Who invented or created the theory of gravity ?

[yevaud]

You are still not describing an outside force. <div class="Discussion_UserSignature"> <p><em>Differential Diagnosis:  </em>"<strong><em>I am both amused and annoyed that you think I should be less stubborn than you are</em></strong>."<br /> </p> </div>

 

[bonzelite]

because there is none. there is no gravity as you assume. only acceleration. and i have given the same examples to describe falling, orbits, grav assists, and grav lensing, as all are the same. <br /><br />we feel the pull of the sun because it is expanding to the earth, and we are expanding to the sun, and being flung around it for the same reasons behind a rock being thrown to the sky then falling back to earth. except our relative speed and motion to sun maintains our orbit, where the rock's human-powered throw is soon overcome by expansion of the larger body. were the human-thrown rock thrown hard enough, it may escape the earth's relative expansion enough to maintain an orbit. or not. orbits are very geometrically specific. <br /><br />

 

[yevaud]

http://www.nrao.edu/pr/2003/gravity/<br /><br />For something that does not exist, it's now been accurately measured.<br /><br />FYI, the expansion of Jupiter would not affect the detected shift mentioned in any way, shape, or form. <div class="Discussion_UserSignature"> <p><em>Differential Diagnosis:  </em>"<strong><em>I am both amused and annoyed that you think I should be less stubborn than you are</em></strong>."<br /> </p> </div>

 

[bonzelite]

that article, as expected, does not prove a damn thing about anything but that jupiter passed in front of a light source and the "gravitational lensing" or bending of radio emissions was present. <br /><br />and their measurements are not ascertained to be anything but vaguely "less than c, but very near it." so they then conclude that speed of g "propagates" at near c. erroneously. <br /><br />there is no gravity that they were detecting. the acceleration of bodies, over billions of years, has <i>far exceeded c by millions of orders of magnitude.</i> <br /><br />there is no c.

 

[yevaud]

And there is no expansion. You have proven nothing, except an interesting verbal game.<br /><br />Two planets, with an observer on each. Each experiences an acceleration.<br /><br />Neither is a force acting between the two that is (and has been) measurable. <div class="Discussion_UserSignature"> <p><em>Differential Diagnosis:  </em>"<strong><em>I am both amused and annoyed that you think I should be less stubborn than you are</em></strong>."<br /> </p> </div>

 

[bonzelite]

<font color="yellow"><br />And there is no expansion. You have proven nothing, except an interesting verbal game. </font><br />the force is the expansion towards the other. i have explained it just recently and in many, many, many prior posts. please re-read them carefully. there is no inherent force that eminates from a body of mass across space. there is only acceleration from a centre of mass. and an interaction of expanding bodies. they either orbit each other, or crash into each other. <br /><br />expansion is proven as all things fall at the same rate. the equivalence principle corroborates gravity as acceleration. mass is not necessary to calculate orbits, as they are geometirc only. orbits have been accurately calculated far before newton ever existed. <br /><br />

 

[yevaud]

<i>orbits have been accurately calculated far before newton ever existed.</i><br /><br />Approximated, not "accurately calculated."<br /><br />Your problem is that you take equivalence as meaning that acceleration is the same as gravitation. You might stop to think about the word "equivalent" first. E.g., having some of the same effects, but not deriving from the same causes.<br /><br />Acceleration is experienced by the direct subjective participants only. Gravity can be experienced by someone or something not directly attached or a part of the onject in question. It is a force. A force propogates at a finite velocity with effects measurable from a distance.<br /><br />Your skew is it's all "apparent." Gravity is a real force, and not apparent at all. <br /><br />Example: you are on an accelerating vehicle, and you pass by me, the stationary observer. <i>I</i> do not feel any gravitational pull from your passage (well I do, but negligible as your mass and my mass are small), although <i>you</i> may experience the equivalent of gravity. <div class="Discussion_UserSignature"> <p><em>Differential Diagnosis:  </em>"<strong><em>I am both amused and annoyed that you think I should be less stubborn than you are</em></strong>."<br /> </p> </div>

 

[bonzelite]

<font color="yellow"><br />Your problem is that you take equivalence as meaning that acceleration is the same as gravitation.</font><br />it is exactly acceleration <i>only.</i> drop the gravity part. <br /><br /><font color="yellow">Acceleration is experienced by the direct subjective participants only. </font><br />it is the only mechanism at work in the entire cosmos. you and i, the sun, the moon, comets, all are experiencing the same mechanism. <br /><br /><font color="yellow">Example: you are on an accelerating vehicle, and you pass by me, the stationary observer. I do not feel any gravitational pull from your passage, although you may experience the equivalent of gravity.</font><br />there is no grav pull from my passage because there is no gravity. you are held fast to the earth as it accelerates, constantly, underneath you. and this expansion supercedes by several orders of magnitude the expansion between you and i. <br /><br />gravity is real in the sense that matter expands from a centre of mass.

 

[eudoxus18]

Perhaps I should've rephrsed this. Sorry, must had my head in the clouds. I was just saying that just because Bonzelite can't refute everything we say doesn't mean his theory isn't true.<br /><br />On a site note, Newton's theories -as stated by him- are completely consistent with Einstein. Later scientists made some assumptions about what Newton said ended up not being true.<br /><br />Though, more to the point, Bonzelite still has not refuted some of my claims.<br /><br />1. This theory cannot account for orbits. I showed this mathematically. The radius vector function of every sattelite in orbit MUST be accelerating. What is causing it to accelerate?<br /><br />2. I showed that NASA's data is consistent with itself. Give me evidence that the gravity is 1/4 somewhere on the moon and I'll believe you.

 

[bonzelite]

<font color="yellow"><br />1. This theory cannot account for orbits. I showed this mathematically. The radius vector function of every sattelite in orbit MUST be accelerating. What is causing it to accelerate? </font><br />yes it does account. i've been telling you how for the entire past few pages. orbits happen exactly because of expansion. you have a stone. you throw it up at a velocity. it reaches apex, it slows down, turns around, falls, gains acceleration more and more until it hits the ground. an orbit is the same thing but it never hits the ground as the orbiting object's speed overcomes the expansion of the parent body. the rock's speed did not --the earth overcame it, expanding up to it. in both cases, there is acceleration. and this is exactly how gravity assists work. <br /><br /><font color="yellow"><br />2. I showed that NASA's data is consistent with itself. Give me evidence that the gravity is 1/4 somewhere on the moon and I'll believe you.</font><br />NASA will never have any data that supports this claim. so forget it. it is a claim of the premise. <br /><br />the centre of mass of the moon is off geometric-centre, nearer the near side of the moon facing us. this region, then, is denser than the far side, making the near side comparatively smaller in expansion --thus 1/6 gravity. the far side is expanding at a greater relative rate compared to the near side, so it's gravity, based upon size of expansion, is <i>more than likely</i> greater. and probably 1/3 --much more than 1/6. <br /><br />you average the two and you get around 1/4; the earth/moon system derive from local disk material. they are one thing.

 

[eudoxus18]

Alright here's a bit of physics review. Acceleration is a chance in either the speed of direction of an object. An object can retain the same speed, but if it's direction is changing, it is accelerating.<br /><br />Now, objects in orbit are accelerating. Their speed is constant, but their direction is constantly changing. For every acceleration there is a force. What is the force that is keeping these objects in orbit? I don't care what model you're using, there -has to be- a centripetal force acting here.<br /><br />1. The radius vector function must be be experiencing a constant acceleration. For every acceleration there is a force. What is the force?<br /><br />2. I have gone through the data personally. I showed you my calculations. NASA is correct. Go through my calculations and show me where I'm wrong.<br /><br />Again, I've yet to see anything except a paltry amount of math. Don't show me words, show me equations.

 

[eudoxus18]

Here's what should make it a little clearer. According to the current model, objects with great enough velocities can orbit the earth because the object is "trying" to get away from the earth, but it is pulled back by gravity. The force acts perpendicular to the object's trajectory, causing it to circle the earth instead of just crash down onto it.<br /><br />Now, -according to your own theory-, as soon as an object leaves the earth, whether it's rock or a rocket, no other force is acting on it. The only thing the object has to go on is it's own velocity (once the fuel has been used).<br /><br />Now, I've calculated (if you want to see them, just ask, but trust me, they're right). That if an object shoots straight up into the air from the ground with initial velocity v, then (on earth) the amount of time it takes for the object to come back down is about t = .204v, that is, if something pops up from the ground at 5 m/s, after about a second it will hit the ground. Now, I don't care how big v is, if no other forces are acting on the object it WILL come back down after (.204)v seconds.<br /><br />If the initial velocity is directed -anywhere except straight up- the amount of time will lessen, no matter what. I can show you this with calculus if you wish. Just ask.<br /><br />Now, if you're right, we can figure out long the sattelites will stay up where they are. We can assume there is no initial expansion velocity, because if there is, then it's the same for both the sattelite and the earth. So all we need to do is figure out if the earth accelerates at 9.8 m/s^2, how long will it take till it overtakes the average sattelite. Since this calculation is rather simple, I'll show it.<br /><br />1. With no inital velocity, d = 1/2 x a x t^2 right? Rearranging, we get:<br />t = sqr. root (2d/a).<br /><br />2. The average sattelite is up about 100 miles, and that's about 161,000 meters.<br /><br />3. a = 9.8. So we plug this into the equation.<br /><br />I got 181 seconds. In oth

 

[eudoxus18]

Bonzelite, in order for ANYTHING to circle ANYTHING there MUST be some centripetal force. And an expansion force doesn't cut it, it must be some attractive force.

 

[bonzelite]

<font color="yellow"><br />Alright here's a bit of physics review. Acceleration is a chance in either the speed of direction of an object. An object can retain the same speed, but if it's direction is changing, it is accelerating. <br /><br />Now, objects in orbit are accelerating. Their speed is constant, but their direction is constantly changing. For every acceleration there is a force. What is the force that is keeping these objects in orbit? I don't care what model you're using, there -has to be- a centripetal force acting here. </font><br />thank your for the review. i know what it is or i'd never be able to carry this debate out. but it's good to sort of reset the dials and reassess and start from the same page. i'm with you. <br /><br />you actually raise very key points to this whole theory. and i'm glad you are being a good sport, and an informative one at that. i appreciate that a great deal. <br /><br />excuse my paltry math, but that is not really my area. i can give it to you in bits. you're very keen on math and can hammer it out like a machine gun. but i'm telling you if you are saying that centripetal force, and the equation for that, and orbits are equivalent, i am proposing that is problematic and erroneous. <br /><br />the equation F=mv^2/R for centripetal force should never have been introduced into orbit equations <i>ever.</i> but it was, and this is how we get the erroneous idea of gravity. where F is the force required to keep a rock and string swinging around. <br /><br />planetary oribts are not at all like this. <br /><br />

 

[bonzelite]

<font color="yellow"><br />Bonzelite, in order for ANYTHING to circle ANYTHING there MUST be some centripetal force. And an expansion force doesn't cut it, it must be some attractive force.</font><br />that is what textbooks say. but it is not proven or true. gravity is not scientifically explained or accounted for. it is just as mysterious a force as the aether. it does not exist. <br /><br />expansion as i have been describing it creates orbits. there is only free-floating conditions with velocity and distances. there is not any such mysterious and unexplained inner force such as gravity to constrain objects into orbits.

 

[eudoxus18]

1. You still haven't addressed the acceleration of the radius vector function. Unless the satellaite is accelerating -outwardly- it WILL fall, and in the span of about three minutes, as I showed you.<br /><br />2. If centripetal force is so erroneous, why does it work so well? It's relatively simple to measure the mass, force, velocity, and radius. Centripetal force is also described by F = mr(omega)^2 where omega is angular velocity, in radians per second. With very simple instruments, it's possible to show this equation is true.<br /><br />And anyway, ICBM's can be fired, and under their own velocity (which is very fast) can curve around the earth (because of gravity) and reach the other side. Under your theory, that is impossible.<br /><br />From an "absolute perspective" let's say you fire a rocket tangent to the earth. Let's say it goes on for a few days under its own velocity. However, from an "absolute perspective" it's trajectory is rigidly straight. To make this easier to visualize, imagine that the after a great distance, the rocket stops. The earth accelerates and catches up to it (imagine that the point at which the rocket stops is simply the distance the rocket has travelled once the earth collides with it). No matter how big the earth has become, it is geometrically impossible for the rocket to be on the other side of the earth.<br /><br />Imagine this: a circle and a line tangent to the circle. The line stays in the same place, but let's say the circle expands. No matter how big the circle gets, the line will never, ever intersect the other side of the circle, no matter what. In other words, there is no point on the rocket's trajectory that reaches the other side of the earth, under your model. But ICBM's and any other rocket can do it.

 

[eudoxus18]

WHen I say anything to circle anything, I don't mean just orbits, I mean EVERYTHING. Centripetal force due to friction is what keeps your car from going straight when it rounds a curve.<br /><br />Perhaps you're centripetal force isn't the same as gravity. When something orbits around something else, there MUST be some force that keeps it orbiting. The force that keeps the object orbiting could be gravity, the coulomb force, tension in a rope (if you're spinning something around your head), or friction (if a vehicle is rounding a bend. And yes, if something is orbiting something, it is of necessity that there is some force there. The object is constantly changing direction, and is therefore accelerating, and for every acceleration there is a force. Here's an example:<br /><br />If a car is rounding a bend, the centripetal force in this case is friction. In other words<br /><br />C = F, where C is centripetal force, and F is friction force.<br /><br />C = mv^2/r, and F = (mu)mg, where (mu) is the coefficient of friction. Setting these two as equal:<br /><br />mv^2/r = (mu)mg<br /><br />M cancels. Isolate v and we get<br /><br />v = sqr. root ((mu)rg).<br /><br />The radius of the circle, and coefficient of friction, and g are all easily measurable. A car must maintain this velocity (or slower) in order to stay driving in a circle.<br /><br />Centripetal force isn't a fundamental force; it's whatever keeps something moving in a circle, again whether it's gravity, coulomb force, tension, or friction.

 

[eudoxus18]

That there is SOME force present when some object is orbiting around something is absolutely undeniable. If something is orbiting, it is constantly changing direction; therefore it's velocity is constantly changing; therefore it is experiencing acceleration; and for every acceleration there MUST be a force.<br /><br />Again, maybe you're spinning a ball around. The force is tension in the rope.<br />You're rounding a bend in your car. The force is friction.<br />Electrons orbit the nucleus of atoms. The force the coulomb force.<br />Planets orbit the sun. The force is gravity.<br /><br />But whatever is causing it to orbit it is called centripetal force for simplicity.

 

[bonzelite]

<font color="yellow">From an "absolute perspective" let's say you fire a rocket tangent to the earth. Let's say it goes on for a few days under its own velocity. However, from an "absolute perspective" it's trajectory is rigidly straight. To make this easier to visualize, imagine that the after a great distance, the rocket stops. The earth accelerates and catches up to it (imagine that the point at which the rocket stops is simply the distance the rocket has travelled once the earth collides with it). No matter how big the earth has become, it is geometrically impossible for the rocket to be on the other side of the earth. </font><br /><br />from what i gather, you are hooked and insist that a centripetal force <i>must be there</i> to constrain anything into an orbit. and this is absolutely untrue. <br /><br />i see your example. i know about tangents to arcs and whatnot, as i am a draftsman. but it as well does not describe reality nor symbolically illustrate what would happen under expansionary conditions. if the rocket stops, it will no longer be accelerating. and as it coasts, it will be overtaken by the earth and crash into the ground. if it suddenly stopped, it would plummet to the ground. it would perhaps coast and glide under aerodynamic forces in the air for a bit, but it would soon crash. what is the problem with that? <br /><br />and nothing traveles in a theoretically perfect straight line. things orbit about each other or curve. a missile can travel across the earth to the other side because it achieves, during it's designed and temporary journey, enough velocity to overcome the rate of the earth's expansion as it's specific relative motion and speed keeps it this way. <br /><br />in an expansionary state, the ICBM would not at all travel for a few days in a straight line. but you must provide more variables than just that. how fast is it going? at what trajectory? how long will it accelerate before it burns it's fuel out? sure, if you set it on a co

 

[bonzelite]

<font color="yellow"> and for every acceleration there MUST be a force. </font><br />yes, it's the force of the expanding atoms pushing out from the centre of mass of the body. <br />

 

[bonzelite]

<font color="yellow"><br />2. If centripetal force is so erroneous, why does it work so well?</font><br />because the equation that incorporates it and is used today is exactly hacked from the orbit equation. when, really, it is entirely unnecessary.

 

[eudoxus18]

Your expansion force is working in the wrong direction, however. You said that the expansion force works the same as gravity but in the opposite direction. Well, that's exactly what's wrong with it. It cannot be a centripetal force because a CF's direction is toward the center of curvature, while yours radiates out from it.<br /><br />And yes, A centripetal force MUST be present for something to circle around something else. You can prove it purely from the definitions:<br /><br />1. Given: Object A is circling around Object B.<br />2. If A is circling around B, it is constantly changing direction.<br />3. If A is constantly changing direction, it is accelerating (acceleration is change is either speed OR direction, or both)<br />4. If A is accelerating, A is experiencing a force.<br />5. This force -is defined to be- the centripetal force. That is, CF is -whatever- keeps something circling around something else.<br /><br />In your model, yes something can travel in a straight line. In current theory, nothing will because every object in the universe will affect its trajectory through gravity. In your model, gravity doesn't exist, which means there's no force that's acting at a distance (through space) on an object in space. This means there is no force exerting on objects floating in space. If force is zero (F = ma = 0) then acceleration is zero (a = 0) and thus (if you integrate that equation), velocity is some constant (v = k). If velocity is constant, neither speed NOR direction is changing, and it is thus travelling in a straight line.<br /><br />So from an absolute perspective, you can think of a rocket having a straight line trajectory with the earth expanding and "catching up to it".<br /><br />Thus, you can think of a rocket's trajectory as a straight line either tangent or secant to a circle (it doesn't really matter). And no matter HOW BIG circle gets, a line that is initially secant to a circle will NEVER intersect the other side, period.

 

[bonzelite]

<font color="yellow"><br />In your model, yes something can travel in a straight line.</font><br />theoretically pure straight line paths do not exist in expansion premise. <br /><br /><br /><font color="yellow">t cannot be a centripetal force because a CF's direction is toward the center of curvature, while yours radiates out from it. </font><br />right. the g vector is reversed in expansion premise. there is no constraining centripetal force. we are in agreement. <br /><br /><font color="yellow">1. Given: Object A is circling around Object B. <br />2. If A is circling around B, it is constantly changing direction. <br />3. If A is constantly changing direction, it is accelerating (acceleration is change is either speed OR direction, or both) <br />4. If A is accelerating, A is experiencing a force. <br />5. This force -is defined to be- the centripetal force. That is, CF is -whatever- keeps something circling around something else. </font><br />per my premise, this is largely correct except for the centripetal force part. you and i are nealy agreeing on everything except that. <br /><br /><br />in the expansion premise, the natural state of objects is to "crowd into" each other. the very nature of objects as they pass each other in space is to approach other objects as they expand. <br /><br />now i think i see what you are saying. you must be referring to absolute momentum of objects in a straight line. unless something acts upon a body, the body will retain a straight line path in it's "absolute momentum." and i'm saying that is non-existent. objects to not possess some absolute momentum. the tangential paths you speak of are moment-to-moment retrained and bent into elliptical paths as the objects expand to meet each other whilst remaining in geometric balance in orbits. their relative motions overcome, <i>in perpetuity</i> the tendency for them to actually touch each other. so, instead, they orbit. <br /><br />the rock thrown up by hand is a very s

 

[derekmcd]

Back to the moon. So what you are implying is that a 200lb person on the near side of the moon would weigh about 33lbs, but if they traveled to the other side, they would weigh about 66lbs? <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[eudoxus18]

Bonzelite, when I say absolute perspective I mean as if someone (immune to the expansion occuring) were to look down on all this. Someone from this perspective would see all this occuring from a true point of view.<br /><br />Your expansion force does not have the "action at a distance" that gravity does. No other forces are directly affecting a floating object in space; there are only 3 other fundamental forces known and they certainly don't affect it. In the absence of gravity, yes it is true that objects travel in exact straight lines. Allow me to prove it to you.<br /><br />Given: F = 0<br />1. F = ma(t) (Newton's 2nd Law, a(t) means acceleration as a function of time)<br />2. ma(t) = 0 (Substitution)<br />3. a(t) = 0 (Divide both sides by m)<br />4. v(t) = k (Integration with respect to time; this statement means the objects velocity as a function of time is some number, may be 2, 3, or a million but it's some number)<br /><br />It's acceleration is zero, which also means it does not change direction. Ever. That's calculus. If you can't believe that, you are hopelessly lost, my friend.<br /><br />Under your view, when you throw a ball at an angle from the earth it is apparently changing direction because you are moving relative to the ball. Not only that, you are accelerating and it is not, and you are accelerating in a different direction. Thus the apparent curvature. The curvature is only there because of relative motion; from an absolute perspective, the ball is moving in a straight line, as I -proved- to you above using calculus.<br /><br />Because all trajectories (under your view) must be straight lines (curvature only appearing because of relative motion) the expanding circle next to a line is a very apt description. And under that the trajectory will NEVER reach the opposite side of the earth.<br /><br />I'm sorry, Bonzelite. Your theory just doesn't work. You can imagine it working in your mind all you want, but your imagination is deceptive. I just show