Question about gravity.

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posted on Oct, 21 2012 @ 01:23 PM
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Originally posted by moebius
reply to post by ImaFungi
 

You are asking a quite fundamental question. Does a mass (or charge) interact with itself (its own field)? This also goes beyond classic physics. But for low velocities this effects can be neglected.

In general you have to work with so called "effective" fields to take self interaction into account. This is above my skill level I fear (would have to look up stuff myself). I recommend to talk to a physicist.

For math inclined a few links:
www.itp.uni-hannover.de...
arxiv.org...


but this i think is how planets form...millions of separate mass in a vicinity that will later be a body of mass because gravity "forces" this mass to interact with one another.....
are the "gnarly" physical conditions at the center of the earth because of gravity?

I know it is different but with a spiral galaxy,, do the star systems that surround the supermassive black hole, contain relatively more mass then the super massive black hole ( like the outer area of a planet, relative to its core, has more mass) ,.., and yet a supermassive black hole is known to have a lot of gravity?

and i dont think its a matter of a mass interacting with itself,, the discussion was if there was a seperate from earth mass at the center of the earth, i was trying to understand why you say that mass would not be affected by earths gravity..
edit on 21-10-2012 by ImaFungi because: (no reason given)




posted on Oct, 21 2012 @ 01:39 PM
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Originally posted by spy66
 


Earths mass will have no effect on the atmosphere inside the tunnel. Because earths mass is not inside the tunnel, it is surrounding the tunnel.

It is not earth's mass that is pulling us down. It is the atmosphere. It is the weight that is above us.

If you fill a bucket with sand and lift it above your head. It will be the bucket of sand + the atmosphere that pull you down. If you step on to a scale you will notise that your weight will be: your weight pluss the bucket of sand.

If you measure the weight at earth center from where you stand. You have to add the column of mass from the center and to where you stand. Than on top of that you have to add your weight + atmosphere.

Because the scale at earths center will read:

1. The column of mass resting on the scale to the surface.
2. Your weight. Because you are standing on surface.
3. The atmospheric pressure of 1 bar. Because it is pushing down on earth surface.



The force applied by the weight of the atmosphere is its pressure. Because air is a gas, it flows around objects. It pushes from all sides equally, so is insufficient in mediating gravitational force.

In fact, the opposite is true. We are actually buoyed up by the air. Low density things (like lighter than air balloons) float up in the air.

You, however, appear to be quite dense.



posted on Oct, 21 2012 @ 02:36 PM
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reply to post by SpearMint
 

As the object passes the center point of the Gravity Well...it encounters the effect of SPACE/TIME CURVATURE...in in the opposite direction of falling...thus...the object is now like a stone thrown into the air...it has the Kinetic Energy of the throw...but Gravitational effect will eventually stop the stone where it will then begin to fall thus gaining Kinetic Energy by Falling.

This is the same with this topics scenario. An analogy is a PENDULUM. The pendulum will continue to swing back and forth until Gravitational effect brings it to a rest dead center of the angle of Gravity's effect...ie...POINTING STRAIGHT DOWN. Split Infinity



posted on Oct, 21 2012 @ 02:43 PM
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reply to post by FooFighter4
 
Your reply is BEYOND wrong. If Gravity had anything to do with an EM Field then a stick of wood would NOT BE EFFECTED! Split Infinity



posted on Oct, 21 2012 @ 02:45 PM
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reply to post by SplitInfinity
 


The pendulum will continue to swing back and forth until Gravitational effect brings it to a rest dead center of the angle of Gravity's effect...ie...POINTING STRAIGHT DOWN.

A pendulum does not come to a stop because of gravitational effects. It comes to a stop because of friction.
If, as in this hypothetical scenario you eliminated friction, the pendulum would swing forever.
edit on 10/21/2012 by Phage because: (no reason given)



posted on Oct, 21 2012 @ 03:58 PM
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reply to post by ImaFungi
 


OK, I thought your question was about mass in motion end the effect of motion on gravity.

An object at center of mass is affected by gravity, but the force is the same in all directions(because we are at the center). Thus the resulting force will be zero.

To guess the mass distribution of a galaxy is difficult because if we count the stars (luminosity) there is not enough of it. An artificial mass is added called dark matter to correct for the missing one. In general the mass increases towards the center. If we assumed an even mass distribution the galactic gravitational force would get lower the closer you are to the center.

There is usually a singularity at the center which will cause a much larger gravitational force when you are close enough to it. But it is not the black hole that pulls the stars to form a galaxy. It is their own mass. The black hole is a local object. Its mass is about 1/1000 of the visible bulge surrounding it afaik.



posted on Oct, 21 2012 @ 05:33 PM
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Originally posted by moebius
reply to post by ImaFungi
 


OK, I thought your question was about mass in motion end the effect of motion on gravity.

An object at center of mass is affected by gravity, but the force is the same in all directions(because we are at the center). Thus the resulting force will be zero.

To guess the mass distribution of a galaxy is difficult because if we count the stars (luminosity) there is not enough of it. An artificial mass is added called dark matter to correct for the missing one. In general the mass increases towards the center. If we assumed an even mass distribution the galactic gravitational force would get lower the closer you are to the center.

There is usually a singularity at the center which will cause a much larger gravitational force when you are close enough to it. But it is not the black hole that pulls the stars to form a galaxy. It is their own mass. The black hole is a local object. Its mass is about 1/1000 of the visible bulge surrounding it afaik.


Remember also that a black hole has less mass than the original star that birthed it.

Despite the steepness to the curvature applied to space-time, a black hole does not reach out "further" than the gravitation of the original star.

The amount of gravity and the "reach" of that gravity are a lot less than the original star.

Similarly, due to relativistic effects, and the extreme distortion to space-time, time dilation would mean that, from the reference frame of an in-falling object, they would be able to watch the whole history of the universe, until the end of time, before they hit the singularity at the core.

From an observational reference frame outside of the Schwarzschild radius (or event horizon), the in-falling object would stop and freeze for all time and never reach the singularity.

For this reason, we can never see the naked singularity at the core of a black hole.

A standard black hole, therefore, never "recycles stars and planets" as was proposed in the OP.

A "white hole" or an exit point for a black hole, despite how obvious it would be, has never been observed.

Black holes do, however, cause Hawking radiation to occur close to the Schwartzchild radius and this is predominantly in the form of Gamma and X-Rays. It is this X-Ray brightness, along with gravitational lensing that allows us to determine black hole candidates.

It has also been theorized that very rapidly rotating black holes may open a wormhole, where it can't fully close the mass down to a singularity, due to centrifugal force. Although this is an appealing idea, I believe that the constraint of velocities to the speed of light would prevent this from occurring and that the black hole would still eventually collapse to a singularity.



posted on Oct, 21 2012 @ 07:35 PM
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Originally posted by moebius
reply to post by ImaFungi
 



An object at center of mass is affected by gravity, but the force is the same in all directions(because we are at the center). Thus the resulting force will be zero.


And the movement and motion of the planet has no effect on a mass at the center of the planet? Or this activity does has an effect at a mass in the center of the earth, but it would not be referred to as gravity?



To guess the mass distribution of a galaxy is difficult because if we count the stars (luminosity) there is not enough of it. An artificial mass is added called dark matter to correct for the missing one.


I think the destruction of stars due to the supermassive black hole,, results in a process of creating "dark matter and dark energy" which then spreads back out through the spiral galaxy.. that would be one of my guesses at least.



There is usually a singularity at the center which will cause a much larger gravitational force when you are close enough to it. But it is not the black hole that pulls the stars to form a galaxy. It is their own mass. The black hole is a local object. Its mass is about 1/1000 of the visible bulge surrounding it afaik.


ok so how do galaxies form,, are there tons of potential stars "swirling" around one another and this increase in attraction and swirling action brings the energy closer and closer and the "black hole" is not itself responsible for spinning the galaxy,, but the black hole is more of a result of the galaxy spinning,, which is potential stars and stars,,, and this activity of them cutting through space individual while traveling as a unit of galaxy through a universal space-time,,, is what keeps a galaxy together,, and a black hole existing and "doing what it is and does"?

but back to what we were talking about,,,, mass in the middle of earth,,,, the earth is rotating and revolving around the sun,,, and traveling with the suns movement,,,, so now,, and mass at the center of the earth would be forced to travel this same exact path along with the earth as a whole,,, would the non touching pressure of the earths mass against the non individually momentus "middle of earth mass" be called gravity,, and wouldnt that force/pressure be existent in reality?



posted on Oct, 21 2012 @ 07:42 PM
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reply to post by Phage
 

Sorry...but what you post is not true. Even if you had a way of having a Pendulum swing in a Vacuum and with ZERO FRICTION of any Mechanical Movement...it would still come to rest from swinging due to Gravity.
THAT IS A FACT! Split Infinity



posted on Oct, 21 2012 @ 08:10 PM
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Originally posted by SplitInfinity
reply to post by Phage
 

Sorry...but what you post is not true. Even if you had a way of having a Pendulum swing in a Vacuum and with ZERO FRICTION of any Mechanical Movement...it would still come to rest from swinging due to Gravity.
THAT IS A FACT! Split Infinity



No, Phage is right. Gravity is what causes the conversion between potential and kinetic energy. Without friction, it would not come to rest.



posted on Oct, 21 2012 @ 08:17 PM
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reply to post by moebius
 


Did you not even read your own source?



(This expression of the force follows from two theorems of Newton, mentioned in the hints; the mass of the part of the Earth above the train has no influence).


Your source is wrong. The mass of the part of Earth above the train DOES have an influence. They fail to include that in their calculations, so their calculations are wrong.

How about YOU explain to ME why you think the mass above the train has no influence. YOU explain to ME why you and your example thinks the mass above the train only has influence when it is past the 1/2 way point.


As of right now, all you have done is copy and pasted an old show of ignorance and incompetence.
edit on 21-10-2012 by illuminated0ne because: (no reason given)



posted on Oct, 21 2012 @ 08:26 PM
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Originally posted by illuminated0ne
reply to post by moebius
 


Did you not even read your own source?



(This expression of the force follows from two theorems of Newton, mentioned in the hints; the mass of the part of the Earth above the train has no influence).


Your source is wrong. The mass of the part of Earth above the train DOES have an influence. They fail to include that in their calculations, so their calculations are wrong.

How about YOU explain to ME why you think the mass above the train has no influence. YOU explain to ME why you and your example thinks the mass above the train only has influence when it is past the 1/2 way point.


As of right now, all you have done is copy and pasted an old show of ignorance in incompetence.
edit on 21-10-2012 by illuminated0ne because: (no reason given)


The mass above the object keeps it from flying out into space at the South Pole and to travel again toward the center, converting that potential energy back into kinetic energy which propels it back to the North pole and vice versa



posted on Oct, 21 2012 @ 08:28 PM
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reply to post by SplitInfinity
 


Even if you had a way of having a Pendulum swing in a Vacuum and with ZERO FRICTION of any Mechanical Movement
We do. It's called an orbit.



posted on Oct, 21 2012 @ 08:47 PM
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reply to post by PurpleChiten
 


So you think the mass above the object only has an effect on the train once it passes the half way point, but no point before that? If so, you are just as wrong as the paper that was linked to earlier.

I think some people are forgetting just how massive the Earth is. We are not talking about a hole drilled through a bowling ball, we are talking about Earth.

Even before the train reaches the half way point, the mass above the object would cause the object to fall slower than calculated thus far. It would NOT act like a pendulum and reach the South pole the same height it was dropped from the North pole. That would only happen if there was no mass above the train at any point.



posted on Oct, 21 2012 @ 08:52 PM
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reply to post by illuminated0ne
 


Even before the train reaches the half way point, the mass above the object would cause the object to fall slower than calculated thus far.
The acceleration would decrease due to both increased mass above and the decreased mass below but the velocity would continue to increase until the ball passes the center at which point the mass "above" would be greater than the mass "below" and the "downward" velocity would begin to decrease. The average acceleration in both "directions" would be the same and so would the kinetic energy acquired. It would continue to oscillate forever (all other factors being equal).
edit on 10/21/2012 by Phage because: (no reason given)



posted on Oct, 21 2012 @ 08:55 PM
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Originally posted by illuminated0ne
reply to post by PurpleChiten
 


So you think the mass above the object only has an effect on the train once it passes the half way point, but no point before that? If so, you are just as wrong as the paper that was linked to earlier.

I think some people are forgetting just how massive the Earth is. We are not talking about a hole drilled through a bowling ball, we are talking about Earth.

Even before the train reaches the half way point, the mass above the object would cause the object to fall slower than calculated thus far. It would NOT act like a pendulum and reach the South pole the same height it was dropped from the North pole. That would only happen if there was no mass above the train at any point.


Nope, read my other posts in the thread



posted on Oct, 21 2012 @ 09:41 PM
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Originally posted by Phage
reply to post by SplitInfinity
 


Even if you had a way of having a Pendulum swing in a Vacuum and with ZERO FRICTION of any Mechanical Movement
We do. It's called an orbit.

Using the example of the Orbit of the Earth around the Sun...the only reason Earth does not plunge into the Sun is that it's Kinetic Force of Motion that is directed at a 90 Degree Angle away from the Sun is compensated for by the Suns Mass which allows the Curvature of SPACE/TIME which keeps the Earth from shooting away into space.

If the Earth was placed the same distance it is now away from the Sun and was placed there without any Kinetic Energy Force that it has now which allows it to move in an orbit....it would be pulled directly into the Sun.
Split Infinity



posted on Oct, 21 2012 @ 09:45 PM
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reply to post by SplitInfinity
 


If the Earth was placed the same distance it is now away from the Sun and was placed there without any Kinetic Energy Force that it has now which allows it to move in an orbit....it would be pulled directly into the Sun.

Yes it would. And if there were a hole in the Sun it pass right through, out the other side to the same distance it started. It would then fall back and continue to do the same thing over and over again. It's an elliptical orbit with a semi-minor axis of 0.

Just as a pendulum would continue to swing if there were no friction.
edit on 10/21/2012 by Phage because: (no reason given)



posted on Oct, 21 2012 @ 09:51 PM
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reply to post by PurpleChiten
 

What you are saying is impossible. It is like saying that if I had a Trampoline and it was in a Vacuum chamber on Earth...and a 10 lbs ball was dropped on the trampoline...that that ball would continue to bounce forever. A Pendulum will not continue swinging forever just because it does not encounter friction if place on earth in a Vacuum Chamber with Theoretical Zero Friction Gears...it will eventually stop swinging. What you state breaks the Rules of THERMODYNAMICS as well as General Relativity and a host of other Physics Rules.

If this was possible then the Human Race would have PERPETUAL MOTION MACHINES ALL OVER THE PLACE as it is possible to use EM Fields for Levitation but one needs a source of Power...ie...a Electrical Receptacle to plug into to power that magnetic Field.

Split Infinity



posted on Oct, 21 2012 @ 10:00 PM
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An elliptical curve is the more likely trajectory in any real scenario especially over time.
First pass through the center of the sun might only be described by a line in an imaginary construct.





 
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