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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.
But the moon is massive, so Earth will be moving relatively quickly in its own orbit around the moon, so if the moon is halted, there's a chance the Earth gets out of the way in time. Probably not. As I said, only Charon has a chance of doing this. I haven't worked out the numbers.
If the planet is moving fast enough, the moon will miss it as it falls and go into that eccentric new orbit.
Most of the time the planet is barely moving. So you stop any typical satellite and it's going to drop pretty much straight down.
In my discussion, I assume it was halted in place in the frame in which it had a circular orbit (the frame of the two-body system). In any other frame, the moon's path was not a circle, but more like a helix.
QuoteQuoteThe 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. We called the orbital velocity V, not S. S is a speed, not a vector..With that terminology exception, your statement is actually correct, and I never said otherwise.
QuoteThe 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.
But keep in mind that these vectors are constantly changing over time for an object in orbit. They're not fixed values, and your argument seems to hinge on them being fixed. There is force on the orbiting thing, so there is always acceleration, and acceleration means all vectors (V, Vf, and Vo) are constantly changing.
QuoteQuoteSo, even if Vf will get to it's maximal velocity, Vo must be zero.If it falls straight down, they yes. It isn't in orbit in any scenario with an impact like that.
QuoteSo, even if Vf will get to it's maximal velocity, Vo must be zero.
48: If the orbit is circular, Vf is zero, which hardly sounds like a 'full match' with Vo.
47: V is orbital velocity whether or not there is this 'full match' between these two component vectors.
"Increase the speed enough, and the projectile will never hit the ground, instead travelling in a circle around the Earth."
Quote from: Halc on 06/12/2019 20:51:34If the planet is moving fast enough, the moon will miss it as it falls and go into that eccentric new orbit. This is incorrect.
Almost Every object in the galaxy orbits around something due to gravity force.
The moon for example is orbiting around the Earth, while it is totally ignore the Sun Gravity.Please be aware that the Moon/sun gravity force is more than twice with regards to the Moon/Earth gravity force.
Please be aware that in our theoretical discussion, we only have reduced the current orbital velocity of the object to Zero.
In that theoretical concept we also didn't break of the orbital object from the current gravity force.So, we didn't break the gravity force between the Moon/Earth.
Quote from: HalcMost of the time the planet is barely moving. So you stop any typical satellite and it's going to drop pretty much straight down.The Earth can move at the speed of light.
As long as the satellite is "gravity bonded" with the earth, it should fall in directly to the center of the earth without any ability to gain any sort of orbital velocity.
Quote from: HalcIn my discussion, I assume it was halted in place in the frame in which it had a circular orbit (the frame of the two-body system). In any other frame, the moon's path was not a circle, but more like a helix.So, please let's focus only on circular orbit and a frame of two-body system.
In my discussion, I assume it was halted in place in the frame in which it had a circular orbit (the frame of the two-body system). In any other frame, the moon's path was not a circle, but more like a helix.
So let's agreeV (orbital velocity Vector) = Vf (falling velocity vector) + Vo (orthogonal vector)
As we discuss on "circular orbit and a frame of two-body system" do you agree that the V, Vf and Vo are fixed?
I hope that you also agree by now that if the moon had totally lost its orbital velocity, while it is still in a two body system with the earth, it must fall in directly to the center of the earth.
Quote from: Halc48: If the orbit is circular, Vf is zero, which hardly sounds like a 'full match' with Vo.This is incorrect.Every orbital object is actually falling in constantly.
"objects in orbit around the earth are in a perpetual state of “falling” toward the earth."
"If the object's velocity were to change (in this case, decrease), then the force of gravity would cause the “falling” to have a greater influence"
resulting in the object's eventual plummet into the atmosphere and toward the surface."
I know that you really dislike those kind of answers from Quora, but I think that they are perfectly correct.
This moon is actually falling constantly directly to the center of the earth, therefore its Vf is not zero.
However, as it also move forward at the orthogonal velocity - Vo, that movement compensate the falling in velocity.Therefore, it keeps itself with the same radius (again, let's assume that the orbital is a cycle with eccentric =0).
So, even as Vf is greater than Zero, due to the full magic fit with Vo we get that magic orbital velocity that keeps it in the orbital cycle (while we might think that Vf is zero).
If there is no full match between the Vo and the Vf there will be no orbital system.
If we can boost the (Vo -orthogonal velocity) of an object to the magic velocity that perfectly fit with its Vf, than this object will set a pure orbital cycles around the main mass.
I don't. My mailbox doesn't. Newton's slow cannon ball doesn't. These objects are subject to gravity force, and yet they don't orbit. Yes, all objects orbit if you restrict 'objects' to things that are in orbit, which is like saying every red object is red.
51 Our own moon is currently stopped in its own frame (by definition), and in that frame, Earth is moving fast enough that the moon misses it and instead goes into an eccentric orbit. Thus my statement is exactly correct.
54 Our moon is gravity bonded with Earth...
In that frame, the planet is always moving, so when you stop the moon, the planet keeps on going.
There is no magic 'match' required for Vo and Vf. If you take a random asteroid from the asteroid belt and give it a random Vo and Vf with no regard to 'matching' them, so long as the sum of them (V) yields a kinetic energy that when added to its potential energy is a negative figure, the object will probably orbit the sun.
For example, Mercury's velocity is currently decreasing, and yet the force of gravity is decreasing because of this, or "causing the 'falling' to have a lesser influence" as that quote words it.