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Posted: 5/21/2003 5:29:43 PM EDT
I was thinking today. (Not a common occurrence) I was thinking about physics, and gravitational acceleration. Well, I believe it all. In a vacuum, if you drop a feather and a bowling ball from the same height, they will hit at the same time. But I think there is another variable left out. It may be small, but it can still be a player. IIRC, I was taught that all objects, no matter how massive, have gravity. A pencil has gravity, however it is not powerful enough to attract anything. Even the moon only has a small gravitational force. So my question is: Wouldn't this variable affect acceleration (i.e. the larger object accelerates faster because it's own gravitational pull is combined with the Earth's.)? Obviously the difference would be so small between a bowling ball and a feather in gravitational pull that no instrument could measure it, but if these principles are applied in astronomical uses, where things happen on a massive scale, it could be a player. Am I thinking with my ass or am I the next Galileo?
Link Posted: 5/21/2003 5:35:45 PM EDT
There is only one way that I can think of to figure this out. First, drop feather over your foot and time how long it takes to hit it. Then repeat the process with a bowling ball. Get back to us on the results.
Link Posted: 5/21/2003 5:36:38 PM EDT
[Last Edit: 5/21/2003 5:50:02 PM EDT by SuperAlpha]
You are right in your thinking. But if you have the same mass (ie: 10kg bowling ball, and 10kg of feathers) then they should drop at the same acceleration. Now here is the real twist: The feather AND the bowling ball are each pulling the earth toward them and the earth is pulling them toward itself. AND to top it all off, the acceleration is based on the distance between the two, so it is constantly changing. If you measure gravity out where the space shuttle orbits, you will find it is about 4.5m/s^2 . This is not zero gravity, but since it is in orbit it counters the gravity. Oh - and do not forget the attraction between the feather and the bowling ball!!!! If you want to put this on paper and do the math, I suppose I could whip up a mathcad or maple simulation...or even MS Excel for that matter.
Link Posted: 5/21/2003 5:43:29 PM EDT
sorry, can't help ya! still working on the apple and Isaac Newton story.. wonder if he invented the Apple flavored Fig Newtons?
Link Posted: 5/21/2003 5:47:44 PM EDT
Link Posted: 5/21/2003 5:52:31 PM EDT
Correct. Center of mass plays a major role here too. also - Guess what the gravity is when you are 2000ft INSIDE the earth.
Originally Posted By TREETOP: Another twist: Gravity from a spherical object pulls in every direction pretty much equally(which is why people in China are also able to walk on the earth's surface). The bowling ball would then have an equal pull towards the top of the vacuum container as the bottom, therefore not speeding it up OR slowing it down.
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Link Posted: 5/21/2003 5:56:08 PM EDT
another thing to think about... quantum effects! Most physics equations and generalizations are for low speed and ignore quantum effects because they seem negligible. There is more to the equation when approaching the speed of light!
Link Posted: 5/21/2003 6:34:35 PM EDT
Originally Posted By SuperAlpha: You are right in your thinking. But if you have the same mass (ie: 10kg bowling ball, and 10kg of feathers) then they should drop at the same acceleration.
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Actually, this is exactly what Galileo disproved. He proved by dropping a 100 pound ball and a 1 pound ball off of the tower of Pisa, that the acceleration due to gravity is independent of mass. They both hit the ground at the same time. The problem on earth with the feather is due to drag caused by the air. In the vacuum of outer space, this problem is eliminated as was shown by the first astronauts to land on the moon.
Link Posted: 5/21/2003 6:43:33 PM EDT
The calculus of trying to figure out gravity 2000 ft inside the planet has get to be evil.
Link Posted: 5/21/2003 6:44:15 PM EDT
if you dropped a 10 lb bowling ball, and a 20 lb ball of the same dimension, the heavier one would hit first, in a vacuum, because there is more gravitational mass between the object being dropped on and the 20lb bowling ball. it would be a very small difference, but it would be there. TXL
Link Posted: 5/21/2003 6:50:14 PM EDT
[Last Edit: 5/21/2003 6:52:06 PM EDT by Halfcocked]
serrada: Actually all objects that have mass have gravity. The greater the mass the greater the acceleration due to gravity. The heavier object will have a greater acceleration, however the difference of the 2 objects compared to the mass of the Earth, or the Moon for that matter, is negligible.
Link Posted: 5/21/2003 6:57:31 PM EDT
Yeah, It's all fun and games until someone throws the earth out of orbit.
Link Posted: 5/21/2003 7:02:56 PM EDT
I will try to sort this out for you. In a vacuum the acceleration due to gravity is 9.8m/2^2 for all objects on earth at sea level. Now you are correct in the assumption of the gravity of the ball and the feather. Of course the feather has less gravity and the bowling ball has more. That number is proportionate to the respected mass of each. Here's the problem. Science has yet figured a way to calculate what is called "Big G" or basically give a numerical equation for gravity in questions such as yours. Suffice to say "Big G" is so incredibly small it is likely we will never be able to measure it although we know it exists. But I digress. Back to your question. There is no measurable difference due to the fact that we cannot figure the value of "Big G" and the fact that if dropped together in a vacuum they would hit the same time and the acceleration would be constant at 9.8m/s^2. Part of the reason that we cannot measure big G lies in the problem that we are affected by the gravity of everything in the universe. As one of the 4 fundamental forces that control everything Gravity is unique in that it is always attractive and it has, theoretically, infinite range. So to measure Big G we would have to account for all the gravity in the universe and eliminate that from the measurements, but easier said than done. There is an unknown amount of dark matter in the universe, there is the constant formation of stars, also because a cloud of interstellar dust strewn across lightyears in the galaxy has less G than the same amount of matter when it is compressed into a star a few hundred thousand miles in diameter. Then you have to create an instrument that can negate the effects of the earths gravity. So it is understood that there is gravity in all mas but here on earth a feather with the mass that is somewhere on the lines of 1/100000000000000000000000000000000000000000000­000000000000000000000000000000000000000000000­000000000000000000000000000000000000000000000­000000000000000000000000000000000000000000000­000000000000000000000000000000000000000000000­000000000000000000000000000000000000000000000­000000000000000000000000000000000000000000000­0000000000000000000000000^10 that of the earth it is neglegable and a bowling ball is not much more significant in the realm of normal physics. This is part of the conundrum presented by the Grand Unified Theory and I am getting way out of my range here. Gravity has mass in the form of "gravitons". They travel at the speed of light. Thus if the sun were to suddenly disappear the earth would remain in orbit for about 8 minutes after it disappeared. But the gavity of the earth is stronger than that of all the individual particals that make it up if they each were measured and summed up alone. That my friend is the big question in physics today. The laws of massive objects seem to be different for small(quantum) particals. Why? We have a few "Theories" but just that!
Link Posted: 5/21/2003 7:06:05 PM EDT
Assuming you manage to find your way to the center of the Earth, you would be weightless. (Of course were ignoring minute variables and the gravity of the sun, moon, mother-in-law, etc). You would be pulled by the gravity of the Earth equally in all directions.
Link Posted: 5/21/2003 7:09:12 PM EDT
Originally Posted By Valkyrie: I will try to sort this out for you. In a vacuum the acceleration due to gravity is 9.8m/2^2 for all objects on earth at sea level. Now you are correct in the assumption of the gravity of the ball and the feather. Of course the feather has less gravity and the bowling ball has more. That number is proportionate to the respected mass of each. Here's the problem. Science has yet figured a way to calculate what is called "Big G" or basically give a numerical equation for gravity in questions such as yours. Suffice to say "Big G" is so incredibly small it is likely we will never be able to measure it although we know it exists. But I digress. Back to your question. There is no measurable difference due to the fact that we cannot figure the value of "Big G" and the fact that if dropped together in a vacuum they would hit the same time and the acceleration would be constant at 9.8m/s^2. Part of the reason that we cannot measure big G lies in the problem that we are affected by the gravity of everything in the universe. As one of the 4 fundamental forces that control everything Gravity is unique in that it is always attractive and it has, theoretically, infinite range. So to measure Big G we would have to account for all the gravity in the universe and eliminate that from the measurements, but easier said than done. There is an unknown amount of dark matter in the universe, there is the constant formation of stars, also because a cloud of interstellar dust strewn across lightyears in the galaxy has less G than the same amount of matter when it is compressed into a star a few hundred thousand miles in diameter. Then you have to create an instrument that can negate the effects of the earths gravity. So it is understood that there is gravity in all mas but here on earth a feather with the mass that is somewhere on the lines of 1/100000000000000000000000000000000000000000000­000000000000000000000000000000000000000000000­000000000000000000000000000000000000000000000­000000000000000000000000000000000000000000000­000000000000000000000000000000000000000000000­000000000000000000000000000000000000000000000­000000000000000000000000000000000000000000000­0000000000000000000000000^10 that of the earth it is neglegable and a bowling ball is not much more significant in the realm of normal physics. This is part of the conundrum presented by the Grand Unified Theory and I am getting way out of my range here. Gravity has mass in the form of "gravitons". They travel at the speed of light. Thus if the sun were to suddenly disappear the earth would remain in orbit for about 8 minutes after it disappeared. But the gavity of the earth is stronger than that of all the individual particals that make it up if they each were measured and summed up alone. That my friend is the big question in physics today. The laws of massive objects seem to be different for small(quantum) particals. Why? We have a few "Theories" but just that!
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SLOWLY!, reach in to your shirt pocket and remove your pocket protector, TWO FINGERS ONLY, AND BACK AWAY FROM THE CALCULATOR!
Link Posted: 5/21/2003 7:09:28 PM EDT
Valkyrie: Your post has twice the mass of anybodies elses and it repels me.
Link Posted: 5/21/2003 7:16:02 PM EDT
Actually Eienstein rejected the idea of a dynamic universe and favored that of one that was static. In his original theories he figured for the affects of gravity and used a function to counter them. Well as we discovered an expanding universe it became evident that he was wrong and he changed his theories. Now we have discovered that he was partially correct because the universe is expanding at an increasing rate rather than slowing down as we once thought it would due to the effects of gravity so there is some repulsive force at work. What it is we have no idea. But it is safe to say that the theory of a big crunch after the beginning of a big bang is no longer valid. There are now even doubts that the universe started at a singularity. This discovery of an expanding universe that is increasing at a proportionate rate turned everything about the big bang upside down.
Link Posted: 5/21/2003 7:19:54 PM EDT
Originally Posted By SuperAlpha: Guess what the gravity is when you are 2000ft INSIDE the earth.
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Acceleration of gravity should be approximatly 32.1636 ft/sec2 at 2000 feet below sea level at the equator. But this is just a guess [:D]
Link Posted: 5/21/2003 7:41:15 PM EDT
Nature abhors a vacuum, so do dogs...
Link Posted: 5/21/2003 8:40:39 PM EDT
Whew, for a second there, I thought that Valkyrie was going to start talking about the zero point field theory. I think I'm gonna need a couple of beers for that one.
Link Posted: 5/21/2003 9:08:54 PM EDT
[Last Edit: 5/21/2003 9:13:44 PM EDT by ColonelKlink]
Link Posted: 5/21/2003 9:20:33 PM EDT
[Last Edit: 5/21/2003 9:21:34 PM EDT by Valkyrie]
Correct but accounting for Big G would need to calculate these forces and negate them her on earth. I think I read someplace physicists are working on a gravitational interferometer. The real grabber is that, if in fact, gravity has a componant called a graviton as theorized that behaves similarly to light than this opens a whole new realm of physics that could lead to anti-gravity, high speed travel, time travel(as gravity causes a warp in space-time), energy production. We just have to figure how we can artificially generate gravity without the mass of matter involved. Do that and we cross into Star Trek for real. It's being studied and theory says it is possible. It may be a reality in the next 50 years!
Link Posted: 5/21/2003 9:35:41 PM EDT
It's threads like this that make me wish I had gone to college.
Link Posted: 5/21/2003 9:37:36 PM EDT
Trying to do that calculation on my 10-key just fried the friggin' thing. I guess you need one of them fancy graphing calculators, huh? [:P]
Link Posted: 5/21/2003 9:38:35 PM EDT
Don't be hard on yourself. I'm a physics major who dropped out junior year! But beleivbe me when I say I'm reaching in the discussion. Theoretical Physics requires mucho math skills and a mind that can think in scale and in the severely abstract. I just am lucky to be good at the abstract and okay at the math. Actually less than okay but better than lousy if that makes sense. There are far better mathmaticians here than I!
Link Posted: 5/21/2003 9:38:39 PM EDT
[Last Edit: 5/21/2003 9:42:22 PM EDT by SuperAlpha]
Most of the physics classes that I took were in High School (Physics AP, junior year - calculus and algebra based) Some physics in college. I was always into Chemistry and Physics and electronics as a hobby...then in college I went into Electronics. And I did not get my math minor because I failed linear algebra....I never showed up! But the highest math I took was Diff Equ 2.
Link Posted: 5/21/2003 10:27:36 PM EDT
[Last Edit: 5/21/2003 10:28:35 PM EDT by zonan]
Originally Posted By Chapman: A pencil has gravity, however it is not powerful enough to attract anything.
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All mass "has gravity" as you say. It most certainly [i]is[/i] enough to attract stuff with exactly the force: F = GMm/r^2, where M is the object's mass and m is the mass of the attracted object.
Even the moon only has a small gravitational force.
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To be accurate, you really need to talk about the moon's [i]gravitational field[/i] not its "gravitational force." This field is merely g = GM/r^2, so the force on some object due to the moon's field is just the mass of that object times g (i.e., F = mg), so it is different for objects of different mass.
So my question is: Wouldn't this variable affect acceleration (i.e. the larger object accelerates faster because it's own gravitational pull is combined with the Earth's.)?
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[b]No[/b], the acceleration of the object is independent of its mass because of Newton's second law: namely, the vector sum of all forces acting on some object is equal to the mass of that object times its acceleration (F = ma). So the gravitational [b]force acting on the object[/b] due to the earth is the object's mass times earth's gravitational field: mg (remember, g is GM/r^2, so the force acting on the object, mg, is equal to GMm/r^2). From Newton's second law, F = ma. Plugging in for F gives: GMm/r^2 = ma, and m (the mass of the object being acted on) divides out, giving GM/r^2 = acceleration. Notice, GM/r^2 is the gravitational field, g, I mentioned above. So: [b]a = g[/b] = GM/r^2. So the acceleration of the object is independent of that object's mass. QED. Now, I believe what you really wanted to know was the following: Would the "heavy object" collide with the earth before the "light object" (ignoring all other forces). The answer to that is [b]yes[/b]. As stated in Newton's 3rd law, every force has an opposing pair. Of course, we already know that the earth generates a gravitational field (let's call it g[e]), g[e] = GM/r^2. So an object passing through experiences force F = mg[e] = GMm/r^2, directed radially inward. But the object itself has mass and thus creates its own gravitational field, g[o] = Gm/r^2. The earth is in this field of course, so the force experienced by the earth is Mg[o] = GMm/r^2 which is the same as F. So the earth accelerates toward the object with a[e] = Gm/r^2. Obviously the object has a higher acceleration toward the earth than the earth does toward the object because GM/r^2 > Gm/r^2. So if you have two masses at some equal distance r from the earth (from its center, so of course we assume r is greater than the radius of the earth), then both objects will have the same acceleration toward the earth (GM/r^2) but the earth will have a higher acceleration toward the one with the higher mass. These effects are negligible unless you are dealing with masses within a couple of orders of magnitude of the earth's mass.
Link Posted: 5/21/2003 10:42:45 PM EDT
Originally Posted By SuperAlpha: Guess what the gravity is when you are 2000ft INSIDE the earth.
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No need to guess. If you assume uniform density in the earth (which is false, of course) then the gravitational field increases proportionally to the distance to the center of the earth. With 2000ft about 650meters, and the radius of the earth about 6500km (we'll round up a little to make it easier), then 650/6500x10^3 or 6.5e2/6.5e6 which is 1e-4, or .0001. So the gravitational field at this depth is .9999g, where g is the gravitational field at the surface. The density of the earth is not uniform (much higher at the center). If you have an accurate model of density as a function of radius then you can integrate it out and get a closer answer.
Link Posted: 5/21/2003 10:46:28 PM EDT
A question what would you 'weigh' at the center of the earth?
Link Posted: 5/21/2003 10:47:05 PM EDT
[Last Edit: 5/21/2003 10:47:24 PM EDT by zonan]
Originally Posted By SuperAlpha: another thing to think about... quantum effects!
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But at these levels we deal only with classical theory (look up the [i]correspondence principle[/i] if interested).
Most physics equations and generalizations are for low speed and ignore quantum effects because they seem negligible.
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Unless I am misunderstanding what you're saying, you are confusing quantum mechanics with relativity. Quantum mechanics is useful in microscopic systems (or extreme pressures/temperatures/etc. like neutron stars, superfluids, black holes, etc) while special relativity deals with objects at high speed.
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There is more to the equation when approaching the speed of light!
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And even when not approaching the speed of light, there is always gravitational time dilation (general relativity). Edit: I'm going to bed now [:D]
Link Posted: 5/21/2003 10:51:22 PM EDT
Originally Posted By Silence: A question what would you 'weigh' at the center of the earth?
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If you are only taking into account the force due to the mass of the earth, then your weight would be exactly zero. Note, this would be easier to prove if the earth were perfectly spherical, but it is still provable and true because of the earth's azimuthal symmetry--i.e., the earth is oblong (larger radius at the equator)
Link Posted: 5/22/2003 3:39:14 AM EDT
Link Posted: 5/22/2003 4:21:53 AM EDT
One of my phsyics professors is dangling the thought that gravity is in fact not a force. This goes along with the existence of gravitrons... oh man, after modern physics py313 im glad i stopped taking that shit...
Link Posted: 5/22/2003 5:26:12 AM EDT
But are you forgetting the mass of the stuff around you, not just the "center of mass of the whole earth", as it does not apply anymore...
Originally Posted By zonan:
Originally Posted By SuperAlpha: Guess what the gravity is when you are 2000ft INSIDE the earth.
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No need to guess. If you assume uniform density in the earth (which is false, of course) then the gravitational field increases proportionally to the distance to the center of the earth. With 2000ft about 650meters, and the radius of the earth about 6500km (we'll round up a little to make it easier), then 650/6500x10^3 or 6.5e2/6.5e6 which is 1e-4, or .0001. So the gravitational field at this depth is .9999g, where g is the gravitational field at the surface. The density of the earth is not uniform (much higher at the center). If you have an accurate model of density as a function of radius then you can integrate it out and get a closer answer.
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Link Posted: 5/22/2003 5:29:30 AM EDT
Actually I meant the relativistic portions are [b]more significant[/b] at higher speeds, and negligible at lower speeds, but they are still there. Just like quantum effects are there.
Link Posted: 5/22/2003 3:59:40 PM EDT
Originally Posted By SuperAlpha: But are you forgetting the mass of the stuff around you, not just the "center of mass of the whole earth", as it does not apply anymore...
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But it does apply! Newton proved that if you are in a sphere, the gravitational forces due to the masses outside your radius cancel out completely. As a side note, that is one of the reasons he invented calculus--to perform that proof. This leads to the interesting fact that if you are inside an empty spherical shell, you will experience no gravitational force due to any of the mass making up the shell. This is true at all points in the sphere--the center, touching the wall at some point, and everywhere else. Note: This is only true because gravity is an inverse square law (that is, the gravitational force is proportional to 1/r^2, where r is the distance). If the force behaved as 1/r, 1/r^3, or anything else, this would not be true, and gravity would not "work" the way it does.
Link Posted: 5/22/2003 4:38:41 PM EDT
Originally Posted By Valkyrie: Science has yet figured a way to calculate what is called "Big G" or basically give a numerical equation for gravity in questions such as yours. Suffice to say "Big G" is so incredibly small it is likely we will never be able to measure it although we know it exists.
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What is this "Big G" you are talking about? Are you referring to the universal gravitational constant, often denoted G? If so, where are you getting the idea that we don't know what it is? It is 6.67x10^-11 N*m^2/kg^2. I don't know how many thousands of physics students have measured this with a torsion balance. Yes, it is very small. That is why the value is only known to a few significant figures (which is more than adequate for every case I'm aware of).
There is no measurable difference due to the fact that we cannot figure the value of "Big G"
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Again, I don't know what you're referring to here.
and the fact that if dropped together in a vacuum they would hit the same time
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The acceleration would be the same, but not the time to impact, as I showed a few posts up.
So to measure Big G we would have to account for all the gravity in the universe and eliminate that from the measurements, but easier said than done.
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Huh? Again, if "Big G" is just the gravitational constant, then it is simple to measure. Gravitational forces [b]superimpose[/b]. This means that you can measure G in the following way: Take two objects of known mass and hang them from two strings of known mass. Isolate one from the other (i.e., bring one far away) and hang it like a pendulum. Mark the way in which it is hanging. Now bring in the other mass and hang it right next to the other one, some known distance away. These masses will disturb each other gravitationally, exhibiting the same force on each other, displacing the hanging masses from the vertical some measurable angle. The displaced angle gives us the force, so: F = GMm/r^2 where all variables are known except G. Solve for G. Of course, since the gravitational force is so small, it will be difficult to measure unless M and m are large and r is small, but that is a matter for experimentalists to figure out (which they have). That's it.
There is an unknown amount of dark matter in the universe
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Which doesn't affect the measurement of G on a local scale to any significant degree.
there is the constant formation of stars
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Which doesn't change the mass of anything (i.e., mass is conserved).
also because a cloud of interstellar dust strewn across lightyears in the galaxy has less G than the same amount of matter when it is compressed into a star a few hundred thousand miles in diameter.
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What is this "G" you are referring to now? Is this different from "Big G?" Is G now the gravitational potential energy?
Then you have to create an instrument that can negate the effects of the earths gravity.
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Why?
Gravity has mass in the form of "gravitons". They travel at the speed of light.
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I have had a brief foray into elementary particle physics, but that was exclusively within the more practical aspects of the standard model. So I have not studied the idea of gravitons in any rigorous way. That said, gravitons are entirely theoretical at this point. And I want to correct one thing in your statement: Gravitons do not have mass--if they did they couldn't travel at the speed of light (special relativity).
But the gavity of the earth is stronger than that of all the individual particals that make it up if they each were measured and summed up alone.
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Please explain this further. This seems to violate the principle of superposition.
Link Posted: 5/22/2003 5:18:30 PM EDT
[Last Edit: 5/22/2003 5:21:10 PM EDT by zonan]
Originally Posted By Valkyrie: I think I read someplace physicists are working on a gravitational interferometer.
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The [i]Laser Interferometer Gravitational Wave Observatory[/i] (LIGO): [url]http://www.ligo.caltech.edu/[/url] We studied this briefly last semester. It is an attempt to detect gravity waves, but again, I don't see what this has to do with your "Big G". Edit: If anyone is interested, I can dig up some information about LIGO to send. Most of what they're doing is far beyond me at this point, but they have some nice pictures. [:E]
Link Posted: 5/22/2003 5:59:08 PM EDT
As a small aside, is anyone here from central maine or has visited Colby College in Waterville? Hidden behind some trees across the road from the admissions building is a monument. A monument dedicated sometime in the 60s to the research into and eventual understanding of anti-gravity. It mentions how someday this research will allow us to reach the stars and make airflight completely safe. Of course such words on such a monument as such a prestigious skule would explain why they planted such a thick copse of trees around it, keeping it from being spotted with a casual glance.
Link Posted: 5/22/2003 6:21:23 PM EDT
[Last Edit: 5/22/2003 6:26:00 PM EDT by Valkyrie]
Zonan, I am refering to a value that can be used in both quantum mechanics and classical theory. Like I said I'm reaching but I had a professor go on a tangent regarding this my last semester at school. I understand your gravitational constant but from what I understand, and I my be mistaken, it doesn't hold for all calculations especially in quantum mechanics. perhaps he was speaking of a value(mass) for a graviton? Gravitons are thought to behave similar to light. Stephen Hawking also speculated about Big G as well as some other guys like Roger Penrose who did a lot of work on black holes. Like I said I'm reaching pretty far past my grasp but that was how I understood it. There is an effort to define unequivically the value of gravity across all mass. As you are aware when we cross into the realm of quantum mechanics we need to incorporate a bunch of stuff to do so, one o fthose things is this Big G. Like I said I'm not a great mathmatics guy but decent in the abstract. I hope I didn't piss into your intellectual Wheaties. I dropped out of school so what do I know? BTW The professor was named Dr Karimi(sp) at Indiana University of Pennsylvania. He listed is in the directory at the college science dept. When I wrote G I was just shortening gravity.
Link Posted: 5/22/2003 6:36:40 PM EDT
Hell, if you want somethign to blow your mind, try figuring this out. 2 people are born at exactly the same time. One upon birth goes to live on top of the highest mountain on earthm the other stays at sea level. They both die exactly at the same instant. Why is the one that live atop the mountain younger than the other?
Link Posted: 5/22/2003 7:06:27 PM EDT
Originally Posted By Valkyrie: I am refering to a value that can be used in both quantum mechanics and classical theory.
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Ohhh, I follow now. I am not very familiar with that whole side of things, but I know there are a lot of people trying to unite quantum mechanics with general relativity, like you say. I think Hawking is in on that. But quantum mechanics is one extreme and general relativity is the other--I don't see how either of these (hence, Big G) matter to determining the problem posed by the starter of this thread. Maybe I'm missing something....
I understand your gravitational constant but from what I understand, and I my be mistaken, it doesn't hold for all calculations especially in quantum mechanics.
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I think you're right. I've heard some mention of this but haven't studied it myself. From what I understand, the weak effects of gravity are negligible on quantum scales, where electromagnetism rules (or the strong force if you are talking about what goes on in a nucleus).
perhaps he was speaking of a value(mass) for a graviton?
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I haven't studied gravitons, either. But as I mentioned in another post, I believe they would have to be massless in order to move at the speed of light--which has been measured recently, in observations of Jupiter and its moons.
I hope I didn't piss into your intellectual Wheaties.
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Not at all. I genuinely did not understand what you were referring to, and I'd rather learn about something I don't know about than just ignore it.
Link Posted: 5/22/2003 7:10:30 PM EDT
There is a silent killer out there that is affecting the lives of every American, young an old. Each year, millions of innocent children are killed and injured due to the ill effects of gravity. Even your neighborhood playground is not safe. Children fall off swings every day causing serious injury. Not even our nation's nursing homes are safe. How many senior citizens' hips are broken every year because of gravity? All that is needed is for Congress to pass laws to deal with this menace, Write your congressional representatives and ask them to take action. Remember, it's for the children.
Link Posted: 5/22/2003 7:12:02 PM EDT
Originally Posted By gunsmithdude: One of my phsyics professors is dangling the thought that gravity is in fact not a force. This goes along with the existence of gravitrons...
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From what I understand, this is what most particle physicists believe. The theory is that all "forces" are due merely to the exchange of particles. E.g., for the weak force this is the W boson. They theorize that it is the same for all forces, and have deemed the particle that theoretically provides for gravity the graviton.
Link Posted: 5/22/2003 7:13:29 PM EDT
Sorry bout' being inconsistant with the terminology. I had a tough time with some of this stuff. Quantum mechanics eludes me all together! Well hope I clarified what I was trying to convey. I think I was trying to say the uncertainty built into Big G makes precise calculations like the one posed at the origin of the thread nearly impossible, if that makes any sense. Like I said the math is tough for me but I can reason a good bit of it without gewtting lost. After that it's time to break out the ol' google search, lol!
Link Posted: 5/23/2003 11:53:15 AM EDT
[Last Edit: 5/23/2003 11:57:17 AM EDT by Chapman]
Originally Posted By Morpheus: There is a silent killer out there that is affecting the lives of every American, young an old. Each year, millions of innocent children are killed and injured due to the ill effects of gravity. Even your neighborhood playground is not safe. Children fall off swings every day causing serious injury. Not even our nation's nursing homes are safe. How many senior citizens' hips are broken every year because of gravity? All that is needed is for Congress to pass laws to deal with this menace, Write your congressional representatives and ask them to take action. Remember, it's for the children.
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Someone get Feinstein on the phone! We will propose GFB (Gravitational Forces Ban) next week to Congress. Also: Ya'll are way above my meager understanding of physics [:E]
Link Posted: 5/23/2003 1:31:30 PM EDT
Originally Posted By rasanders22: Hell, if you want somethign to blow your mind, try figuring this out. 2 people are born at exactly the same time. One upon birth goes to live on top of the highest mountain on earthm the other stays at sea level. They both die exactly at the same instant. Why is the one that live atop the mountain younger than the other?
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IIRC, general relativity says that time is slower in stronger gravitational fields relative to weaker ones. If that's true then wouldn't the twin on the surface be younger?
Link Posted: 5/23/2003 1:50:34 PM EDT
Link Posted: 5/23/2003 1:59:46 PM EDT
Link Posted: 5/23/2003 2:06:32 PM EDT
I LIKE THE POLL! Truthfully I feel the more I learn the more I understand I really don't know anything in the grand sceme of the universe, if that makes sense.
Link Posted: 5/23/2003 2:22:41 PM EDT
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