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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.

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

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

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

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

QuoteTherefore, 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

(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.

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."

QuotePlease 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.

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

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).

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.

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 itPlease look at the following experiment for Newton ball:

Quote from: Halc30: 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.

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.

Newton has told us that if the current orbital velocity (red) it too low, "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..

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?

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

QuoteQuoteIn 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 5Smaller average radius means higher velocity, yet you continue to assert this:

QuoteIn 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.

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

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.

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

Dear HalcIt seems that you have totally got lost.Quote from: Halc on 30/11/2019 06:10:27You'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 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.

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.

A question to Kryptid - Did I understand your example correctly?If so, That actually is a perfect match to Newton explanation.

In this Newton example the cannon ball is fired vertically to the ground.

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, the Satellite in Krypid example MUST fall in to Earth.

It can't convert its potential energy to ORBITAL kinetic energy.

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

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

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?

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.

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.

Quote from: Halc on 02/12/2019 05:05:55We'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.ThanksSo 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"

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:quora.com/What-is-the-work-done-by-the-force-of-gravity-on-a-satellite-moving-around-the-earthWe 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?

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.

QuoteQuoteI 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.Look at comets, which have almost the exact sort of orbit you're describing here. They start out coming almost straight in (nearly pure 'falling' velocity as you put it), and suddenly as they pass the sun, the high velocity vector doesn't change much, but the sun is suddenly off to the side and that high velocity vector is now completely orthogonal to the force vector. The force vector (representing 'down') rotates around quickly but the velocity vector doesn't so much.

QuoteI 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.

QuoteQuoteSo, how could it be that this falling in vector can set or be transformed to any orthogonal velocity?By rotation of the vectors. It is falling fast, and suddenly 'down' is a different direction as it passes by Earth. The direction of 'down' is changing all the time. Takes month to go all the way around, but less time in this new orbit you've given it.

QuoteSo, how could it be that this falling in vector can set or be transformed to any orthogonal velocity?

Comets are irrelevant to our discussion

Please look at ...:1. Free Falling Object Dropped From a Known HeightWe see that the final velocity (at the collision point) is v = √(2gh)

It is also very clear that this velocity vector is horizontal to the Earth.

So, how this Horizontal falling in velocity vector could be transformed to orthogonal velocity vector (orbital velocity)?

Would you kindly use Newton formula & mathematics to prove that unbelievable idea?

They are completely relevant since the orbits of the objects you describe (the ones that don't involve impacts) will orbit forever in their new highly eccentric orbits.The ones we see happen just like you describe: A reasonably circular orbit is suddenly altered, stopping the comet in place. It thus begins to fall into the inner solar system, achieving far higher velocities than it ever had before. This is exactly what I've been describing.

The picture does not depict an orbit. Find a picture of a comet orbit to see a real example of this so called unbelievable idea. Here's one:http://www.khadley.com/Courses/Astronomy/ph_205/topics/pluto/images/cometorbit.jpgThe comet moves clockwise. Draw a velocity vector at say the 1985 mark and notice it is pointed almost the same direction as the force vector, not perpendicular at all. So the comet is gaining speed as it fall nearly directly towards the sun. It reaches its highest speed as it swings behind the sun near where the word 'Mars' is written. At that perihelion point the velocity is entirely perpendicular to the force vector. It is no longer 'falling in' at all, but is moving faster than any other point in the orbit. That is because the velocity vector has rotated only about 80° from the 1985 point to the perihelion point, but the force vector has rotated around 170°, pointing nearly the opposite direction as it was in 1985. That's the vector rotation I'm talking about.

What velocity vector? The picture you asked me to look at shows a guy dropping a rock from a small height. The (implied, not depicted) velocity is totally vertical in the picture.

We see that the final velocity (at the collision point) is v = √(2gh)Has nothing to do with collision point. That formula assumes a uniform gravitational field g, which means it only works for short distances. Gravity force on our moon varies by its distance from the planet, so that formula doesn't work. Use the PE formula, and convert that figure into KE. That gives an accurate final velocity for dropping something from a large height.

QuoteQuoteWould you kindly use Newton formula & mathematics to prove that unbelievable idea?Use the formulas I indicate above, which give accurate velocity values at any point in the orbit. Those formulas do not give eccentricity for a given scenario, but they give what you asked: velocity resulting from dropping an object from altitude X to altitude Y.

QuoteWould you kindly use Newton formula & mathematics to prove that unbelievable idea?

Hence, there is no change in the potential energy or orbital kinetic energy due to a direct collision.

Well, if we can show or prove that the comet had completely lost its orbital velocity due to collision, but now it regain it as it falls in, than, yes that could be a good example.

So, do we have any sort of evidence for current immediate outcome due to comet collision?

Why do you claim: "A reasonably circular orbit is suddenly altered, stopping the comet in place."Why the comet had stopped in place?

Don't you agree that we actually see a comet in a normal orbital cycle that just got to its maximal radius and then comes back?

That activity doesn't give any indication for sudden orbital velocity lost.

So, if you don't have a clear evidence for a comet collision (and the direct outcome from that collision) than the comet is just irrelevant for our discussion.

In the article it is stated clearly:"Comets are cosmic snowballs of frozen gases, rock and dust that orbit the Sun".So, it is just in a constant and stable orbital cycle around the Sun

while its eccentric could even be close to one.

Therefore, the comet should come again and again to the same minimal radius and the same maximal radius after every full orbital cycle.

Again - we try to understand the impact of collision that force the orbital object to lose suddenly its orbital velocity.

That could be a normal explanation for a any orbital cycle with eccentric close to one.

Therefore, that Vector rotation is a very normal outcome from any orbital system at any sort of eccentric

So, do you agree that the velocity of an object that falls in (to a planet) must be represented by a totally vertical velocity vector (with reference to the planet)?

However, the exact formula for the vertical falling in velocity is not so important.

It is very important that any falling in must be vertical to the planet.

This answer isn't clear. How do you convert the "totally vertical" velocity vector due to falling in, to orthogonal velocity vector?

QuoteQuoteThis answer isn't clear. How do you convert the "totally vertical" velocity vector due to falling in, to orthogonal velocity vector?The answer I gave yields orbital speed S, not component speeds. Orbital velocity V is the velocity with speed S tangential to the orbital path. If you want to break that velocity into your two components at any point in the orbit, find the angle between the acceleration (or force) vector and the velocity (or momentum) vector. What you call the 'falling in vector' Vf is Scos(θ) in the direction of the primary. The vector orthogonal to Vf (Vo) is Ssin(θ) in the direction perpendicular to Vf. It's that easy.

QuoteThis answer isn't clear. How do you convert the "totally vertical" velocity vector due to falling in, to orthogonal velocity vector?

QuoteSo you agree that the small orbital object that had completely lost its orbital velocity should collide with the main object.Most of the time, yes.

So you agree that the small orbital object that had completely lost its orbital velocity should collide with the main object.

But there is a change of potential and orbital kinetic energy due to the immense change in orbital radius of the comet over time.

(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.

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.

In a stable orbital cycle as the eccentric is greater than zero and less than one, that change in the energy (potential/kinetic) is actually a temporary change.

S = Orbital velocity

Vf (falling in velocity vector)= Scos(θ)Vo (Ortogonal velocity vector) = Ssin(θ)That could be correct as long as we monitor the velocity vectors.

However, in reality Vf is a direct outcome from gravity force

while Vo is the first moment orthogonal velocity of the object (as was explained by Newton cannon ball.

So, just if there is a full match between the Vf and Vo (orthogonal) we can get the magic velocity that we can call Orbital velocity.

Therefore Vf in reality has no impact on Vo (orthogonal) and vice versa.

The formula should be as follow:S (orbital velocity Vector) = Vf (falling velocity vector) + Vo (orthogonal vector)So, the orbital velocity is the outcome between the Vf and Vo.

Let's go back to Krypptid example.Please remember that we have stated the due to the collision the orbital object had totally lost its orbital velocity.

At that moment, as there is no orbital velocity, R is actually represents H.

So, just after the collision, the object is located at H, its orbital velocity (Vo) is zero

its falling in velocity vector Vf is zero and also its orthogonal velocity is zero.

So, We have the main object (let's assume that it is a planet),

You have stated that if the orbital object is small enough, most of the time it should collide with the planet.I need to understand why do you claim "only most of the time" and how it could gain any orthogonal velocity as it falls in if its orbital velocity just after the collision is zero.

In order to understand that, let's assume that we have an object with mass of 1,000Km.

Let's drop it at a high of 10,000Km above the planet.

So at that moment, H = 10,000Km, Vo =0 , Vf =0.I assume that we all agree that (if we ignore the air) it should fall horizontally to the planet.

Therefore, as it falls in it accelerates, its falling velocity - Vf (horizontally to the planet)

is increasing over time, while its Vo should be zero.

So, even if Vf will get to it's maximal velocity, Vo must be zero.

Now, if we take much heavier object and set it at a higher distance from the planet,Do you see any possibility that it will increase its Vo as it falls in?

We want to understand what will be the real impact due to a collision with the orbital object.

I claim thatS (orbital velocity Vector) = Vf (falling velocity vector) + Vo (orthogonal vector)Let's look at Newton cannon ball explanation:

The Vf is a direct outcome due to gravity. therefore, Vf is in a directly towards the centre of the field (or the center of the planet if you wish)

It is stated clearly that in order to "travelling in a circle around the Earth" there must be a full match between Vf and Vo.

So, when you think about that ellipse orbital cycle (with eccentric greater than zero, less than one), the orbital object must get a boost in its orthogonal velocity.It is also stated:" if the speed is increased enough, the trajectory becomes hyperbolic. At this point, the projectile has enough velocity to leave the gravitational field."At that moment the eccentric is greater than one.

So, your both assumption that the object can set even one full orbital cycle after loosing significantly it orthogonal velocity (or even zero) is totally wrong.