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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.
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.
No. Gravitational potential energy is related to gravity, but there are other kinds of potential energy.
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).
QuoteQuoteTherefore, 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.
QuoteTherefore, gravity by itself represents a direct falling in velocity vector.
QuoteQuoteSo, 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.
QuoteSo, if gravity force represents a direct falling in velocity vector, (the blue line in the diagram),
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.
Ow ow No!Elliptical orbits put the planet at one of the foci, not in the middle like that.
(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).
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!!!
A falling in acceleration can't increase any current Orbital velocity.
QuoteQuote from: Dave Lev on Yesterday at 18:31:43As 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.
Quote from: Dave Lev on Yesterday at 18:31:43As the requested velocity had been dropped from v to 3/4v that satellite MUST collide with the main object.
Gravity can't increase the orbital velocity.
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???
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?
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.gifIn 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.
Newton has already answered this question.
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.
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.
QuoteQuote from: Dave LevThere 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.
Quote from: Dave LevThere is NO WAY for it to increase its orbital velocity due to increased gravity force!!!
Quote from: Dave Lev on 28/11/2019 08:38:41Newton has already answered this question.Yes he did.
If energy cannot be created, then the energy later must be the same as the energy now.
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?
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)
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.
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.
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.