[Bored chemist]

It is for free.

There is no such thing as free energy in science.
So, once again, are you a foll who believes in in it, or a troll who pretends to?

[Bored chemist]

We have to ask Newton why the gravity force doesn't need any energy for its work

Gravity doesn't usually do work, so it does not need energy.
If you want to  use the energy of a rock rolling don a mountain, you have to start bu putting the work in to take the rock to the top of the mountain.


This is fundamental, high-school physics you do not understand.

[Dave Lev (OP)]

It is for free.

I didn't ask if it was for free or not, I asked where it came from.

The mass of an object causes the space around it to essentially bend and curve.
https://www.universetoday.com/75705/where-does-gravity-come-from/
Albert Einstein explained how gravity is more than just a force: it is a curvature in the space-time continuum. That sounds like something straight out of science fiction, but simply put, the mass of an object causes the space around it to essentially bend and curve."
Hence, there is no need for any sort of energy to set that gravity force. Therefore, the gravity force is for free.

Quote from: Dave Lev on Today at 06:55:24
It is for free.

There is no such thing as free energy in science.

As the gravity force is for free, then the work due to gravity force is also for free.
In our real Universe, in order to set a work, energy is needed.
As the Gravity force comes for free then the gravity work also comes for free and therefore, the energy for that work comes for free.

Gravity doesn't usually do work, so it does not need energy.
If you want to  use the energy of a rock rolling don a mountain, you have to start bu putting the work in to take the rock to the top of the mountain.

You use the word - Usually.
I prefer to use - sometimes.
Hence, sometimes, if there is already a rock at the open space (asteroid) and let it fall in the direction of Earth, it would set a work due to gravity force.
We can easily calculate the energy at the moment of the collision. That energy is for free as it is due to a free gravity force.
However, I fully agree that once the rock is in the valley, we won't gain new energy by taking it to the top of the mountain.
Theoretically, we can gain energy from orbital objects free of charge.
If we could set a wire all the way to the moon, then as the moon orbits around the earth, the movement of this wire could generate electricity - free of charge.
Therefore, any energy that comes due to gravity is a free energy.
Hence, as the tidal heat comes due to gravity force, it is a free energy.

[Bored chemist]

You use the word - Usually.
I prefer to use - sometimes.

Well, let's go with the word "Always" .
Because it is always the case that someone, or something had to put the rock up there.

Hence, as the tidal heat comes due to gravity force, it is a free energy.

In that case, the energy isn't "free" it comes from whatever created the Moon.

This is fundamental, high-school physics you do not understand.

[Kryptid]

Therefore, the gravity force is for free.

Again, I didn't ask if it was for free or not. I asked where it came from. "Free" is not a location. You can't find "free" on a map.

[Bored chemist]

Theoretically, we can gain energy from orbital objects free of charge.

No,
You "harvest" the kinetic energy of the object and change its orbit.

[Bored chemist]

Therefore, the gravity force is for free.

Yes.
But energy is not.
Do you understand the difference?

[Dave Lev (OP)]

Quote from: Dave Lev on Yesterday at 20:29:13
You use the word - Usually.
I prefer to use - sometimes.

Well, let's go with the word "Always" .
Because it is always the case that someone, or something had to put the rock up there.

Are you sure about "Always"?
Lets assume that this rock is located at a star that orbits around the SMBH.
Based on your explanation as this star falls inwards, it should break down and its temp should be increased. As its matter gets to the accretion disc, the temp would be increased to 10^9c and the orbital velocity would be close to the speed of light.
If I understand you correctly, that added energy is due to gravity force.
However, the same gravity force that pulled inwards that star, breaks it down and increases so dramatically the energy in that matter, would change the falling direction of the matter and suddenly push outwards most of the heated plasma.

Hence, based on the wisdom of our scientists the gravity force around the SMBH works as follow:
It grabs star that orbits around it.
It breaks it down to its particles, increases the temp, orbital velocity and the energy.
As the matter gets very close to the event horizon, the gravity force makes a unique movement.
Instead of just dump that ultra energetic matter into the SMBH, it suddenly push it outwards as the observable UFO.

Quote from: Dave Lev on Yesterday at 20:29:13
Therefore, the gravity force is for free.

Yes.
But energy is not.
Do you understand the difference?

So, you fully confirm that the gravity force is for free.
However, you claim that the gravity force can't be converted into energy.
How can you use double standard when it comes to the conservation law?
When you need an extra energy free of charge - then the gravity force is there to contribute unlimited energy for free.
When you don't need it - then the same gravity force can't do so.

Is it real?
Are you sure that this forum is all about real science?

[Dave Lev (OP)]

Free gravity force - Free of charge energy.
Theoretically, we can gain free of charge energy from a free gravity force.
All we need is to come close enough to the SMBH.
Dump a cold rock into the SMBH and get it back as a supper heated plasma at 10^9c.
Use the energy in that plasma and then dump it gain into the SMBH.
Hence, free gravity force offers free of charge energy.

[Kryptid]

Do you plan on answering this?

Again, I didn't ask if it was for free or not. I asked where it came from. "Free" is not a location. You can't find "free" on a map.

[Bored chemist]

If I understand you correctly,

You don't understand anything.

that added energy is due to gravity force.

The energy was stored there by the BB.

As the matter gets very close to the event horizon, the gravity force makes a unique movement.

No
I already explained this.

NEVER and EVER.

We actually use this effect.
We send a spacecraft falling towards the Sun in order to launch it into interstellar space/.
It's called "gravity assist"
And it doesn't matter if you use capital letters, it is still observed to be true.


Exactly the same orbital mechanics could launch a small fraction of the incoming material away from a BH.

Why do you not remember things you are taught?

So, you fully confirm that the gravity force is for free.
However, you claim that the gravity force can't be converted into energy.
How can you use double standard when it comes to the conservation law?
When you need an extra energy free of charge - then the gravity force is there to contribute unlimited energy for free.
When you don't need it - then the same gravity force can't do so.

Is it real?

No.
It's not real, its tosh you made up.

Only you think that there is unlimited energy.
We scientists know better.

If you don't believe that, please show a post where anyone but you has said that there's unlimited energy.

[Dave Lev (OP)]

Do you plan on answering this?

Again, I didn't ask if it was for free or not. I asked where it came from. "Free" is not a location. You can't find "free" on a map.

I have already did:

The mass of an object causes the space around it to essentially bend and curve.
https://www.universetoday.com/75705/where-does-gravity-come-from/
Albert Einstein explained how gravity is more than just a force: it is a curvature in the space-time continuum. That sounds like something straight out of science fiction, but simply put, the mass of an object causes the space around it to essentially bend and curve."
Hence, there is no need for any sort of energy to set that gravity force.

I can't offer an answer better than Einstein.

We send a spacecraft falling towards the Sun in order to launch it into interstellar space/.
It's called "gravity assist"

Gravity assist doesn't explain the accretion disc.
https://en.wikipedia.org/wiki/Gravity_assist
A gravity assist around a planet changes a spacecraft's velocity (relative to the Sun) by entering and leaving the gravitational sphere of influence of a planet.
https://en.wikipedia.org/wiki/Gravity_assist#/media/File:GravAssis.gif
*
Therefore, Gravity assist is all about eccentricity of the orbital shape.
as eccentricity is close to one than the orbital shape is ellipse and the orbital object get closer to the main mass:
https://astronomy.swin.edu.au/cosmos/O/Orbital+Eccentricity
If the eccentricity is higher than 1 the orbital shape is hyperbola.
That hyperbola represents the Gravity assist.
In that case, the "falling object" can't set even one circular orbital cycle around the main mass.
It just comes close enough to the main object and then boosted away into the space.
However, the accretion disc has a circular orbital shape.
We clearly see it in the image of M87 accretion disc.
Hence, any person that claims that an object from the bulge could fall inwards into the accretion disc, once it get near the event horizon it would sets many circular cycles and after some time it would be ejected outwards due to Gravity assist., doesn't represents real science.
It is just science fiction.
If you still hold the imagination of the Gravity assist to help you in the activity at the accretion disc, then please explain how an orbital object with high eccentricity would suddenly change its eccentricity almost to zero as it gets to the event horizon, and after many circular orbits at that radius - it will change its eccentricity back to over than one.

[Dave Lev (OP)]

Eccentricity.

Let's make it clear.
Orbital objects do not fall in.
It is all about eccentricity.
If the eccentricity is zero than the orbital shape is circular.
Hence, the object keeps exactly the same radius in full orbital cycle.
If the eccentricity is higher than zero but less than one, then the orbital shape is ellipse.
In each full orbital cycle there is a point when the orbital object comes closer to the main mass.
However, it is a fatal mistake to call it as "falling object" when it comes closer to the main object.
It is not falling. It is orbiting.
The last issue is that the orbital object does not change its eccentricity so dramatically in one moment.
Hence, an orbital object with high eccentricity (close to one) would never change its eccentricity to zero (or close to zero) as it gets to the event horizon of the main mass.
In the same token, an orbital object with eccentricity of almost zero at the accretion disc, would never ever change its eccentricity to over than one in one moment..
Our scientists don't have a basic clue how the accretion disc really works.
Its time to set their imagination of falling stars deep in the garbage.

[Kryptid]

I have already did:

Quote the part of your post that states the location where gravity got the energy from.

[Bored chemist]

Gravity assist doesn't explain the accretion disc.

Nobody said it did.
But it does explain the thing you actually asked about.

Did you know that, if you have forgotten what question you asked, you can look back through the thread and read it again?

the gravity force makes a unique movement.
Instead of just dump that ultra energetic matter into the SMBH, it suddenly push it outwards as the observable UFO.

The small amount of stuff that followed the right path got flung out by a mechanism that looks like a gravity assist.

[Bored chemist]

Let's make it clear.
Orbital objects do not fall in.

One of the moons of Mars is doing.
Why do you post something when you know it is wrong, and then say everyone else is a liar?

[Bored chemist]

I can't offer an answer better than Einstein.

It seems you can't offer any answer at all.
Why is that?

[Dave Lev (OP)]

The small amount of stuff that followed the right path got flung out by a mechanism that looks like a gravity assist.

This is not realistic due to the following:
1. If that was the case, than we have to observe also that matter as it comes in.
There is no way that due to gravity assist we can observe the matter as it goes out but we can't see it as it comes in.
2. . Eccentricity for gravity assist
You ignore my explanation about eccentricity.
I have already proved that in gravity assist the eccentricity is higher than one.
Therefore, in this case the object doesn't orbit around the main mass.
So, it is not realistic to claim that an orbital star with eccentricity lower than one would "fall in" (let's say at eccentricity of 0.7) , as it gets near the event horizon it would change its eccentricity to almost zero (in order to set the circular orbital shape of the matter over there) and after some time it would shift its eccentricity to higher than one just to be ejected outwards.
Sorry - this is pure imagination.
3. Collision
The accretion disc is very thin and located at a very specific location which is orthogonal to the magnetic poles of the SMBH. The chance that the orbital plan of a random star is identical to the accretion disc plan is almost zero. Hence, the chance that an orbital object (or imagination falling object)  would fall exactly at that orbital plan of the accretion disc and collide with the plasma over there is also almost zero.
Even if a falling star would fall exactly at the accretion plan and collide with the plasma, then we have to observe it as it falls/comes in and we also have to observe the impact of the collision at the accretion disc.
Therefore, a collision is also imagination.

Let's make it clear.
Orbital objects do not fall in.

One of the moons of Mars is doing.
Why do you post something when you know it is wrong, and then say everyone else is a liar?

Well, it is absolutely incorrect to claim that Phobos is falling inwards.
It is expected that you would know by now that Phobos is spiraling inwards and not falling inwards and also understand the difference between the two.
That moon orbits around Mars at eccentricity which is close to zero. However, it spirals in by only 2 meter per year. Therefore, it should take it 40 million years to come closer to mars. There is big difference between spiraling in activity (at eccentricity close to zero) which could take 40 Million years and is very observable to a falling in (at eccentricity higher than one) which is very fast but should also be observable.
In our case the falling matter isn't observable at all. Therefore, it isn't spiraling inwards and even not falling inwards.
You can continue to dream on your imaginary falling stars but you would NEVER & EVER observe any star in the entire Universe that falls into the SMBH' accretion disc. With all the advanced technology that we have, we didn't find it in the last 50 years and we won't find it in the next million years.
But you can continue to dream on falling stars forever and ever.
Dreaming is free of charge (even more than gravity force).

[Bored chemist]

It is expected that you would know by now that Phobos is spiraling inwards and not falling inwards and also understand the difference between the two.

Only you imagine that there's a difference.
If I throw a ball and it traces a parabola on its way to the ground, do you say it isn't falling because it's not straight down?
Since you don't understand that the words mean the same thing, it isn't likely that you understand much else.

Why do you refuse to learn science?

If that was the case, than we have to observe also that matter as it comes in.

How many times do I have to explain this?
You do not see it because it's dark.

You will only see any of it in unusual circumstances.

You ignore my explanation about eccentricity.

I ignore a lot of things that are wrong; you just added one more to the list.

So, it is not realistic to claim that an orbital star with eccentricity lower than one

Do you understand eccentricity?
https://en.wikipedia.org/wiki/Orbital_eccentricity

If it's greater than one, you don't have what most people would call an orbit; you have a "flypast" or a miss.
Between zero and one you get the elliptical orbits- essentially everything "in orbit" is in the category.
And if the eccentricity is zero you get a perfectly circular orbit.
But as soon as the thing in orbit gets hit by so much as a photon (or the recoil from emitting a photon) that orbit will become elliptical.

So everything we are concerned with is in the category


What you seem to miss is that many of the things that fall in are not single solid bodies (indeed,nothing is "solid" when it faces the gravity of a BH).
So, when a lump of rock, (which will typically be spinning about its own axes) falls in and breaks up, some bits will be sent into orbits with eccentricities grater than one.
And those are the ones which, having just been kicked out of a shattering rock, are hot enough to glow- so we see them.

I already explained this to you ages ago.
Why do you not learn?

would fall exactly at that orbital plan of the accretion disc and collide with the plasma over there is also almost zero.

Nobody said it would.

The accretion disc is very thin and located at a very specific location which is orthogonal to the magnetic poles of the SMBH.

Yes and no.
The magnetic field and the accretion disk are both aligned (parallel or perpendicular) to the angular momentum of the object.

And if you drop something in at some random angle you will change the orientation slightly.

So the plane is not a requirement for stuff to hit it edge on.
The plane of the accretion disk is just the average of the angles at which everything has hit it in the past.

If something comes in at the "wrong angle"- as almost everything will, it either misses completely (or falls into orbit) or it is caught and "hits" the ring but the ring just wobbles, sheds the extra energy as thermal radiation and settles down again on a slightly different angle.

The rings  ring, just like Saturn's.
https://www.space.com/saturns-rings-ring-like-bell.html

There is big difference between spiraling in activity (at eccentricity close to zero) which could take 40 Million years and is very observable to a falling in (at eccentricity higher than one) which is very fast but should also be observable.

There's a very big difference between Mars and a SMBH, isn't there?
Also, there's a better mechanism for matter in the accretion disk to shed energy and fall in.

[Dave Lev (OP)]

So, when a lump of rock, (which will typically be spinning about its own axes) falls in and breaks up, some bits will be sent into orbits with eccentricities grater than one.
And those are the ones which, having just been kicked out of a shattering rock, are hot enough to glow- so we see them.

Ok
Let me focus on that Lump of Rock.
I have a perfect example for that Lump of Rock. It is called comet and it orbits around the Sun
Halley’s comet has a semimajor axis of about 18.5 AU, a period of 76 years, and an eccentricity of about 0.97.
Evey 76 years it comes very close to the sun and it also has a dust tail.
Please look at the following image:
https://en.wikipedia.org/wiki/Comet#/media/File:Cometorbit01.svg
We clearly see that the tail is kept outwards from the orbital cycle.
This is very important issue.
It proves that nothing from this comet really falls inwards into the sun even if that Lump of Rock comet arrives very close to the Sun.
You can also claim that this comet seems as it falls into the Sun but in reality it does not fall. (even if it comes very close to the Sun). Not the comet itself and not even one tinny dust from that comet tail
The comet is not going to change its eccentricity to zero as it comes to that minimal distance in order to set a circular path over there.
Therefore, any object that orbits around the SMBH at high eccentricity would keep its eccentricity.
Our scientists even claim that in future it should be ejected from the Sun.
So, it is expected that the eccentricity would be increased in the future.
Once the eccentricity would be increased just by above 0.03 (from 0.97 to one) this comet would be ejected into the open space.
Therefore, as this comet would never ever decrease its eccentricity at the minimal radius, S2 star that orbits around the SMBH would do exactly the same.
It would come very close to the SMBH, but it won't change its eccentricity to zero at the minimal distance.
Even if it would break to Lump of rock and carry a dust tail, the matter in the tail would be kept outwards from the orbital cycle.
As the tail goes with the comet wherever it goes, then if there was a tail for S2 it would go with it and keep the eccentricity of the orbital path.
We clearly see the tail of the comet as it comes into the direction of the Sun and as it goes out.
In the same token If S2 has a tail we should observe this tail as it comes in and as it goes out.
So, please from now on
Do not claim again that the orbital star with high eccentricity (let's say higher than 0.2) around the SMBH falls in.
They do not Fall in – They orbit!!!.
They could come very close to the main object (if their eccentricity is close to one) but they do not fall in.
NEVER and EVER.