Who invented or created the theory of gravity ?

[bonzelite]

the idea works right away if there is no relative motion, per your belief. so this is basically what you are saying, when adding relative motion of objects, if i understand you correctly:<br /><br />to measure the distance from the objects centres to the centre of the earth: in your belief system, <i><b>theoretically,</b></i> the earth's gravity will accelerate a lighter object with acceleration <i>g,</i> and the light object's gravity will accelerate the earth with an acceleration <i>g'</i>, which is <i>very, extremely, small</i> (since the earth is so massive). <br /><br />so <i> g'</i> = g*m/M, where m is the mass of the object and M is the mass of the earth. as the earth and the object are approaching each other, the distance between them is decreasing with an acceleration of <i>g+g'</i>, which is only very slightly greater than <i>g.</i> the smaller object attracts or perturbs the earth inasmuch as the earth attracts/perturbs the smaller object. with <i>g'</i> being nearly zero. <br /><br />the heavier object will give the earth an acceleration <i>g"</i>, which is larger than <i>g'</i>, although still much much less than <i>g</i>. therefore, the earth and the heavier object will approach each other with an acceleration of <i>g+g"</i>, which is slightly greater than g+g'. this is all about mutual attraction due to <i>mass.</i><br /><br />this is how you are thinking you are correct, yes? <br /><br />

 

[eudoxus18]

No, I mean what force is causing the expansion? In current theory, space-time is curved (in a fourth-dimensional sense) which is what causes gravity. Electrons are involved in electro-magnetism, but as far as we know there is no "graviton", though scientists intuitively believe there must be some particle involved there. Mass itself causes space to curve to light itself isn't really bending, it's space that's bending, which is why light apparently follows a curved path.<br /><br />I'm not trying to be insulting, but it seems this "expansion force" is far more mysterious than gravity.

 

[bonzelite]

yes, i'm not hostile either. we're just talking.

 

[eudoxus18]

Bonzelite: What I meant was this: if objects are indeed expanding, then it seems to me there would have to be only two options:<br /><br />1. Space itself is also expanding at the same rate.<br />2. The objects are accelerating away from each other.<br /><br />Let's assume that 1 is true. This would mean that, basically, the universe is undergoing a size change (geometrically defined) according to the function S(t) = kt^2, where S(t) is the size of the universe, k is some (unknown at this point) constant, and t is time. In other words, the size of the universe at t = 1 (second, let's say) was k. At 2 seconds, it is 4k, at 3 it is 9k, etc. First of all, note it MUST be of this form. If it was a true exponential function of time (mathematically defined), the force we would feel on earth would -not- be constant; but the force we feel is constant, because the acceleration is constant.<br /><br />Now, what this all boils down to is we have some expansion force that behaves slightly differently from gravity (it's expanding matter, not attracting it) but does -exactly- the same thing. This is like what I said before, that essentially what you're doing is calling gravity by a different name. Now you -say- it's actually expanding the universe, but it ends up doing the exact same thing. I too could say that there are "gravity fairies" that have no mass and no energy and attract matter to matter with constant acceleration, but that would be a little weird.<br /><br />But let's assume 2 is true, that matter is accelerating away from other matter (which is why planets don't crash into each other). By your own admission, this is happening at an atomic level, and space is not made of atoms, there no "expansion force" should be present in space, which means planets are actually not accelerating away from each other, and therefore the earth and all matter has crashed into itself. But if that's happened we're all screwed anyway.

 

[bonzelite]

to address MentalAvenger's challenge:<br /><br />a point left out: it is in official science, too, that the inertial differences of objects of differing masses <i>counteracts</i> the force of <i>varied</i> gravitational attraction. in this manner, all objects remain falling at the same rate.<br /><br />in other words, while it is true that an object twice as massive as another, for example, will have twice the <i>theoretical</i> pulling force upon it --it will be <i>twice as difficult to accelerate.</i> F=ma. <br /><br />therefore, the differing inertia equalises or cancels out the gravitational force differences. and all objects fall at the same rate; this is widely known. <br /><br /><br /><br />

 

[eudoxus18]

And I think we all owe a thanks to Bonzelite. Crazy theories like this is what keeps people thinking and keeps us questioning. Who knows, maybe a theory like this will eventually become "fact" and he has done a good job of reminding us that our pursuit is for science and not grant money.

 

[bonzelite]

your idea is very similar to the premise i am positing. you pretty much "get it." but the premise i am debating is not predicated upon the actual space itself expanding to create the acceleration. it is the matter within it. but your idea is pretty cool <img src="/images/icons/wink.gif" /> i feel you. <br /><br />insofar as point 2, i did not contend or imply that because there is acceleration during orbits, lets say, that the planets are being cast away from each other. in the premise, their orbits are a byproduct, if you will, of their expansion <i>into each other.</i> and it is their balance of <i>relative motions, velocities, trajectories,</i> that prevent full-on encounter and smashing. and as matter expands in proportion, into an infinite sea, there is perpetual space to be occupied. <br /><br />by the way, i like your ideas. you're a good sport. i appreciate that. remember, we're only ping-ponging theories. nobody is the enemy here. i'm trying to maintain a civil tone and just have fun with this. and i want to thank Calli and Yevaud and Mental Avenger for putting up with me and guiding the posts. <br /><br /><br /><br />

 

[bonzelite]

thank you, Eudoxus. that means a lot to me.

 

[eudoxus18]

And let me reiterate what I meant, Bonzelite. When I said "science is only for scientists" that does not mean that science is only for a select group of people and others "aren't allowed" to participate. Everyone is welcome to be a part of science provided they know their stuff. In the same way that one has to learn how to paint well in order to be a painter, you must learn how to distinguish good science from bad science to be a scientist.<br /><br />One question: I stated that the rate of expansion of the universe would have to be a power function. I recall in a previous post that you said under your theory it would expand "exponentially". What precisely do you mean by this? I gave you the benefit of the doubt and assumed you were using it coloquially, but answering the question of precisely how fast it is expanding is of vital importance before we go any further.<br /><br />NOTE TO EVERYONE: Keep in mind just because Bonzelite can't answer every objection does not mean his theory is bunk. Every modern theory in every branch of science has holes in it somewhere, it's just that it can take years to find it. Newton's own theory stood for a few centuries before it was debunked by Einstein.

 

[eudoxus18]

Okay, so we've established the following things, assuming this theory is correct:<br /><br />1. Gravity does not exist.<br />2. Massive objects (i. e. those with mass) are expanding at a rate with constant acceleration<br /><br />First, let me flesh out #2. In order for the force we feel on Earth to be constant, the acceleration must be constant (because I'm assuming the mass to be, er, constant.) Let v(t) be the expansion velocity. a(t) = k (an as yet unknown constant), which (by integration) means v(t) = kt (we're assuming that at t = 0 the expansion velocity was also zero, a reasonable assumption IMO). Further this means that x(t) = (kt^2)/2, with x(t) being the position function; it represents how far the universe has expanded. If mass is constant, the function MUST be of this form, no matter what. It's simple math and there's no way around it.<br /><br />So, a few questions<br /><br />1. Does mass also expand at the same rate? i. e., does mass and size increase. If mass is constant, the size must increase at the rate described above. If mass is not constant, at rate does it increase. <br /><br />Let me put it this way. Let's say at t = 1, a proton has a mass of 1 Dalton. At t = 2 is its mass 4 times the mass of the proton at t =1? But then the size is also increasing to counteract it?<br /><br />2. I still don't quite get how orbits can occur. Do the planets push each other out of the way (from a distance obviously) and thus never collide?

 

[bonzelite]

Eudoxus, no problem. i understand. <br /><br />expansion rate is determined by using the earth's own rate as a means to discover the "actual" rate for all matter. and since this is basically a talk about <i>spherical expansion from a centre of mass,</i> we must derive this rate using the earth's radius. <br /><br />in a prior post i related this derivation, so for your benefit, i will do it again:<br /><br />take d=1/2at^2, for example. this is distance traveled due to constant acceleration. in 1 second (time as "t"), then, the earth expands at a certain rate. this is: <br />d=1/2(9.8)(1)^2 = 4.9metres. <br /><br />so the earth's expansion is 4.9m/sec. this is corroborated as it takes 1 sec for any object to fall to earth dropped from 4.9m in height (disregarding wind resistance). but the object is not really falling to earth. the earth is going up to meet the object. <br /><br />as well, objects expand outwardly from their centres (of mass). this would establish an expansion in all directions out from the centre --a radial expansion. <br /><br />once we have established the earth's specific expansion rate from it's centre of mass, which we have done: 4.9m/s, we can then arrive at finding how quickly everything expands, as the earth's expansion is in relative proportion to everything else (meaning, all things maintain their 'sizes' as they expand, because it is all happening at once). <br /><br />the "actual" rate of expansion, then, can be derived as a fraction of the earth's own radius relative to it's own expansion rate ---- />we can take the radius of the earth and divide it into 4.9 --and this gives us the expansion rate of "everything." that is, we can solve for the "constant rate of expansion of all things" by knowing the expansion rate and radius of the earth alone. <br /><br />the radius of the earth is approximately 6,371,000 metres. <br />4.9/6,371,000 = .00000077/s^2 <br /><br />all matter expands at this "actual" rate. and the relative expansions of these things, all thin

 

[eudoxus18]

Slight modification. The earth expands at a rate of v(t) = 9.8t. Remember, with constant acceleration, you get velocity that's constantly increasing.<br /><br />Beyond that, I still need to know if both mass and size are expanding, or just size, and at what rates.

 

[bonzelite]

<font color="yellow"><br />1. Gravity does not exist.</font><br />correct. per this <i>theory,</i> yes. no gravity <i>as we know it.</i> the premise reverses the gravity vector. this premise posits the earth's surface "falling up" as it accelerates. gravity is the result of an acceleration <i>only.</i> <br /><br /><font color="yellow">2. Massive objects (i. e. those with mass) are expanding at a rate with constant acceleration </font><br />correct. the expansion is from any given objects's <i>centre of mass, at rate in relative proportion to other objects dependent upon size.</i><br /><br /><br /><font color="yellow">1. Does mass also expand at the same rate? i. e., does mass and size increase. If mass is constant, the size must increase at the rate described above.</font><br />correct. <br /><br /><font color="yellow">But then the size is also increasing to counteract it? </font><br />yes. conditions maintain proportion.<br /><br /><br /><font color="yellow"><br />2. I still don't quite get how orbits can occur. Do the planets push each other out of the way (from a distance obviously) and thus never collide?</font><br />imagine you have a stone you picked out of the river. you hold it's smooth surface, look skyward, and hurl the rock across the riverbanks to the other side. and you make the arc of the trajectory such that you hurl it quite high up, like 80degrees to vertical, and very hard. it sails up and out over the river. <br /><br />at a point in it's trajectory, it will begin to slow down as it "peaks." now imagine that is the point at which the MRO craft, as it was approaching and then overshot Mars to insert into orbit, begins to turn back around from being flung into space as it was "captured." it turns around. it goes back to Mars. everyone cheers at JPL. <br /><br />MRO, as it began to turn back to Mars upon orbit insertion, slowed a bit, turned back around, and then accelerated again faster and faster, doing TCM burns a

 

[eudoxus18]

Oh my goodness! I've got it, it's been staring me in the face the whole time but I didn't think of it till now. I haven't used my brain in some time I guess.<br /><br />To put it simply your theory requires that all bodies are expanding at the same rate for their relative sizes to remain the same. In other words, in the time the Earth takes to double its size, in that time the Moon must also double in size.<br /><br />BUT the downward acceleration the Earth exerts on objects on its surface is g. On the moon its about g/6. This is an observable fact and would require that the Moon and the Earth are expanding at different rates.<br /><br />In other words, for your theory to make sense, the Earth and Moon must not only expand at the same rate and at the same time expand at different rates.<br /><br />If you want more more mathematical precision than that, I'd be glad to bring it out.<br /><br />Sorry, Bonzelite, it was a good effort but it doesn't look like your theory holds much water.

 

[bonzelite]

rate is simiar to currency devaluing due to inflation at: (currency value)/(1+x) <br /><br />where x = inflation rate. <br /><br /><br />

 

[bonzelite]

<font color="yellow"><br />Oh my goodness! I've got it, it's been staring me in the face the whole time but I didn't think of it till now. I haven't used my brain in some time I guess. <br /><br />To put it simply your theory requires that all bodies are expanding at the same rate for their relative sizes to remain the same. In other words, in the time the Earth takes to double its size, in that time the Moon must also double in size. </font><br />eureka. you got it. <br /><br /><br /><font color="yellow">BUT the downward acceleration the Earth exerts on objects on its surface is g. On the moon its about g/6. This is an observable fact and would require that the Moon and the Earth are expanding at different rates. </font><br />that would <i>appear that way from standard theory and the accepted notion that the gravity is 1/6. but it cannot possibly be that.</i> <br /><br />mass of bodies cannot be determined by current gravitational models for mass. the moon should be 1/4 earth's gravity. but is observed to be 1/6. this is probably because the moon's centre of mass is off-centre. that is, the centre of mass of the moon is not the geometric centre. <br /><br />this has huge implications. this means that the density of the moon is varied from near to far side, and the average readings of 1/6 are way off. when the moon, then expands from it's centre of mass (that being off centre and more than likely located off-center <i>nearer the earth side</i> --thus making the near side <i>denser</i>)--the denser near side at the measured 1/6 (as said to be averaged for the whole moon), is offset by the <i>less dense</i> and "larger" far side at a <i>greater gravity value as that side is expanding more as it is larger, farther away from it's centre of mass, and thus accelerates at a faster rate than the near side.</i> all things expand from their centres of mass. <br /><br />the moon has varying gravity and is not uniformly 1/6 earth gravity. but averaged between near and f

 

[eudoxus18]

What do you mean that the moon's gravity "should be" 1/4? That needs a citation.

 

[bonzelite]

the moon is just a bit over 1/4 the size of earth. it's grav should be, then 1/4 of earth. but it is commonly believed to be <i>less dense</i> than earth at a measured 1/6 g. this gives credence to the cataclysmic foregin impactor idea for the moon's formation that i have never believed. i have always thought the moon to be of the earth, ie, coalesced from the same disk of material. <br /><br />the earth and moon are of the same local material. and under this premise, the moon's gravity at it's surface is not uniform. the far side of the moon should be, and may in time be measured to be, <i>greater than the denser nearside.</i>

 

[eudoxus18]

As far as not being able to find the mass of a planet, well there your just patently wrong. Here's how:<br /><br />1. F = GmM/r^2 (Universal Law of Gravitation, F is force the two objects exert on each other, G is the universal gravitational constant, m is the mass of the object, M is the mass of the planet, and r is the separation between them; if on the surface of a planet r is the radius of the planet)<br /><br />2. So both objects exert equal forces on each other. F = ma too, so ma = GmM/r^2. m cancels, and if you rearrange the equation you get M = gr^2/G. In other words, the mass of a planet is equal to it's gravitational acceleration times the square of its radius divided by the universal Gravitational Constant, which Cavendish measured by observing minute changes in the positions of two suspended objects placed closely together<br /><br />We could have said F = Ma and solved for the mass of the object, but we already know what it is; it's easily measurable.<br /><br />Besides that, I don't think NASA would be so unwise as to launch shuttles out in space and -onto other planets- if they didn't have a very good idea of how the planet's gravity would affect the craft. <br /><br />And I've heard it said that scientists have measured G and often gotten vastly different results. It is true scientists have had a tough time pinpointing G's value, but all of them agree it's about just above 6. Further, some scientists may have done slipshod work. I could measure it with very crude instruments and claim I got 13, but that doesn't mean I'm a good scientist and suddenly scientists can't know G's value very precisely.

 

[eudoxus18]

What do you mean by "size"? Do you mean it's radius is 1/4 earth's or it's volume is 1/4 earth's?

 

[bonzelite]

<font color="yellow"><br /><br />Besides that, I don't think NASA would be so unwise as to launch shuttles out in space and -onto other planets- if they didn't have a very good idea of how the planet's gravity would affect the craft. </font><br />yes. their calculations are <i>near approximations.</i> and they are generally near enough to send craft out. but the moons 1/6 g is merely an <i>approximation, as that body's centre of mass is more than likely off-centre from it's geometric dead-centre, rendering actual gravity for it's surface to <b>vary</b> from region to region.</i>

 

[bonzelite]

moon is physically about 1700km in radius. earth's radius is about 6300km. that makes the moon a bit over 1/4 size of earth.

 

[eudoxus18]

Yes, earth's radius is 4 times the moon's, but you're forgetting that if you quadruple the radius of a sphere you multiply its volume by a factor of 64! Using only the following information I calculated the ratio of earth's gravitational acceleration and the moon's.<br /><br />Radii of earth and moon<br />Density of earth and moon<br /><br />My answer was 6.07. Here's a basic rundown:<br />Let G, M, R, V, and D be the gravitational acceleration, mass, radius, volume, and density of the earth. Let g, m, r, v, and d be the moon's.<br />Let H be the universal gravitational constant in this case.<br /><br />1. We are trying to find the ratio G/g, which should be 6 (even though you say using calculations it's 4).<br />2. According to the Universal Law of Gravitation, G = HM/R^2; further g = Hm/r^2. Since we're trying to find G/g, we divide these two equations, rearrage and get<br />G/g = Mr^2/(mR^2)<br />3. The equation for density is D = M/V (d = m/v). If you input the volume of a sphere equation into, rearrange, and solve for the mass you get<br />M = 4(pi)DR^3/3 (m = 4(pi)dr^3/3).<br />4. If we plug the equations for the masses of the earth and moon into the equation in (2), cancel the 4(pi)/3 and the radii, you get<br />G/g = DR/(dr)<br />I plugged in the values for the densities and radii that I found at<br /><br />http://www.spacegrant.hawaii.edu/class_acts/MoonFacts.html<br /><br />and got 6.07.

 

[eudoxus18]

In other words, the gravitational acceleration of a planet is a function of radius AND density, not radius alone. Sure the ratio of the radii is 4, but this is balance by the densities to make it 6.

 

[bonzelite]

i think your on the right track. and <i>in great likelihood,</i> the moon's density varies as it's centre of mass is located off-centre, positioned within the moon on the <b>near side</b> (more compacted side) facing earth. and the greater accelerating far side (an area more vast and removed from the moon's centre of mass) is <i>less dense.</i> expansion occurs from a body's centre of mass. <br /><br />density plays a key role, as well, in the perceived "gravity" of a body. a less dense body <i>of the same size</i> as a greater dense body will have less "gravity." why? because the expansionary forces <i>within the body itself</i> will be absorbed somewhat by the less-dense, less-rigid, state, as the particles push outward to the surface from it's centre of mass. <br /><br />a more rigid, more dense body, will not so easily absorb the acceleration forces expanding outward. and that body will have greater "gravity," ie, acceleration force, at it's surface. <br /><br /><br /><br />