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Question about gravity.

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posted on Oct, 23 2012 @ 12:43 AM
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reply to post by chr0naut
 





At the North Pole, where the 10kg mass is released, the force is all in potential energy and the initial velocity is zero.

Falling towards the center of mass, the object accelerates and the velocity increases. The potential energy is being converted into momentum.



I want to ask you a question.

If the 10kg mass was resting on a slide inside the vacuum tunnel before it was released.

How much would the 10kg mass weigh on the slide before being released?

If the mass is measured to be 10kg in atmosphere. Will the mass weigh 10kg inside the vacuum tunnel?




posted on Oct, 23 2012 @ 01:34 AM
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reply to post by spy66
 


If the mass is measured to be 10kg in atmosphere. Will the mass weigh 10kg inside the vacuum tunnel?
No.
A balloon of helium which weighed 10kg in atmosphere would weigh more in vacuum. In principle, with an accurate enough scale, any 10kg mass would weigh more in vacuum because the buoyancy effect of the atmosphere would be absent.

But the object would still mass 10kg. There is a difference between weight and mass.



posted on Oct, 23 2012 @ 03:33 AM
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Originally posted by SplitInfinity

Originally posted by Phage
reply to post by SplitInfinity
 


There is loss of energy! Every swing travels a shorter distance as it is forced to contend with Gravity.
The swings get smaller because of friction. Gravity does not extract energy from the system.


Again I ask you...how can you account for the effect of Gravity not slowing the upswing of the Pendulum?
I answer again. Gravity converts the momentum of the pendulum into potential energy.


I have stated that even in a complete Theoretical Non-Friction Environment as well as in a TOTAL VACUUM...GRAVITY will effect the Pendulum's swing and slow it to an eventual stop.

What you have stated goes against the rules of THERMODYNAMICS....how do you account for this?
Split Infinity


Phage is right my friend. Without friction gravity converts to and from potential energy in a 1:1 ratio. It is the work of friction that would cause the object to eventually stop. If the object was placed dead center it would never move. However far from the center the object was placed, it would move that exact distance in the opposite direction.



posted on Oct, 23 2012 @ 03:49 AM
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Originally posted by illuminated0ne
reply to post by Phage
 


So you understand that the acceleration would decrease, but you don't understand that the velocity would decrease as well? The mass above the train would convert the kinetic energy of the train back into gravitational potential energy significantly before it even reaches the center. There would hardly be enough kinetic energy in the train to pass the center by any significant amount.

The center of the Earth, and the center of this gravity train track, would have a significant near equal gravitational pull in ALL directions. Not just an equal pull vertically, but also horizontally, and all directions between, which is a very significant thing to take into account. Every time this gravity train reaches the center, the train wants to be pulled in all directions which would convert part of the train's kinetic energy into gravitational potential energy that can never be used because it is fixed on a vertical track.

Comparing a gravity train to a pendulum is highly misleading, because pendulums deal with a vertical gravitational pull only. Add a horizontal gravitational pull to a pendulum and watch it come to rest in equilibrium in the center.


The pull would be equal and cancel out, having zero net effect. Unless you are talking multiple sources of gravity, you are wrong.



posted on Oct, 23 2012 @ 04:05 AM
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reply to post by Phage
 


Ok let me ask you this...

Since I think my previous posts were not clear enough I have decided to modify this image (please forgive the quality) to illustrate my point...



When the gravity train is in the center of Earth, gravity is pulling it in all directions. It seems to me that most all calculations regarding the gravity train are only considering the vertical axis (up/down arrows) of gravity. Do you not concur that the horizontal axis (left/right arrows) of gravity would effect the velocity of the gravity train every time it passes the center of Earth? Keep in mind, this image is only 2 dimensional. There is also a 3rd axis of gravity (forward / backward) that I did not illustrate which would seem to also effect the train. Would not the gravitational pull in those directions other than vertical effect the train as it passed the center?

I believe it would.

I believe the center of Earth would slow the train down every time it passes. I think every time the train passes the center, a small amount of kinetic energy would be converted into gravitational potential energy that it can't use. Over time the train would get stuck in the center of Earth with only gravitational potential energy and no kinetic energy.

I think it is easy to calculate the gravity train's acceleration and velocity when you only consider the vertical axis (z) of gravity. However, in the center of Earth, there is two more axes of gravitational pull (x and y) acting on the object. I believe these are going ignored.

edit on 23-10-2012 by illuminated0ne because: (no reason given)



posted on Oct, 23 2012 @ 04:07 AM
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reply to post by OccamsRazor04
 


It would cancel out motion in those directions, but not force.

In the center of Earth there are multiple sources of gravity. Up, down, left, right, forward, and backwards are sources of gravity.
edit on 23-10-2012 by illuminated0ne because: (no reason given)



posted on Oct, 23 2012 @ 04:14 AM
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reply to post by Phage
 



Ok.

If you have a vacuum tunnel. Like in the image under. Will the red mass sink to the bottom?

NB. Earths mass is no longer surrounding the tunnel.

How would the mass know which way to fall when there is vacuume above and below?




posted on Oct, 23 2012 @ 04:22 AM
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Originally posted by spy66
If you have a vacuum tunnel. Like in the image under. Will the red mass sink to the bottom?
NB. Earths mass is no longer surrounding the tunnel.


The mass would only move if there was a force acting on it.


How would the mass know which way to fall when there is vacuume above and below?


It would not move if no force was acting on it.



posted on Oct, 23 2012 @ 04:31 AM
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reply to post by spy66
 


If both objects are floating in dead space, a pure vacuum, with no outside gravitational fields reacting on it.... Then I believe the red mass would move to the center of the vacuum chamber. Both the vacuum chamber and the red object have mass, and I believe both of their masses would bend space and time, and they would align themselves to the center of their mass.



The red mass would only move to the bottom of the vacuum chamber if there was a gravitational field below it.
edit on 23-10-2012 by illuminated0ne because: (no reason given)



posted on Oct, 23 2012 @ 06:19 AM
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Originally posted by illuminated0ne
reply to post by OccamsRazor04
 


It would cancel out motion in those directions, but not force.

In the center of Earth there are multiple sources of gravity. Up, down, left, right, forward, and backwards are sources of gravity.
edit on 23-10-2012 by illuminated0ne because: (no reason given)


No, it would cancel out FORCE that is why there is no motion. It's a very easy experiment to conduct, a simple exercise in logic will give you the answer.

X force = enough force to move the object at 5mph.

Apply X Force going > < v. How fast would the object move and in what direction?

In case you have trouble the answer is 5mph moving in the v direction. The only way the other forces have any bearing is if friction is in play.
Apply



posted on Oct, 23 2012 @ 06:21 AM
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Originally posted by spy66
reply to post by Phage
 



Ok.

If you have a vacuum tunnel. Like in the image under. Will the red mass sink to the bottom?

NB. Earths mass is no longer surrounding the tunnel.

How would the mass know which way to fall when there is vacuume above and below?



If that vacuum is in range of Earth's gravitational pull, the mass would move towards whatever direction the Earth is in relation to it. Why do you think Vacuum makes gravity not function?




posted on Oct, 23 2012 @ 07:27 AM
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Originally posted by OccamsRazor04
No, it would cancel out FORCE that is why there is no motion.


Just because there is no motion, doesn't mean there is no force acting on it.

Just because you are sitting in a chair, not falling, doesn't mean the force of gravity is not acting upon you.


Originally posted by OccamsRazor04
It's a very easy experiment to conduct, a simple exercise in logic will give you the answer.

X force = enough force to move the object at 5mph.

Apply X Force going > < v. How fast would the object move and in what direction?

In case you have trouble the answer is 5mph moving in the v direction. The only way the other forces have any bearing is if friction is in play.
Apply


Actually, that is not logical at all...

If you apply 5mph of force to the right > and 5mph of force to the left < that is equivalent to 10mph of force locking the object in place, and it won't move up nor down. If you then apply 5mph of force down V there wouldn't be enough force to break the lock of the 10mph force that is locking it in place.


edit on 23-10-2012 by illuminated0ne because: (no reason given)



posted on Oct, 23 2012 @ 02:04 PM
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Originally posted by illuminated0ne

Originally posted by OccamsRazor04
No, it would cancel out FORCE that is why there is no motion.


Just because there is no motion, doesn't mean there is no force acting on it.

Just because you are sitting in a chair, not falling, doesn't mean the force of gravity is not acting upon you.


Originally posted by OccamsRazor04
It's a very easy experiment to conduct, a simple exercise in logic will give you the answer.

X force = enough force to move the object at 5mph.

Apply X Force going > < v. How fast would the object move and in what direction?

In case you have trouble the answer is 5mph moving in the v direction. The only way the other forces have any bearing is if friction is in play.
Apply


Actually, that is not logical at all...

If you apply 5mph of force to the right > and 5mph of force to the left < that is equivalent to 10mph of force locking the object in place, and it won't move up nor down. If you then apply 5mph of force down V there wouldn't be enough force to break the lock of the 10mph force that is locking it in place.


edit on 23-10-2012 by illuminated0ne because: (no reason given)


Gravity is not a rope, or string, which acts from a point source either side of the object. The rope conception is really just an analogy or metaphor to help describe the way the gross forces work.

Gravity is a curvature to space-time and therefore has a much less distinct source location. Generally, we consider that the "source" of gravitation is at the center of mass, but really all of the mass bends space-time in a cumulative manner.

Although the forces are in balance, the object is not "locked" in place. The tiniest force that would unbalance the action of the balanced forces would cause the object to begin to move.

Objects like moons, stars and planets can be "tidally locked", like with the rope analogy, but even then, tiny imbalances would cause the locked object to move.

In your revised diagram the left-right components (vectors) would always cancel each other out, they would negate each other perfectly and so can be removed from any calculation. In truth, even this is a simplification as all of the forces from all of the mass of the Earth in many directions would be acting upon the 10kg mass. We just simplify this down to a couple of vectors which themselves are the vector sums of many other vectors. This simplification (the center of mass vector sums) will give the same answer as the more complex system and allows for easier mathematical manipulation and conception of the forces involved.

It is really just a case of the simple vector addition of all the forces acting upon the object.


edit on 23/10/2012 by chr0naut because: I erroneously used a stop rather than a comma! Gaak!



posted on Oct, 23 2012 @ 02:27 PM
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Originally posted by illuminated0ne
reply to post by spy66
 


If both objects are floating in dead space, a pure vacuum, with no outside gravitational fields reacting on it.... Then I believe the red mass would move to the center of the vacuum chamber. Both the vacuum chamber and the red object have mass, and I believe both of their masses would bend space and time, and they would align themselves to the center of their mass.



The red mass would only move to the bottom of the vacuum chamber if there was a gravitational field below it.
edit on 23-10-2012 by illuminated0ne because: (no reason given)


As Illuminated has shown, the red mass would initially be attracted towards the center of mass (in the absence of any other forces).

This would be very slow as the forces of gravitation for such low mass objects are very slight.

The thing is that with a loss-less environment like a vacuum, the energy would never be depleted from the system and so the red mass would move (very slowly) from one end to the other in simple harmonic motion, forever - perpetual motion!



posted on Oct, 23 2012 @ 07:11 PM
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Originally posted by illuminated0ne
reply to post by Phage
 


Ok let me ask you this...

Since I think my previous posts were not clear enough I have decided to modify this image (please forgive the quality) to illustrate my point...



When the gravity train is in the center of Earth, gravity is pulling it in all directions. It seems to me that most all calculations regarding the gravity train are only considering the vertical axis (up/down arrows) of gravity. Do you not concur that the horizontal axis (left/right arrows) of gravity would effect the velocity of the gravity train every time it passes the center of Earth? Keep in mind, this image is only 2 dimensional. There is also a 3rd axis of gravity (forward / backward) that I did not illustrate which would seem to also effect the train. Would not the gravitational pull in those directions other than vertical effect the train as it passed the center?

I believe it would.

I believe the center of Earth would slow the train down every time it passes. I think every time the train passes the center, a small amount of kinetic energy would be converted into gravitational potential energy that it can't use. Over time the train would get stuck in the center of Earth with only gravitational potential energy and no kinetic energy.

I think it is easy to calculate the gravity train's acceleration and velocity when you only consider the vertical axis (z) of gravity. However, in the center of Earth, there is two more axes of gravitational pull (x and y) acting on the object. I believe these are going ignored.

edit on 23-10-2012 by illuminated0ne because: (no reason given)


A horizontal force has no effect on vertical motion. If you notice in your drawing, the force to the right is the same as the force to the left, that keeps the object in equilibrium. Also, at the particular point at which you placed it, the upper and lower forces are also congruent so there is no "net force" on the object. However, the object had already been moving because the lower force was greater than the upper force so it has velocity. Even though the net force is zero, it has kinetic energy because it was moving. It is going to continue moving until that Kinetic energy is transformed to potential energy. As the kinetic energy decreases and the Potential energy increases (it goes down), the upper force is going to become greater than the lower force until there is no more kinetic energy and the upper force (the cause of the potential energy) pulls it back "up" changing that potential energy back to kinetic energy at the center and then the kinetic energy back to potential energy from the center to the top in an endless cycle.

Since there is no friction in the theoretical model or any other loss of energy, it will travel back and forth indefinitely from the North to the South pole. You can argue that it wouldn't happen in reality, but of course in reality, we're not going to create a tunnel from the North to the South pole and put a 10 kg weight in it anyway. This is a theoretical situation and in the theoretical situation, there is no other energy loss due to the vacuum. So, it will travel up and down and up and down and up and down from one pole to the other forever and ever.



posted on Oct, 23 2012 @ 07:33 PM
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reply to post by PurpleChiten
 


would taking into consideration the earths rotation,, revolution around the sun,, and movement through interstellar space as a part of the solar system,,, change the outcome?



posted on Oct, 23 2012 @ 08:00 PM
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Originally posted by ImaFungi
reply to post by PurpleChiten
 


would taking into consideration the earths rotation,, revolution around the sun,, and movement through interstellar space as a part of the solar system,,, change the outcome?


To a great extent, the 10kg mass was already in orbit (with the Earth) at the start of the experiment.

There are some minor things that would be likely to perturb the linear path with respect to the Earth (as i mentioned in an earlier post) but these are minor and would only be likely to cause the 10kg mass to keep hitting the walls of the tube (which would deplete energy).



posted on Oct, 23 2012 @ 08:45 PM
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OK...Although many of you have provided some very compelling concepts and math and physics about why you believe the object dropped would continue to move back and forth in a perpetual state of motion based on the premise that Gravity in a Frictionless Environment in a Vacuum due to a 1 to 1 ratio of transfer of Kinetic Energy based on applied Gravitational Force....THIS is why you are WRONG.

1. Gravity...is NOT a FORCE. It is an EFFECT which occurs in the presence of any MASS.
2. Gravitational EFFECT that is created in a Celestial Body such as Planet Earth...creates a SINGULARITY at the center point of the entirety of that Celestial Body. Thus we have the greatest amount of Gravitational Space/Time Warping of expression of One Dimensionality or Singularity dead center of Earths Mass...thus that is why there is only a form of Liquid Metallic Core essential surrounding that center point and not anywhere within the area that is Earths Mantle.
3. Gravitation Compression is at it's Highest Level of Effect at Earths Center...which is the Center of Earth Gravity Well.
4. Gravity does not push or pull any object or even Quantum Particle/Wave Form of Energy...it simply FALLS. A FALLING OBJECT that is placed into a state of Fall is not actually MOVING but rather it is EXISTING AT DIFFERENT INFINITE POINTS ALONG A PERPENDICULAR LINE TO THE CENTER OF A GRAVITY WELL DEPENDENT UPON OUR PERCEPTION OF TIME. The object has no FORCE or QUANTUM PARTICLE FIELD ATTRACTION or DIRECTED KINETIC FORCE setting it into a state of FALL.
5. Since standard models of Kinetic Force Transfer cannot be used to describe Gravitational Effect...Math and Physics uniform Models of Cause and Effect are not applicable in this Topics Question.
6. Since it is SPACE/TIME that is being effected to place an object into a state of FALLING...Math and Physics representations that are based on Newtonian Energy Transference and Standard Kinetic Energy Applications will not be able to describe what the true outcome will actually be in this topics question.
7. An perfect analogy to describe what I have been stating is impossible as only an imperfect analogy is possible since the Human Mind cannot Mentally Picture Space/Time Curvature of Gravity being an expression of Singularity or One Dimensionality.
8. If I was WRONG about what I have been stating then Celestial Bodies such as the Planet Mars would never have lost it's EM Field due to that Planets Liquid Metallic Core ceasing to Spin as it Cooled Down as it's Gravitational Compression at the center of that Planets Gravity Well which is Direct Center of it's Mass...eventually slowed to a stop the Spinning Core.
9. Earths Spinning Liquid Metal Core will also eventually stop spinning as Gravitational Compression will bring it to a halt.
10. The earths MOON is slowly drifting away from Earth and eventually will break orbit with the Earth and move into an Orbit around our Sun that is separate from the Earth. The reason for this is that the Kinetic Force that the Moon has obtained as well as it's current distance and velocity the Moon orbits the Earth at is just slightly greater than the Gravitational Effect keeping the Moon in Earths orbit. If the Moon was orbiting any slower or was just slightly closer than it's current orbit...would mean that the Moon would eventually impact Earth. The Moon is orbiting in a Frictionless State and it's orbit and eventual fate is solely determined by it's Velocity and distance from Earth.
11. I am Just TOO COOL to be WRONG! LOL! Split Infinity



posted on Oct, 23 2012 @ 08:49 PM
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Originally posted by chr0naut

Originally posted by ImaFungi
reply to post by PurpleChiten
 


would taking into consideration the earths rotation,, revolution around the sun,, and movement through interstellar space as a part of the solar system,,, change the outcome?


To a great extent, the 10kg mass was already in orbit (with the Earth) at the start of the experiment.

There are some minor things that would be likely to perturb the linear path with respect to the Earth (as i mentioned in an earlier post) but these are minor and would only be likely to cause the 10kg mass to keep hitting the walls of the tube (which would deplete energy).



ok,, i see,,,,

could this be the reason that we cant make a perfect vacuum in the first place?

also if the column through earth was a perfect vacuum,,, the vacuum itself would still be effected by outside forces and motions? and if the 10 kg mass was place inside the vacuum,, the mass would be effected by the outside forces,, pretty much only,, because it being in a vacuum would be there would be nothing in the column itself to effect it?



posted on Oct, 23 2012 @ 09:09 PM
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reply to post by SplitInfinity
 


10. The earths MOON is slowly drifting away from Earth and eventually will break orbit with the Earth and move into an Orbit around our Sun that is separate from the Earth. The reason for this is that...

Out of all the other nonsense you've spouted I will address this one because it's the last.


In order for an object to move to a higher orbit it must obtain energy. Where is the Moon obtaining that energy from?

The Moon's orbit is widening because of tidal effects. The tidal effects of the Moon on Earth are slowing Earth's rotation and transferring that energy (conservation of energy, remember?) to the Moon's orbit, moving it into a higher orbit. Eventually the Earth will become tidally locked with the Moon just as the Moon is tidally locked with Earth. At that time the orbit of the Moon will be fixed, the movement from the Earth will stop because the orbital period of the Moon and the rotation of the Earth will be synchronized.


Now lets' back up a bit. Please explain how the laws of thermodynamics are violated by a frictionless pendulum swinging forever.

edit on 10/23/2012 by Phage because: (no reason given)



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