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Not according to the law of conservation of energy, it doesn't. Do you think you are some kind of higher authority than the laws of physics themselves?You are right in the sense that gravity is converting gravitational potential energy into kinetic energy.
Therefore, as Ek is not equal to Ep any change in the radius also set a change in Et.
If the total rotation energy is changing due to radius change - where do you get the idea of law of conservation of energy in gravity force???
Do you see any error in this simple example?
In any case, let me remind you that we have started this discussion as I have stated that there is no energy transformation in Gravity.
I have used the moon example -It had totally lost its rotate ability as it is fully locked with the Earth.
Even so, any object that has a wish to orbit around the moon must fully obey to the same Ep, Ek and Et as I have used.So, with the ability to rotate/spin or without it, the outcome due to gravity force is the same.
Therefore, there is no need for transformation of energy from the rotate/spin energy into the orbital rotation energy and vice versa.
This idea is a fiction and I have just proved it!
How those scientists that set laws of physics can't get that simple outcome?
Right, but the radius doesn't change spontaneously. Energy either has to be added to the orbit or lost from it in order for such a radius change to occur.
So it still obeys conservation of energy.
An increase in the orbital radius requires a net input of energy,
The total energy remains the same both before and after the orbit change.
and that required amount of energy comes from the black hole's spin
The Moon is still rotating, it just takes the same amount of time to complete one rotation as it takes to complete one orbit around the Earth.
ThanksSo, you agree that as the object is changing its orbital radius, "Energy either has to be added to the orbit or lost from it"That is a key element. It proves that there is no conservation of energy in the total rotation energy
If you mean that there is a conservation of energy after adding new energy - that is fully clear.But again - There is no conservation of energy in the total rotation energy!
So, we fully agree that once you change the radius you need to add energy or decrease energy.
If you add or decrease an energy to or from the total energy - than how can you still think that the total energy remains the same?
Do you agree that based on this idea, in order to decrease an energy of the total rotation energy, the black hole's spin rotation should be increased?
In any case, you claim for a direct linkage between the Total rotation energy of the orbital object around the main body, to the spin of the main body.
If you set there one billion objects that orbits clockwise or the other side - do you agree that all of them will fully obey to Newton law whithout any ability for the moon to transfer even one bit of its spin energy (especially not to the one of the orbital direction). So the moon's spin would continue to rotate at the same fixed velocity while all the Billion objects will fully obey to Newton law and orbital velocity whithout getting any extra energy from the Moon's spin.
Hence, there is virtually no ability for any real energy transformation from the Moon spin to the requested orbital objects.Therefore, If the moon doesn't transfer any energy from its spin velocity to any orbital object (while it's Kinetic energy + potential energy will fully obey to Newton law) - could it be that even the SMBH won't need to transfer any energy from its spin velocity?
Please - So far you couldn't offer any real formula (By Newton or Einstein) that directly links the spin energy/velocity of the main body to the orbital energy of the orbital object.
I have a question for you:Why are you so sure that "an increase in the orbital radius requires a net input of energy"?
Therefore, an increase in the orbital radius requires less and less input of energy.
However, it seems to me that there is a rang limit for that potential energy.
However, if the object is too far away from the Earth or even from the sun, the gravity force is decreasing dramatically and I wonder what is the real meaning of the potential energy while the gravity force is almost gone to zero.
I'm not fully sure about it, but it seems to me that for any orbital system up to a certain radius - we can claim for sure that: "an increase in the orbital radius requires a net input of energy.
Now, we know for sure that the Oort cloud is drifting outwards constantly.So, without any need for external energy - those objects at Oort cloud are increasing their radius over time.
1. Is it a Normal activity at gravity that too far away orbital objects are drifting outwards over time?
2. Could it be that the gravity force is decreasing over time for orbital objects that are located too far away?
3. How can we find that radius range that converts the " an increase in the orbital radius requires a net input of energy" to "an increase in the orbital radius over time requires no net input of energy - It is a normal activity of Gravity force"?
4. Could it be that this range is set by finding the radius when The Kinetic energy is equal to potential energy?In other words - if Kinetic energy is greater than Potential energy - an increase in the orbital radius requires a net input of energy, however if the potential energy is greater that the kinetic energy than " an increase in the orbital radius over time requires no net input of energy"?
potential energy reaches its maximum (and finite) value at infinity.
An object won't change its orbital radius unless it is gaining energy from an outside source or losing it to an outside source. A spacecraft won't spontaneously increase its orbital radius from the Earth, for example. If the astronauts want to increase the orbital radius, they will have to add energy by firing the engines.
Sorry, I totally disagree with that.
Sorry, I totally disagree with that.As we increase the radius, we decrease the gravity force by r^2.However the potential energy is increasing only by r.So, the decreasing in the gravity force is dramatically higher than the increasing of the potential energy.If the radius is long enough, the gravity force should go almost to Zero.At the infinity, the gravity force is virtually ZERO. So, what is the meaning of Potential energy at a distance with zero (or virtually zero) gravity force?
Therefore, there is a meaning for potential energy ONLY as long as the gravity force is high enough.
You have set the calculation of the potential energy of the Moon/Earth while the radius is 1000 LY and the outcome was quite impressive Potential energy.However, at that distance - the gravity force between the Moon/Earth should go down to virtually zero.So, do you really think that the moon will be affected by the earth gravity (which is virtually zero)?
Therefore, that high potential energy at a distance of 1000 LY is useless in the Erath moon system - as the gravity force goes to almost zero at relatively high radius.
Why our scientists do not add the potential energy when they look at the spiral arm?
If you think that the Earth/moon potential energy has still an impact at long distance of 1000Ly, than why the potential energy of nearby stars (at a range of less than 100 Ly - there are 512 stars at that range) have no impact on the Sun orbital movement?
Please be aware that due to their massive mass and fairly shorter distances - the gravity forces between the nearby stars in the spiral arms are not so neglected.So, the potential energy should have an impact on nearby stars at the spiral arms.
Why our scientists totally ignore the potential energy when it comes to fairly short distances at the spiral arms (only 100Ly- when gravity force is still quite high) while we claim/hope that potential energy still has an impact even at the infinity (when gravity force is virtually zero)?
Why do you think the moon orbits around the earth instead of the Sun.
We have already found the gravity force of the Sun/moom is more than twice than the Moon/Earth.
Therefore, the only answer for that is: Gravity Works Locally!
A spacecraft changes its orbital radius from the Earth over time (without operate the internal rockets).Normally it falls in. So it reduces its orbital radius over time.
However, that is correct as long as the spacecraft is close enough to the Earth.
Try to set a spacecraft at a significantly further from the Earth - and you would surly find that it should increase its orbital radius over time.
I wonder why our scientists do not add the "over time" impact to the gravity formula.
You would probably claim that it isn't due to due to tidal or spin energy.
QuoteIf you think that the Earth/moon potential energy has still an impact at long distance of 1000Ly, than why the potential energy of nearby stars (at a range of less than 100 Ly - there are 512 stars at that range) have no impact on the Sun orbital movement?They do. The Sun is far from sitting still, you know.
Because the strength of Earth's gravity as felt by the Moon is higher than what it feels from the Sun. That's because distance matters as much as mass does. If the Sun was heavy enough, the Moon would feel it pulling on it more strongly than the Earth does. In that case, the Moon would indeed orbit the Sun.
There is nothing "useless" about it. The problem is that 1,000 light-years is well outside of Earth's Hill sphere, so the Moon could not actually orbit at that distance because there are so many other stars and planets around that would be pulling on the Moon much more strongly than the Earth. If the Universe was empty of all matter except for the Earth and Moon, however, such an orbit could indeed exist because there would no longer be such a Hill sphere.
I wonder why our scientists do not add the "over time" impact to the gravity formula.You can if you want to, but no amount of waiting will change the total orbital energy until energy is lost or gained by the object in orbit. Again, energy cannot be created or destroyed. You can't win an argument against nature.
Quote from: Dave Lev on 20/10/2019 12:54:04Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?There are 133 stars visible to the naked eye within that radius, which is less than 10% of the stars in total. Map of the 133 here: http://www.icc.dur.ac.uk/~tt/Lectures/Galaxies/LocalGroup/Back/50lys.htmlIt says there are an estimated 1800 stars in 1300 systems (so much for binary being prevalent).Your problem is that you get your data from hand-selected google pages instead of where scientists get it: From looking at the stars.
Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?
Would you kindly advice what our scientists really understand from that nearby Potential energies/gravity forces?
Let's verify the observations:At a radius of 100Ly around the sun there are 512 Stars.That volume represents 8 times the volume of 50Ly.So, if we try to calculate the average stars density per radius volume of 50 Ly the outcome will be:512 / 8 = 64 stars per 50Ly.Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?That shows that the density of stars in the Orion arm is exactly 64 Stars per 50Ly (or 512 Stars per 100Ly)This is not a random activity!!!A density wave can't generate that kind of fixed density.
Let's verify the observations:At a radius of 100Ly around the sun there are 512 Stars.That volume represents 8 times the volume of 50Ly.So, if we try to calculate the average stars density per radius volume of 50 Ly the outcome will be:512 / 8 = 64 stars per 50Ly.Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?That shows that the density of stars in the Orion arm is exactly 64 Stars per 50Ly (or 512 Stars per 100Ly)This is not a random activity!!!A density wave can't generate that kind of fixed density.The whole idea of density wave is that in some arias there is higher density of stars while in other arias there are less density of stars.This is not the case in the Orion arm and not in any other spiral arm.I would love to know the real 50 Ly radius volume densities of stars per distance from the galactic center - in the arm.Based on my theory, we should find that as we go closer to the galactic center the density is higher.As we go further away from the galactic center - the density is lower.However - at any spiral arm - at any distance from the galactic center, the density of stars is FIXED.Outside the arm - the density MUST be ZERO!However, the arm might have any kind of shape and there are bridges between the arms.I can promise you that as we cross the edge of the arm - there will be ZERO stars there!!!So, at the nearby aria around the sun - the density could be: or 64 per 50Ly radius or ZERO. There is another issue which is a direct outcome of theory D.As we go outwards from the galactic center the arm gets thinner and thinner.If we go further enough from the galactic center - the stars starts to be disconnected from the arms at the galactic disc.That sets the end point of the spiral arms.Our scientists don't have a basic clue how the spiral arm really works as they totally ignore the real impact of that density and why the arm is thinner and thinner as we move outwards.If they would try to understand the real impact of the Potential energy/gravity force in short range - less than 500 or 1000Ly, they should get to the same conclusions which I have got.If we could go back in time - let's say 100 M years ago - we would find that all the nearby stars are still there.So, any star that is moving away from us today - should come back in the future.While any star that is moving in our direction, should move away in the Future.How is it possible?The density wave doesn't give an answer for all of that.However, the Potential energies and the gravity forces between nearby stars give a perfect solution.They are all gravity bonded IN THE ARM itself!!!
So please - show me how our scientists translate that nearby potential energy/gravity force into real understanding about the structure of spiral arms.
2. Rotational energy conservation/transformation after the BBTYou claim that stars can't just decrease or increase their orbital radius without external energy.
If that is correct - than how the whole galaxies had been formed after the Big bang while there were no rotational energy at all?
no kinetic Energy
At the early days - 380M years after the Big Bang we have only got matter.No stars, No galaxies, No orbital objects no main body, no kinetic Energy No rotation energy - only matter everywhere.So, if we take a cube of one billion LY - in that cube there was no Kinetic energy, no spin energy, no rotation energy - Nothing)However, Today at the same volume cube there is so high rotational energy. It is everywhere. Try to take the same cube today and calculate the total rotation energy (of all the stars, galaxies) in that cube.From Zero rotation energy in age of 380 M years we have got so high total rotation energy today.So, how your idea of rotational energy conservation could be correct if in the early days there was no rotation energy at all???
Please - If you offer a random activity
Please - If you offer a random activity, than why the same idea for random gravity force/Energy that creates the whole stars, planets, moons, BH, SMBH, and even unlimited no of galaxies without any request for early rotational energy transformation/conservation in the Early days, can't work today even for just one pair of particle???
How can anyone believe that somehow in the early days of our universe the whole galaxies have been formed without any need for rotational energy transformation but today even for one pair of particle you insist for rotational energy transformation?
So please - would you kindly show how the whole orbital activities in the early galaxy had been formed out of nothing (no orbital activities after the matter creation - age of 380 MY after the Big bang), while we have to assume that the same rotational energy conservation should works also for those early days.
So, you don't agree that the Sun/Moon gravity force is stronger than the earth/moon force?https://www.universetoday.com/116158/why-doesnt-the-sun-steal-the-moon/"If you’re up for some napkin calculations, you little mathlete, by using Newton’s law of gravity, you find that even with its greater distance, the Sun pulls on the Moon about twice as hard as the Earth does.
However, I have asked why the Moon had started to orbit around the Earth at its first day?Our scientists claim:https://www.space.com/29047-how-moon-formed-earth-collision-theory.html"The formation of the moon has long remained a mystery, but new studies support the theory that the moon was formed from debris left from a collision between the newborn Earth and a Mars-size rock, with a veneer of meteorites coating both afterward."I claim that this story is incorrect because:a. Meteorites can't form new Moon. They only can set some sort of a ring, as we see around Saturn.b. If one of the meteorite was as big as the moon, it had to follow the gravity force of the Sun at the moment of the impact (as at that moment the gravity force of the sun was stronger than the earth).
Hence, as we monitor the rotational energy rather than gravity force we might find the answerfor why gravity works locally and why the moon orbits around the earth instead of orbiting directly around the Sun.
Well, I think differently
Every orbit has a shape of spiral.In the nature nothing comes back exactly to the same point (same orbital radius).It might fall in or it might drift out.
Unfortunately, our scientists have missed that key element of "Over time".
I can promise you that by 100% all the stars in the spiral arms are drifting outwards over time.After every cycle around the galactic center - the sun is drifting outwards.
If we could come back in one billion years from now we might find that our sun had been ejected from the arm and from the galactic disc.
QuoteLet's verify the observations:At a radius of 100Ly around the sun there are 512 Stars.That volume represents 8 times the volume of 50Ly.So, if we try to calculate the average stars density per radius volume of 50 Ly the outcome will be:512 / 8 = 64 stars per 50Ly.Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?That shows that the density of stars in the Orion arm is exactly 64 Stars per 50Ly (or 512 Stars per 100Ly)This is not a random activity!!!A density wave can't generate that kind of fixed density. That density isn't fixed. The stars are moving relative to each other. It’s also a very bad assumption to calculate the density of stars within 100 light-years of the Sun and extrapolate that to the density of all of the other stars in the spiral arm. What does that have to do with anything, anyway?
There is another issue which is a direct outcome of theory D.As we go outwards from the galactic center the arm gets thinner and thinner.If we go further enough from the galactic center - the stars starts to be disconnected from the arms at the galactic disc.That sets the end point of the spiral arms.Our scientists don't have a basic clue how the spiral arm really works as they totally ignore the real impact of that density and why the arm is thinner and thinner as we move outwards.
We will continue our discussion about the Energy.
Quote from: Kryptid on 06/10/2019 07:04:23I'm putting a hold on all of the other matters for the moment and focusing on conservation of energy because that is the crux of the problem right now. I'm not moving on until that issue is solved first.Yes, fully agree.
I'm putting a hold on all of the other matters for the moment and focusing on conservation of energy because that is the crux of the problem right now. I'm not moving on until that issue is solved first.
Again, you have it backwards. Adding energy from an outside source will increase the orbital radius. You can't say, "I want the orbital radius to increase, so the energy required to do that pops up out of nowhere". The radius won't increase unless the object in orbit is supplied with energy from another source that already existed. Look at my example with the spaceship.
Our scientists have developed some ideas/assumptions about gravity and energy.
However, spacecraft is not a star.
We need to verify the stars activity in the galaxy and see if those assumptions are correct.The only way to verify that is by our observation on the galaxy.
So, please show me those stars that were captured by the SMBH?
Did we ever see one single Hypervelocity Star that was pushed inwards to the SMBH in the Milky Way?
Those Hypervelocity Star were not ejected due to the Binary affect.
They were ejected from the galaxy as they have moved to far away from the requested density in the arm/galaxy
Our sun is located near the edge of the Orion arm. If we will dare to move even 30 LY away from the edge of the arm, we will lose the gravity bonding of all our nearby stars.
At that moment we will be boosted outwards from the galaxy and will never come back again.
All the stars and matter that we see in the galaxy had been created by our mighty SMBH.
So, please - would you kindly answer the questions about the thickness and G stars density in the spiral arms?How the "density wave theory" can still be valid under those observations?
Gravitational sling-shotting speeds one body up at the expense of slowing another body down. If you knew how conservation of energy worked (which you have abundantly demonstrated that you don't), you would have known that. Both stars can't make each other go faster as that would violate conservation of energy. If one speeds up, that is because it has extracted energy from the other star via gravitational interactions, which makes the other one slow down
Your model is automatically wrong because it violates conservation of energy. Is gravitational energy a form of energy? Yes. Therefore, it must obey the law of conservation of energy.
So you claim that: "Both stars can't make each other go faster as that would violate conservation of energy."We know that the average orbital velocity of the stars around the galaxy is about 200 Km/sI have found a runaway star at 1200Km/s:https://www.space.com/28737-fastest-star-galaxy-strange-origin.htmlThe runaway star, US 708, is traveling at 745 miles per second (1200 km/s) — that's 26 million miles per hour (43 million km/h) —making it the fastest star in the Milky Way ever clocked by astronomers, according to the new research"It is also stated:"The monster black hole at the center of the Milky Way has the gravitational muscle to fling a star on a one-way-track out of the neighborhood, and many other hypervelocity stars are thought to originate from there. But US 708 didn't start its journey near the galactic center, the new research shows."Based on your advice it must come from a binary star in the galactic disc (as it is far away from the SMBH).However, you have stated that: : "Both stars can't make each other go faster as that would violate conservation of energy."So, how can we convert the energy of 200Km/s to 1200Km/s?
The fastest-known star in the Milky Way is on a path out of the galaxy, and new research suggests it was a supernova that gave it the boot.
Theory D is the only valid theory that gives a perfect explanation for everything we see.It meets all the observations by -100%.
Actually, there is a simple way to verify which theory is real.You can tell me what is your expectation based on your theory/understanding, and I will tell you the expectation by theory D.Let's start by the nearby G stars density.I can tell you by 100% that at any 50LY radius in the nearby aria in the Orion arm - the density MUST be 64 G stars.If you go out the arm - the density must drop to zero.So, there are only two possibilities for star density - or 64 per 50Ly or Zero. Nothing in between (unless we are located exactly at the edge of the arm or the bridge).Would you kindly advice what would be your expectation for the G star density around us (take radius of 2000 Ly?If we will cross the Orion arm, what might be the star density?Why don't we ask NASA to verify my expectation?If we will discover that theory D is correct, do you agree to reconsider your objection against this theory?
A force can generate new energy.
Evey orbit of a star consume energy.
If I will give you a rope with a ball at the end, don't you need to generate new force/energy in order to keep it orbiting around you?
In the same token - Every orbit of a star consumes Force and energy.
However, as the gravity force comes for free (Due to Newton , there is no need to fuel the gravity force or set any energy transformation), the energy due to this force also comes for free.
So, far you couldn't show any formula which links the gravity force into energy transformation.
In gravity force - As Newton and Einstein didn't specify any request for gravity energy transformation - your assumption is totally incorrect.
Therefore, the statement that gravity force can't generate new energy is a severe violation of Newton law
The answer is in that very link:"The fastest-known star in the Milky Way is on a path out of the galaxy, and new research suggests it was a supernova that gave it the boot".
I did give you such an equation. The gravitational potential energy equation shows exactly that: how much energy is released by allowing an object to fall into a gravitational field or how much energy is consumed moving an object up against a gravitational field.
Both Newtonian and relativistic physics allow energy to be converted from potential to kinetic by a gravitational field, so what you are saying is false: https://en.wikipedia.org/wiki/Gravitational_energy#Newtonian_mechanics
Nowhere do any of Newton's laws state that gravitational fields can generate new energy.
When you are in space twirling your test mass around on a string both your hand and the test mass will revolve around their mutual centre of gravity if you make an effort to maintain your hand away from the centre of gravity it will absorb energy from the system there will also be a small amount of energy radiated away in the form of gravitational waves.
If we set a ball on atomic bomb, is there a chance that the atomic bang will rock the ball to the moon?
Sorry, a supernova near a star should break it to pieces.
Our scientists were positively sure that all of those hyper star had been ejected from the galaxy due to binary star.
They didn't mention any word about supernova.
That proves that they really don't have a basic clue how gravity really works in the galaxy.
If they have stated that it is due to binary star and the observation contradicts this statement - It is expected that they should set this idea in the garbage.
You can't just invent new idea every time that new observation contradicts your current assumption.
If our scientists give an explanation for something and this explanation can't meet the observation - they should say: Sorry we have an error in our assumption. Clear and loud!
We see again and again that our scientists have a fatal error in understanding how gravity really works in the galaxy.
I have proved that the density wave is a pure fiction.
So, how can you claim that your current understanding in gravity is the ultimate one?
You can't just disqualify my understanding without offering real alternative.
Please show how your current theory about gravity meets the density and thickness of the spiral arms
However, once we deal with orbital kinetic energy - there is no energy conservation between the potential energy to the orbital kinetic energy.
I have already proved that:If you decrease the distance/radius between the objects by twice you actually increase the Potential energy by twice and get only half of the orbital kinetic energy.So, we see clearly that the sum of the total orbital energy changing as we change the radius.You actually have agreed with that.
Based on my understanding, the orbital kinetic energy is decreasing over time.Therefore, the radius/distance between the orbital objects is increasing over time.
So, the increased radius between the earth/moon systems is not due to tidal but it is due to that "over time" phenomenon.
Again - this doesn't cover the orbital kinetic energy.
He also didn't say that the orbital kinetic energy needs a transformation of energy.
If the orbital is new, than the orbital kinetic energy is new!!!
Newton didn't ask for any transformation in orbital kinetic energy.
The question then becomes: can enough kinetic energy be imparted to the star to speed it up to 1,200 kilometers per second from 200 kilometers per second without giving it enough energy to destroy it in the process? The mass of the star is estimated at 9.9425 x 1029 kilograms, which we put into the kinetic energy equation Ek = 0.5mv2. So we get Ek = (0.5)(9.9425 x 1029)(1,000,000)2 = 4.97125 x 1041 joules. It's close, but this demonstrates that the star can indeed acquire the needed energy from the supernova to speed up to 1,200 kilometers per second without being destroyed. Much of the star probably was blown off in the process, actually, since O-type subdwarfs are thought to be the result of red giants having their outer layers removed by some powerful process.