gravity is stronger closer to the main object.

Let me explain:
In order to gain a stable orbital cycle - (with eccentricity = 0), there must be a full match between the radius to the orbital velocity. The requested velocity is called by Newton - "Magic velocity"
If you decrease the radius - then you actually MUST increase the orbital velocity.
Gravity can't increase the orbital velocity.
1. If the orbital velocity is too low it must collide with the main mass:
Please look at the following experiment for Newton ball:
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=6000.gif
"In this experiment Newton visualizes a cannon on top of a very high mountain. If there were no forces of gravitation or air resistance, then the cannonball should follow a straight line away from Earth. If a gravitational force acts on the cannon ball, it will follow a different path depending on its initial velocity. If the speed is low, it will simply fall back on Earth."
So, just by decreasing the velocity of the orbital object - you won't get any sort of stable orbital cycle.

2. If the Orbital velocity it too high it should be escape from the main mass.
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=10000.gif
"In this experiment Newton visualizes a cannon on top of a very high mountain. If there were no forces of gravitation or air resistance, then the cannonball should follow a straight line away from Earth. If a gravitational force acts on the cannon ball, it will follow a different path depending on its initial velocity. if the speed is very high, it will indeed leave Earth"

3. In order to gain the eccentric orbit you actually must increase the orbital velocity (by only some percentage - I assume)
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=8000.gif
In this experiment Newton visualizes a cannon on top of a very high mountain. If there were no forces of gravitation or air resistance, then the cannonball should follow a straight line away from Earth. If a gravitational force acts on the cannon ball, it will follow a different path depending on its initial velocity. If the speed is higher than the orbital velocity, but not high enough to leave Earth altogether (lower than the escape velocity) it will continue revolving around Earth along an elliptical orbit."

So, yes there is a possibility to get an elliptical orbit - but at this case it seems to me that you need to INCREASE the orbital velocity by some low percentage. If it is too high - it will be ejected from the main mass.
However, based on Newton experiment it is clear that if the orbital velocity is too low, the ball/orbital object must fall in and collide with the main mass.

Actually, you could ask NASA how do they set a satellite in a stable orbital cycle around the Earth.
It is quite clear that they have to bring the satellite to the requested radius and give it the requested orbital velocity.
If they won't do so, the satellite could fall in or be ejected to the open space.

Therefore, it is impossible mission to get a stable eccentric orbit by just random falling in.
You must have full control on the engine in order to set a satellite in a stable eccentric orbit.

Asteroids can be used as excellent example
The Erath is crossing their path from time to time.
We see them as meteorites while they fall in to our planet.
So, in total Million over billions of asteroids have cross the Earth path.
How many of them had been trapped in a stable orbital cycle around the Earth.
Do you agree that the number is ZERO???
So, just falling in by random velocity won't set any sort of stable orbital cycle.

[Kryptid]

Gravity can't increase the orbital velocity.

Then you need to answer this question:

If not gravity, then what force is it that increases the planet's orbital velocity as it moves towards perihelion?

How many of them had been trapped in a stable orbital cycle around the Earth.
Do you agree that the number is ZERO???

The average asteroid travels at more than twice Earth's surface escape velocity, so of course you'd expect the vast majority of them to evade capture. Then doesn't mean all of them have gotten away, though:  https://www.cnn.com/2016/06/16/us/nasa-asteroid-circles-earth/index.html

[Dave Lev (OP)]

Then you need to answer this question:
If not gravity, then what force is it that increases the planet's orbital velocity as it moves towards perihelion?

Newton has already answered this question.
Please read again my following reply:

3. In order to gain the eccentric orbit you actually must increase the orbital velocity (by only some percentage - I assume)
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=8000.gif
In this experiment Newton visualizes a cannon on top of a very high mountain. If there were no forces of gravitation or air resistance, then the cannonball should follow a straight line away from Earth. If a gravitational force acts on the cannon ball, it will follow a different path depending on its initial velocity. If the speed is higher than the orbital velocity, but not high enough to leave Earth altogether (lower than the escape velocity) it will continue revolving around Earth along an elliptical orbit."

So, yes there is a possibility to get an elliptical orbit - but at this case it seems to me that you need to INCREASE the orbital velocity by some low percentage. If it is too high - it will be ejected from the main mass.

Again - In order to get the elliptical orbit - you must INCREASE the orbital velocity or the orbital kinetic energy (at a given moment).
Therefore, by the decreasing of the orbital velocity – or the kinetic orbital energy, we should get a clear collision.

[Kryptid]

Newton has already answered this question.

Yes he did. His answer is gravity. But what is your answer? You already said that it wasn't gravity. What force do you say causes a planet's orbital velocity to increase as it approaches perihelion?

We are getting away from conservation of energy. If energy cannot be created, then the energy later must be the same as the energy now. Recall my bucket analogy. This is why your black hole cannot increase the total amount of energy in the Universe over time. Increasing the total amount of energy requires new energy to be created. Conservation of energy doesn't allow that.

[Dave Lev (OP)]

So instead let's consider a rock at the summit of Mount Chimborazo.  No cannon.  Suddenly all of Earth is compressed into a Neutron planet and has a radius of only 10km.  The only thing left behind is this one rock with no change in its initial velocity sitting up there where the mountain used to be. It begins to fall.  It has a speed well below this magic speed at which it would have a circular orbit.  So it goes into an elliptical orbit and misses the surface of Earth (not by much) and come right back to where it started in about a half hour.

No, this is a fatal Error!!!

Alternatively, if there were a tower built on that mountain 16,500 km tall, a rock dropped from it would be in a stable orbit without the need for Earth to shrink to get out of the way.  Venus would need a considerably taller tower.

No No!
If that rock is falling down, (without any starting orbital velocity) it must directly collide at  the center with that 10Km Neutron planet.

Quote from: Dave Lev
There is NO WAY for it to increase its orbital velocity due to increased gravity force!!!

This is clearly false, and violates conservation of energy, so you know it's false.
If I drop a rock off a bridge, it gains speed from zero to enough speed to splash into the water.  That is empirical evidence refuting your statement.  Yes, the rock is in orbit, at least until the water gets in the way.  It should continue its orbit because you said no friction.

Secondly, the radius has dropped to a lower value, so by the formula for PE, the PE is less. You assert the velocity cannot increase, so this asserts that somehow the kinetic energy cannot increase.  That would be a net loss of energy, violating conservation of energy, contradicting your own statement. Your assertion is wrong.

You have to distinguish between falling in acceleration to orbital velocity.
As the radius has dropped to a lower value, the falling kinetic energy is increasing. But that velocity or acceleration is in a direct line between the center of the object mass to the center of the main mass. - Please look again at the Blue arrow in the diagram.
That falling in kinetic energy (or acceleration) must set a clear collision between the two objects.
As I have already stated - Potential energy can't generate any sort of Orbital energy or orbital velocity!!!
Please let me know if you still don't understand that kind of basic falling in acceleration due to gravity.

Newton has already answered this question.

Yes he did.

As stated by Newton:
"If the speed is low, it will simply fall back on Earth."
Hence, there is no way to increase the orbital velocity or the orbital kinetic energy over time. So, the potential energy CAN'T increase the orbital energy at all.

If energy cannot be created, then the energy later must be the same as the energy now.

There is no way for an object to get inwards and increase its orbital velocity over time.
That must be clear for all of us!

His answer is gravity. But what is your answer? You already said that it wasn't gravity. What force do you say causes a planet's orbital velocity to increase as it approaches perihelion?

Sorry.
My answer is fully correlated with Newton.
YES - it is all about gravity!!!
However, in our discussion we only focus on real changes in the orbital radius or in the average orbital kinetic energy & average potential energy.
Therefore, If the orbital path is a pure cycle or elliptical, there is no change in the average radius over time.
In the same token there is no change in the average orbital velocity or the average Orbital kinetic energy and the potential energy.
If you insist to understand how the energy is working in elliptical orbit, than you have to ask Kepler about it. 
In any case, we focus only on real changes in the average orbital radius over time.
Newton has proved that by decreasing the potential energy as the object is falling in, there is no way to gain extra orbital velocity.
Therefore: "If the speed (magic velocity) is low, it will simply fall back on Earth."
In other words : By decreasing the potential energy we gain extra kinetic falling in energy (or falling in acceleration). However, that falling in kinetic energy doesn't contribute even 0.0..01 j to the current orbital kinetic energy.

[Kryptid]

Hence, there is no way to increase the orbital velocity or the orbital kinetic energy over time. So, the potential energy CAN'T increase the orbital energy at all.

And yet that's exactly what happens in an eccentric orbit. A planet's orbital velocity increases as it approaches perihelion. If the orbital velocity has increased, then its orbital energy has increased as well.

There is no way for an object to get inwards and increase its orbital velocity over time.

This is what happens in an eccentric orbit.
This is what is happening to Phobos and Deimos.
This is what is happening to the Hulse-Taylor binary neutron star system: https://en.wikipedia.org/wiki/Hulse%E2%80%93Taylor_binary

That also avoids the issue I was bringing up. Even if you were correct and gravitational potential energy could not be converted into orbital kinetic energy, that still would not change the fact that the total energy in the system cannot increase over time. As such, your black hole cannot create new energy.

Sorry.
My answer is fully correlated with Newton.
YES - it is all about gravity!!!

So now you are contradicting yourself. You said earlier that gravity cannot increase the orbital velocity. Now you are saying that it can. Which is it?

However, in our discussion we only focus on real changes in the orbital radius

An eccentric orbit does have a real change in orbital radius. For Pluto, that orbital radius change is particularly extreme (perihelion is 4.4 billion kilometers and aphelion is 7.4 billion kilometers).

Therefore, If the orbital path is a pure cycle or elliptical, there is no change in the average radius over time.

If you think that a planet in an elliptical orbit doesn't change its orbital radius over time, then you don't know what "elliptical" means.

If you insist to understand how the energy is working in elliptical orbit, than you have to ask Kepler about it. 

Yes, and Kepler says that the orbital velocity in an eccentric orbit increases when the planet approaches perihelion.

[Dave Lev (OP)]

Therefore, If the orbital path is a pure cycle or elliptical, there is no change in the average radius over time.

If you think that a planet in an elliptical orbit doesn't change its orbital radius over time, then you don't know what "elliptical" means

At any point in the elliptical orbit cycle the radius is the same after full cycle.
So, there is no change in the average radius from one full cycle to the next one.
Therefore there is no change in the average radius per cycle over time.
In the same token there is no change in the average orbital kinetic energy from one full cycle to the next one.
Therefore, there is also no change in the average orbital kinetic energy per cycle over time.
In our discussion we focus only on real changes in the average radius or in the average energies (from one full cycle to the next one).
Please see your following example:

(1) The satellite in orbit remains in a region around the planet of constant gravitational field strength. Its velocity is constant.

(2) The satellite collides with something in space, such as an asteroid. This collision causes its orbital velocity to decrease somewhat. Since the satellite is now going slower than it was before, it can't follow its original path around the planet and instead begins to fall closer to the planet. Since it is traveling closer to the planet, it is passing through regions where the gravitational field strength is higher than it was in the original orbital radius. A stronger field means more force means more acceleration. This is what causes the satellite to accelerate to an average velocity in excess of its original orbital velocity. This is how gravitational potential energy gets converted into orbital kinetic energy when some event causes the satellite's original velocity to slow down (whether it is a collision, tidal deceleration, gravitational wave emission, or whatever else).

(3) The new orbit is closer to the planet than before, with the satellite's velocity having increased because gravity is stronger at this radius than it was at the original orbital radius.

In this example you start with a satellite which is in orbit around the planet of constant gravitational field strength. Its velocity is constant.
Hence, it's average radius is constant from cycle to cycle and also its average velocity (or orbital kinetic energy) is constant from cycle to cycle.
However, due to the collision with asteroid the satellite had lost some of its orbital velocity.
"(2) The satellite collides with something in space, such as an asteroid. This collision causes its orbital velocity to decrease somewhat.
Let's stop here and try to understand the scenario:
1. The satellite is still located at the same average orbital radius - r
2. Due to the collision, the average velocity (or orbital kinetic energy) had been dropped.
In order to understand the outcome - we only need to ask Newton about it:

Please look at the following experiment for Newton ball:
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=6000.gif
"In this experiment Newton visualizes a cannon on top of a very high mountain. If there were no forces of gravitation or air resistance, then the cannonball should follow a straight line away from Earth. If a gravitational force acts on the cannon ball, it will follow a different path depending on its initial velocity. If the speed is low, it will simply fall back on Earth."

It is stated clearly:
"If the speed is low, it will simply fall back on Earth"
So, that satellite Must fall in and collide with the Earth as Newton had clearly stated.

Please look again at the Blue arrow in the diagram.
That falling in kinetic energy (or acceleration)

30: equivocation of energy and force. The blue arrow is force in that picture.  Kinetic energy is not a vector, but is a function of velocity, which is represented by the red arrow, not the blue one.

Somehow you still insist that the potential energy must be transformed to Orbital kinetic energy.
This is a severe mistake.
Newton has told us that if the current orbital velocity (red) it too low, there is no way to increase it by any sort of potential energy:
again:
"If the speed is low (red), it will simply fall back on Earth"
If you both still don't agree with Newton and you still hope that the somehow the potential energy can increase the orbital velocity, than would you kindly show Newton why his message is incorrect..

[Kryptid]

So now you are contradicting yourself. You said earlier that gravity cannot increase the orbital velocity. Now you are saying that it can. Which is it?

I'm still waiting for an answer to this question.

[Dave Lev (OP)]

Dear Halc
It seems that you have totally got lost.

You're switching from average to momentary radius and kinetic energy to average and back again.

1. Momentary radius - the current radius in the orbital cycle. In elliptical orbit it can change from maximal to minimal. However, in a stable elliptical orbit the momentary radius doesn't change the average radius or the average orbital velocity per cycle. Therefore it is none relevant to our discussion.
2. Average radius - It means the average radius PER cycle. Therefore, if there is no change in the average radius per cycle there is also no change in the average orbital velocity per cycle or orbital kinetic energy per cycle. We only focus on that radius in our discussion. Therefore, we only verify a real change in the average velocity from one full orbital cycle to the other one.
3. In order to get better understanding, let's assume that we only focus on a pure orbital cycle (eccentric =0). Therefore, please ignore the issue of Momentary radius or average radius. Just radius in a pure orbital cycle & pure orbital velocity

Quote
In this example you start with a satellite which is in orbit around the planet of constant gravitational field strength. Its velocity is constant.
Hence, it's average radius is constant from cycle to cycle and also its average velocity (or orbital kinetic energy) is constant from cycle to cycle.
However, due to the collision with asteroid the satellite had lost some of its orbital velocity.

See what I mean?  You're talking about it losing its orbital velocity, but above you said you wanted to concentrate on the average velocity and radius.  It has not done this.  Lower orbits have higher average velocities, so this new orbit is going to have higher average speeds than before.  Look at the average velocities of each of the planets: (km/s) 47 35 30 24 13 10 7 5
Smaller average radius means higher velocity, yet you continue to assert this:

4. In that example by Kryptid, we discuss on a pure orbital cycle. Kryptid have stated that the satellite had only lost some of its orbital velocity due to collision with asteroid. However, it is also clear that at the same moment of the impact there was no change in the radius.
Therefore: the radius is R before the collision and stay the same just immediately after the collision. However, the orbital velocity that was V before the collision, had been dropped to V' immediately after the collision.
The question was - how that decrease in the Orbital velocity could affect the satellite, while the radius had not been change - again, immediately after the collision?
4. Kryptid had estimated that as the average orbital velocity had been dropped, than the satellite can't keep on with its current radius. therefore, he assumed that the satellite must fall inwards. We can see it in the following diagram.

 Orbit2.jpg (30.47 kB . 390x390 - viewed 218 times)

A question to Kryptid - Did I understand your example correctly?
If so,
That actually is a perfect match to Newton explanation.
"Please look at the following experiment for Newton ball:
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=6000.gif
"In this experiment Newton visualizes a cannon on top of a very high mountain. If there were no forces of gravitation or air resistance, then the cannonball should follow a straight line away from Earth. If a gravitational force acts on the cannon ball, it will follow a different path depending on its initial velocity. If the speed is low, it will simply fall back on Earth."
In this Newton example the cannon ball is fired vertically to the ground. Therefore, it's first moment speed represents an orbital velocity.

Newton's ball example is fired into the ground, violating the condition that we're ignoring friction.  No satellite orbits right at the surface of a star or planet.  The same under-speed cannon shot would be in orbit if the friction was ignored.

Therefore, you clearly don't understand how Newton ball really works.
The meaning of the first moment speed is the first moment orbital velocity!!!
Therefore, When Newton claim for : "If the speed is low, it will simply fall back on Earth"
He means : If at the first moment the orbital velocity is too low (comparing to the requested "magic orbital velocity"), the cannon ball must fall to earth.
Please see again the following image:
https://en.wikipedia.org/wiki/Newton%27s_cannonball#/media/File:Newtonsmountainv=6000.gif
Therefore, the Satellite in Krypid example MUST fall in to Earth.
It can't convert its potential energy to ORBITAL kinetic energy.

His message is entirely correct.  Nowhere does it say that velocity will not increase as it falls to Earth.

Yes, as it falls in it must gain higher total velocity.
However, the potential energy is converted kinetic energy, but this energy creates higher falling in velocity vector. Let's call it Vf. That falling in velocity is vertically to the ground. I hope that we all agree that when something is falling (due to potential energy) it must fall vertically to the ground. I'm not aware about any object that can fall horizontally to the ground.
Therefore, the total velocity vector is the sum of the current orbital velocity V' vector -horizontally to the ground, and Vf vector - vertically to the ground (due to the potential energy that is converted to kinetic energy).
So, I agree that as the potential energy is converted to kinetic energy it increases the total velocity, however it doesn't increase the orbital velocity V' which is horizontal velocity vector - in red)).
Therefore, the total velocity could be higher than the requested "magic orbital velocity", but as it is not horizontally to the earth, it must lead the satellite to a direct collision with the earth - As was expected by Newton explanation.
I hope that by now it is fully clear.

[Kryptid]

A question to Kryptid - Did I understand your example correctly?

I'll answer that once you've clearly answered this:

So now you are contradicting yourself. You said earlier that gravity cannot increase the orbital velocity. Now you are saying that it can. Which is it?

[puppypower]

The mass of spiral galaxies decrease with time due to the mass burn within stars. It is estimated our sun burns the mass equivalent of the earth every 70,000 years. This mass burn times hundreds of millions of stars per galaxy results in the gravitational force decreasing with time.

The fusion in stars creates energy, while the second law states that the entropy of the universe needs to increase, with entropy increasing by absorbing energy. The entropy increase from fusion  is expressed in the complexity of higher atoms. It is also expressed as solar flares where confined and compressed materials lower pressure and temperature and have more freedom of expression; radical and stable chemical states.

The solar wind then goes into space where temperatures can approach absolute zero causing the chemical entropy to fall. Since the second and first law is still in effect, energy is released from entropy and recycled into other forms of entropy with a net increase in universal entropy; diffusion and super conductivity. and magnetism.

Our sun is a second generation star, formed from chilled gases and dust compressed by gravity. This lowered the entropy since it reverses super conducting states. This gives off energy trapped as entropy. This is expressed as a very low energy background universal background signal.that randomizes the galaxy.

If gravity was a force it would give off energy as the potential lowers. This exothermic output is related to the lowering of material entropy which is proportional to the gravitational work. The summation of all the forming stars adds energy into entropy resulting in layers of rotations.
. .

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[Dave Lev (OP)]

If you're saying it cannot increase the orthogonal velocity, you're wrong.  Look at Kryptid's elliptical orbit in picture 2.5.  The orbital velocity is initially V' at the top.  At the bottom of the orbit it is much closer to the primary (lower potential energy).  If it was still moving at the same speed as at the top, its kinetic energy would be the same, so the total energy would have gone down, violating energy conservation.  Vf is zero at both points since motion is horizontal at both of them.  So you're asserting that total energy has gone down, a violation of energy conservation.  The orthogonal velocity vector (which is the same as the total velocity at those two points) is much larger in magnitude at the bottom to account for the gained kinetic energy needed to balance the loss of the potential energy.

Simple question:
Please focus on Kryptid example.
Let's assume that due to the collision between the satellite and the asteroid, the Satellite had totally lost its orbital velocity (orthogonal velocity = 0)..
So, just one moment after the collision, that satellite is still located at a radius r from the planet and its orthogonal velocity is Zero.
Hence, as the Satellite must fall in due to gravity, do you see any possibility that it should gain any orthogonal/orbital velocity and restart to orbit the planet at a lower radius?

[Dave Lev (OP)]

We're assuming  a small object orbiting a large one, so yes, in the situation you describe, the impact will probably take place before the orthogonal component of its velocity regains the magnitude of the original orbital velocity.

Thanks
So you agree that the small orbital object that had completely lost its orbital velocity should collide with the main object.
But why do you claim: "before the orthogonal component of its velocity regains the magnitude of the original orbital velocity" and why do you highlight the issue that it is just because the orbital object is very small?

If the orbital object was a moon that falls in to Earth, do you see any possibility that it will regain the magnitude of the original orbital velocity?
Please look at the following diagram:
https://www.quora.com/What-is-the-work-done-by-the-force-of-gravity-on-a-satellite-moving-around-the-earth
We see that the orbital velocity vector V (or orthogonal velocity) is in green is zero.
The gravity force (vector Fg) is in red.
Let assume that at one moment the moon had suddenly lost completely its orbital velocity.
So, V =0.
In this case, do you agree that the moon will fall directly to Earth - in the green vector of Fg?
So, how could it be that this falling in vector can set or be transformed to any orthogonal velocity?
Even if we "they were both point masses (or sufficiently small)":

If they were both point masses (or sufficiently small), then no impact would result, and yes, there would be a point where the orthogonal component of the new orbit would vastly exceed the original orbital speed, as it must as it must convert potential energy into kinetic energy.

I really don't understand why do you see any possibility to convert the in falling velocity (or falling kinetic energy) to orthogonal velocity (or orbital velocity) at any sort of orbital objects.