Duration Posted March 24, 2009 Posted March 24, 2009 I've always wondered what the significance of gravity is to extremely small objects. What kind of gravitational force does a proton feel from a neutron or another proton? I thought that since very small distances between two protons' centers of gravity are so small that maybe gravity plays a pretty huge role in the nucleus. Even though the masses of two protons are so extremely small, their distances between one another are so extremely tiny, and since distance is inversely proportional to the force of gravitational attraction, gravity might just hold two protons together. Using this formula I disproved my theories, coming up with a tiny gravitational attraction. Has anyone ever dabbled in this sort of gravitational attration on a small scale that can give me some help in coming up with a different answer. I was hoping for an answer that would prove gravity to be stronger than the nuclear forces trying to pull protons apart. Try calculating the net force divided by the distance. Gravity is caused by a net force. The greater the net the greater the acceleration, inversely proportional to the distance. Let's just ASSUME the Earth has the mass of 5,974,200,000,000,000,000,000,000 kg. We are going to compare impact times of two different objects when dropped from an exact height of 16.087 feet. Object A has a mass of 1 kg Object B has a mass of 10 kg Using the formula A=(L-S)/R2 Object A has a "A value" of 371,368,185,491,390,563,809,286.93976503 Object B has a "A value" of 371,368,185,491,390,563,809,286.38030708 A previous test was done with object A. It was determined that object A took exactly 1 second to impact the ground when dropped from a height of 16.087 feet, which is an acceleration of 32.174 ft/sec^2. That means an “A value” of 1, has an acceleration of 32.174/371,368,185,491,390,563,809,286.93976503= .0000000000000000000000866363928224699541361186580 02141 ft/sec^2 If you multiply that by Object B’s “A value”, you find an acceleration of 32.173999999999999999999951530582 ft/sec^2 for object B. Now, let’s look at the time it takes for each object to hit the ground, when dropped from the 16.087 feet. Object A- 1.0000000000000000000000000000000 seconds Object B- 1.0000000000000000000000007532389 seconds
ajb Posted March 24, 2009 Posted March 24, 2009 mmmm, strange mix of units. So what are you trying to show? We have [math]F =G\frac{ Mm}{r^{2}} = ma[/math] for the magnitude of the instantaneous force on the mass [math]m[/math]. Here [math]M[/math] is the mass of the Earth. Thus [math]a = \frac{G M}{r^{2}}[/math] Note no dependence on the mass [math]m[/math]. Thus I am a little confused about what you are doing.
Duration Posted March 24, 2009 Author Posted March 24, 2009 mmmm, strange mix of units. So what are you trying to show? We have [math]F =G\frac{ Mm}{r^{2}} = ma[/math] for the magnitude of the instantaneous force on the mass [math]m[/math]. Here [math]M[/math] is the mass of the Earth. Thus [math]a = \frac{G M}{r^{2}}[/math] Note no dependence on the mass [math]m[/math]. Thus I am a little confused about what you are doing. I'm showing that gravity is caused by a net force, and is inversely proportional to the distance. Acceleration rates are different for different mass objects in free fall. All objects do not fall at the same rate.
ajb Posted March 24, 2009 Posted March 24, 2009 ah ok. What is a net force? Acceleration as I have shown, in the absence of air resistance or any other external forces does not depend on the mass. As such, objects (ignoring air resistance) do fall at the same rate, i.e are subject to identical acceleration. This is quite well backed up by experimental results. You need to explain yourself much clearer.
Duration Posted March 24, 2009 Author Posted March 24, 2009 ah ok. What is a net force? Acceleration as I have shown, in the absence of air resistance or any other external forces does not depend on the mass. As such, objects (ignoring air resistance) do fall at the same rate, i.e are subject to identical acceleration. This is quite well backed up by experimental results. You need to explain yourself much clearer. No, past experiments have a common problem, they don't have the proper equipment to measure a difference of .0000000000000000000000000001 seconds. I already posted the facts that say different mass objects fall at different rates. A net force is the difference between one force and another force. The greater the net the greater the acceleration, as shown above, the more massive objects takes longer to impact.
ajb Posted March 24, 2009 Posted March 24, 2009 I'm not getting involved in this any further. I'm fed up of trying to explain well established classical physics to people on here. Frankly, I have better things to do. 1
Sayonara Posted March 24, 2009 Posted March 24, 2009 I'm not getting involved in this any further. I'm fed up of trying to explain well established classical physics to people on here. Frankly, I have better things to do. Well Duration has written about more physical stuff in this thread than you have so I don't think you are even a real person.
Sisyphus Posted March 24, 2009 Posted March 24, 2009 (edited) 1) Define your terms. 2) State clearly why you think [math]F =G\frac{ Mm}{r^{2}}[/math] is false. 3) State clearly why you think your idea is correct. Edited March 24, 2009 by Sisyphus
Duration Posted March 24, 2009 Author Posted March 24, 2009 I'm not getting involved in this any further. I'm fed up of trying to explain well established classical physics to people on here. Frankly, I have better things to do. Just one question: What time device do they use to measure .000000000000000000000000000000000000000000000000000000011111111 seconds?
Sayonara Posted March 24, 2009 Posted March 24, 2009 Just one question: What time device do they use to measure .000000000000000000000000000000000000000000000000000000011111111 seconds? This is basically a symptom of what I mentioned in the other thread. You have primed yourself to expect certain objections and as such you overlook more fundamental problems with your approach. Then you have the nerve to question the methodology of the actual physicists. Don't you think it's about time that you tried something new? I mean, after the first few massive failures most people realise that they are making some kind of procedural mistake and change their method based on what failure taught them. I don't remember the last time I saw a living thinking human fail to learn from the same mistake hundreds of times in a row.
Sisyphus Posted March 24, 2009 Posted March 24, 2009 Again: 1)Define your terms. 2) State clearly why you think [math]F =G\frac{ Mm}{r^{2}}[/math] is false. 3) State clearly why you think your idea is correct.
mooeypoo Posted March 24, 2009 Posted March 24, 2009 Again: 1)Define your terms. 2) State clearly why you think [math]F =G\frac{ Mm}{r^{2}}[/math] is false. 3) State clearly why you think your idea is correct. You mean the 99,224,125 other times he repeated his mantra weren't enough?
Duration Posted March 24, 2009 Author Posted March 24, 2009 You mean the 99,224,125 other times he repeated his mantra weren't enough? Thank You.
Sisyphus Posted March 24, 2009 Posted March 24, 2009 You mean the 99,224,125 other times he repeated his mantra weren't enough? He's repeated himself a lot, but as far as I know he hasn't done any of those things. So, Duration, put up or shut up. Those aren't outrageous requests; they're the absolute bare minimum you need to even be coherent, let alone be taken seriously. If your next post on this thread isn't at least an attempt to do all three, I'm locking it.
Duration Posted March 24, 2009 Author Posted March 24, 2009 He's repeated himself a lot, but as far as I know he hasn't done any of those things. So, Duration, put up or shut up. Those aren't outrageous requests; they're the absolute bare minimum you need to even be coherent, let alone be taken seriously. If your next post on this thread isn't at least an attempt to do all three, I'm locking it. Do a search for: Motor Daddy Traveler Duration Read all those thousands of posts, in all those threads, and then, when you still don't understand it, come see me. I'll answer any question I can.
mooeypoo Posted March 24, 2009 Posted March 24, 2009 Do a search for: Motor Daddy Traveler Duration Read all those thousands of posts, in all those threads, and then, when you still don't understand it, come see me. I'll answer any question I can. He's even proud of the fact he trolls with his hands on his ears.
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