Gravity Doesn't Exist

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TheShadow

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MOD claims: <i> So by that logic, the Sun would produce 1,000's of times more gravity than say, the Earth. </i><br /><br />The gravity on the surface of the Sun at the equator (273.95 ms<sup />2</sup>) is 27.9 times that on the surface of the Earth at the equator. That is because, although the Sun is 332,950 times more massive than the Earth, the Sun is 109 times the diameter of the Earth, and the force of gravity is inversely proportional to the square of the distance from the center of mass of an object. For our Sun, that most of that mass lies within about 0.7 radii of the center. When you talk about the force of gravity, you must reference it to the distance from the center of mass of the object.<br /><br />MOD claims: <i> A gravitational field that strong would eventually suck everything into it (like a black hole). It would constantly draw orbits of planets, comets, etc. closer to the sun. </i><br /><br />Not exactly. As long as the planets, asteroids, and other bodies remain in orbit around the Sun, they will not fall in. In a perfect vacuum, the planets would continue to orbit the Sun forever, if the Sun did not lose mass. Two things are at work here. The Sun loses mass continually through the Solar Wind. Therefore the gravity of the Sun will continue to get weaker. The effect of the Solar Wind is that it puts a constant steam of particles in the path of the planets and other orbiting bodies. This creates some resistance which tends to slow the orbits of these bodies. These two contradicting forces currently balances out. However, as the Sun ages, and is transformed into a red giant, that will all change.<br /><br />I was giving a serious response, then I saw your comments about people being upside down and sideways. Until then I didn’t realize this was just a stupid joke.<br /> <div class="Discussion_UserSignature"> <p> </p><p> </p><p> </p><p> </p><p><font size="1" color="#808080">Who knows what evil lurks in the hearts of men, the Shadow knows. </font></p> </div>
 
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thepiper

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<font color="orange">Explain how the moon affects tides if it is due to acceleration and not mass. </font><br /><br />Why do you believe the moon is what affects the tides? How does it somehow cause a high tide on both the near and far side of the Earth through gravity alone?<br /><br />According to numbers obtained from a page on Wikipedia we get the following:<br /><br />the Sun's mass = 333,000 Earths<br />the Moon's mass = 0.0123 Earths<br />and so Earth mass = 81.3 Moons<br /><br />The Moon is at an average distance of 384,400 km from Earth.<br />The Sun is at an average distance of 150,000,000 km from Earth.<br /><br />The sun is thus 27,072,900 times the mass of the Moon, which means it would have just as strong a <b>gravitational</b> effect on Earth as the Moon does if it is 27 million times farther away.<br /><br />Yet the Sun is a mere 390,625 times further away from Earth than the Moon is. If the gravitation-only model was correct, the Sun should then exert 69 times the gravitational effect on the Earth's tides that the Moon does. <br /><br />Even Jupiter, which has 318 times the mass of the Earth (25,850 times the mass of the Moon) and is only 1530 times farther away than then Moon, should have a stronger gravitational effect on our planet. <br /><br />The way I see it, the Earth is moving through space and is spinning on its axis. The tides are the "sloshing" caused by this movement. The direct gravitational effect of the Moon or even the Sun is minimal.
 
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bonzelite

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<font color="yellow"><br />Explain how the moon affects tides if it is due to acceleration and not mass.</font><br /><br />that is mass-related, but not due to overall mass of the moon. it is the <i>center of mass-distribution between the earth and the moon.</i> the center of this mass, between the double-planet, is actually within the earth at an off-center point to earth's rotational axis. this creates a wobble, monthly, as the moon orbits the earth. oceans are therefore flung out over the axis of this wobble --that which intersects with the earth/moon system, acting as one "body." <br /><br />in other words, the moon and the earth <i>both</i> act to create the tides as opposed to the idea that the moon "tugs" on the oceans. it's not really doing that. it is the monthy periodic wobble-axis of the <i>earth</i> that flings the oceans out and back, ebbing and flowing.
 
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thepiper

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<font color="orange">The fact that dinosaurs were humungous (muscular), that would support a increased gravity, in my opinion, since the solar system was more massive 60-million years ago, or was it?</font><br /><br />I don't think the solar system was more massive, though it may have had more/less planets.<br /><br />A lower gravity would allow creatures to grow to humongous proportions and still be able to move around. Today an elephant is near the upper limit of what can walk around, and the heaviest bird that can still fly is about 65lbs (some pterosaurs had wingspans in excess of 33 feet).<br /><br />There are also cave paintings which show Mammoths in full running flight, while elephants can barely attain a trotting gait today.
 
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TheShadow

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The effect of gravity decreases as the <i>square</i> of the distance from the object. Recalculate all your figures to take that into account. <div class="Discussion_UserSignature"> <p> </p><p> </p><p> </p><p> </p><p><font size="1" color="#808080">Who knows what evil lurks in the hearts of men, the Shadow knows. </font></p> </div>
 
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derekmcd

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My next question was going to be about the wobble of stars and how we determine if a planet is orbiting it. You've already hit on that subject. So what you are saying is mass IS directly related to gravity. As for tides, they are daily and can be precisely predicted based on the moons location. <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>
 
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thepiper

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My mistake, which means gravitational effects are even more minimal for all concerned.<br /><br />How does the conventional view explain this lack of the Sun's gravitational effect on the Earth with its keeping the Earth and the outer planets in orbit around it using only gravity?
 
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bonzelite

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<font color="orange"><br />My next question was going to be about the wobble of stars and how we determine if a planet is orbiting it. You've already hit on that subject. So what you are saying is mass IS directly related to gravity. As for tides, they are daily and can be precisely predicted based on the moons location.</font><br /><br />we're almost on the same page. mass is directly related to acceleration, ie, the nature of the acceleration <i>effect.</i> the acceleration effect on people is that they are stuck to the earth, mass of the earth negligible in this case (size of the earth and it's subsequent relative expansion is what matters). the acceleration effect upon a center of mass akin to the earth/moon system, that with oceans, is that the earth's spin, the earth's acceleration outwards, the moon's acceleration outwards, and the center of mass between the earth and the moon, as they both accelerate to each other, results in oceanic tides. <br /><br />gravity does not enter the picture.
 
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derekmcd

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We can't almost be on the same page... your last post made absolutely no sense to me. I'm trying to "think outside the box". Either I lack the ability, or.... <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>
 
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kmarinas86

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<font color="yellow">The way I see it, the Earth is moving through space and is spinning on its axis. The tides are the "sloshing" caused by this movement. The direct gravitational effect of the Moon or even the Sun is minimal.</font><br /><br />Actually, the moon orbits the earth, causing the tides to change. The sun does not orbit the earth, and the acceleration due the sun's gravity on the earth's "sphere" of oceans is relatively uniform, meaning that these oceans are essentially orbiting the sun. On the other hand, these same oceans are not orbiting the moon. Any pull of the Sun due to gravity causes a seasonal effect on the tides on Earth.<br /><br />http://www.gateway-to-the-universe.org/brent/tourist/ptides.htm
 
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bonzelite

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there is no pulling. gravity does not pull in this setup. it is created as an effect only due to perpetual and constant acceleration of expanding bodies. relatively small objects sitting upon the earth, when dropped, are free-floating as the earth slams up to meet these objects, regardless of their masses (to an extent). when an object of considerable size, such as the moon, has a relative expansion that is large enough, it will orbit the larger body around a center of mass. i believe this is called a Lagrange point. the earth/moon Lagrange point is within the earth. this point creates a wobble within the earth, distinct and separate from the center of it's axis. this perpetual wobbling creates the tides. <br /><br />again, earth/moon spin around a center of mass that is within the earth itself, off-center to the actual daily axial spin. they are also mutually accelerating outward, constantly. this total action "sloshes" the oceans, as a prior poster suggested.
 
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jatslo

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My thinking outside of the box implies that we take everything we know, and reverse it. For example, if gravity is like a vacuum, then bonzelite's interpretation, would imply that gravity is like an inflationary vacuum, I think? Bonze is just changing direction, if that makes since. Where does +(F = ma) fit; -(F=ma) ?!?!
 
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bonzelite

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^^^you're starting to nearly exactly hit on it. yes. in a manner speaking, reverse what is considered to be "gravity." and, instead, per einstein's elevator experiment, replace "g" with "a," as in d=1/2at<sup>2</sup> ---and the result is exactly the same, minus the unwieldy and un-necessary concept of "gravity" that has myriad problems and contradictions associated with it. <br /><br />"gravity" can be purely geometric, as it was to begin with before Newton came up with the un-needed concept of an endemic force possessed by all objects of mass. there is only free-floating conditions in a vacuum, or there are accelerating/expanding bodies relative to each other. gravity is no longer necessary.
 
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jatslo

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Yeah, I get you, I think; the Sun is pushing as it inflates, and this inflation is accelerating with the rest of the universe, right? You just flip flopped everything is all. However, you are accounting for space-time as a substance, and if it isn't, your hypothesis is doomed. <img src="/images/icons/wink.gif" /><br /><br />Like I said, I follow the impression that space-time is nothing, or not a fabric, and treatments in that regard are only measuring the fabric *IN* space.
 
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jatslo

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The space-time that is my monitor is finite, so space-time is finite, and space-time is infinite.
 
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derekmcd

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<font color="orange">again, earth/moon spin around a center of mass that is within the earth itself, off-center to the actual daily axial spin. they are also mutually accelerating outward, constantly. this total action "sloshes" the oceans, as a prior poster suggested.</font><br /><br />The sloshing idea is close. The centrifugal force as the earth and moon orbit each other create a high tide on the exact opposite side of the moon. With that said, there is also a high tide in the direction of the moon meaning 2 high tides per day. How would you account for that? <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>
 
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bonzelite

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<font color="orange"><br />The sloshing idea is close. The centrifugal force as the earth and moon orbit each other create a high tide on the exact opposite side of the moon. With that said, there is also a high tide in the direction of the moon meaning 2 high tides per day. How would you account for that?</font><br /><br />yes. it is the centrifugal force (or centripetal?) of the spinning (and wobbling) earth creating the sloshing bulges we observe on opposite sides of the world. the opposite-sided high tides, in exact alignment with the moon, are created by the center of mass' wobble within the earth being in <i>constant and static alignment with the moon,</i> as it faces the earth constantly in a tidal lock, literally coinciding with the tidal bulges geometrically.
 
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derekmcd

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Centrifugal (or lack of centripetal) force acting upon the earth in no way will create a bulge on both sides. The tidal lock of the moon is a result of the rotational energy dissapating... actually has nothing to do with the creation of tides on a daily basis. Tidal lock is an effect, not a cause. Your explanation was really hard to follow. <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>
 
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siarad

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There are 1, 2 or 4 tides a day depending on place & some lock to the sun not moon.
 
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siarad

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I don't understand whether you mean the earth is <i>actually</i> expanding but as gravity, or whatever you say it is, is an <b>acceleration</b> then the expansion is exponential so how long before the earth is infinite in size.<br />d = at<sup>2</sup>/2
 
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mars_or_die

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Okay, Im gonna explain somethings to you guys that I left out when I started this post. I was hoping that someone else would FIGURE OUT the REST OF THE STORY, but sadly no one has.<br /><br />So here goes:<br /><br />The EFFECT we call GRAVITY is observed because of a.) an object's mass, b.) the objects accelleration through space and c.) the type of accelleration the object is exhibiting. Now I will explain this.<br /><br />If you take a massive comet or asteroid that has no spin, but it is traveling at .5c it will exhibit a very minute amount of gravity. That amount of gravity is directly relative to the mass of the object and the speed it is traveling. However, you can take a very low mass object (like a 10 ton space ship) and send it thru the solar system at .05c with a rotation of 1 rotation per hour and get a measurable gravity close to 1g. This is a proven scientific fact which for some darn reason main stream scientist conveniently over look.<br /><br />Now what this proves it that a.) Gravity is a result of accelleration and b.) different types of accelleration produce different levels of gravity. In order to produce 1g of gravity (regardless of the mass of the object) you would have to either accellerate the object forward at incredible speeds (probably greater than c) and/or accellerate the object forward with spin. The only aspect of gravity that is relative to mass is the composition of the mass. The composition of the mass determines how much accelleration and/or spin is needed to produce 1g. A mass of 1 trillion tons that is comprised entirely of non-metallic materials like paper would require drastically more accelleration and spin to produce 1g than an object of mass of 1 billion tons that is comprised of highly charged materials like molten metal. Additionally an object that is spinning with a core that is also spinning in the same direction as the object around it will require even less accelleration and spin to produce 1g. However, it still boils down to
 
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derekmcd

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<font color="orange">As I stated above the moon rotates on its axis 1 time every 2 Earth days.</font><br /><br />Huh? The moon takes 27.3 days to orbit the earth... hence it rotates on its axis once every 27.3 days. Not sure if I even want to pick apart the rest of this post. <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>
 
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yevaud

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***/Oh the Hell with it/*** <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>
 
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