Gravity Doesn't Exist

[derekmcd]

<font color="orange">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.</font><br /><br />They overlook statements like this because they make absolutely no sense. First, gravity has nothing to do with 'speed'. It's acceleration... 2 completely different things. The sensation of gravity on a ship has absolutely nothing to due with it's mass... That is centrifugal force.<br /><br /><font color="orange">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.</font><br /><br />Whaa?? 1 'g' force here on earth is only 9.81 m/s^2.<br /><br /><font color="orange">. The only aspect of gravity that is relative to mass is the composition of the mass</font><br /><br />Where do you get these ideas from? Mass is merely the amount of matter in a finite space. What it is made of is of no consequence.<br /><br /><font color="orange">If you take any amount of mass, with any composition and stop it from moving in anyway it will have 0 gravity.....Period! </font><br /><br />Absolutely not true. If the earth stopped spinning, i would likely weight more (though very insignificant). The spinning of the earth actually wants to throw you off due to the lack of centripital force. Gravity keeps you here.<br /><br />I'm not even going to bother with the moon and tides. <br /><br />I, certainly, hope some 6th grader <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>

 

[bonzelite]

let me start again, saying it in another way. <br /><br />the Lagrange point of the earth/moon is inside the earth. the earth/moon system revolves around this point. this point is off-center of the earth's literal central axis of daily spin, creating a wobble of the earth. the Lagrange point of earth/moon is the center of mass between the bodies. <br /><br />i will stop there so that i am clear, and that you hear that part. are we in agreement on what i just said, before i continue?

 

[derekmcd]

I believe you are talking about the Barycenter and Not Lagrange point, however, I'm still with ya <img src="/images/icons/smile.gif" /><br /> <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>

 

[bonzelite]

ok. i am not completely hip to correct terms. please accept my apology for that. it's where the masses of the two bodies are balanced. i thought that was a Lagrange point. but you follow me so far, yes? <br /><br />

 

[bonzelite]

continuing with my point slowly....<br /><br />if, for example only, the center of mass between the earth and the moon were exactly between the bodies, then they would actually orbit each other. is this correct or not correct? <br /><br />in other words, they'd be like two balls placed exactly equally across from each other on a spinning merry-go-round, right? <br /><br />i'm going to make a point in a bit, but i'm asking some questions right now. it actually helps me to understand better, as well <img src="/images/icons/wink.gif" />

 

[derekmcd]

Exactly correct. <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>

 

[bonzelite]

ok. cool. neat. <br /><br />ok. so that the barycenter is <i>NOT between the earth/moon exactly at center,</i> and is instead WITHIN the earth itself, this center of mass point within the earth creates a wobble in the earths motion. <br /><br />in other words, instead of the "wobble" being an exact center point of orbit of the other body, as two balls equidistant on a merry-go-round would be encircling each other exactly around a center point, this "center point" is constrained, if you will, to being <b>within</b> one of the bodys, ie, the earth. therefore, the merry-go-round is very "lopsided," as the earth/moon are not a perfectly centered and balanced merry-go-round. <br /><br />do you accept this?

 

[derekmcd]

Accepted. I understand how the moon and earth interact. <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>

 

[bonzelite]

i know you do. but others may not. and i am trying to be clear so that i do not post difficult to follow things. that's all i mean by this pedantic explaining style <img src="/images/icons/wink.gif" /><br /><br />you're probably way smarter than me, by the way. i'm not a scientist or engineer.

 

[derekmcd]

And considering you accept the premise of how the barycenter creates a 'wobble', then you must come to the conclusion that mass affects gravity. <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>

 

[derekmcd]

"i'm not a scientist or engineer."<br /><br />Nor am I. <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>

 

[bonzelite]

moreover, do you accept that the sloshing bulges we observe on opposite sides of the world; the opposite-sided high tides, are in exact alignment with the moon?<br /><br />and are created by the center of mass' wobble within the earth being in constant and static alignment with the moon?

 

[bonzelite]

<font color="yellow"><br />And considering you accept the premise of how the barycenter creates a 'wobble', then you must come to the conclusion that mass affects gravity.</font><br /><br />in a manner of speaking, yes. but i want to first let you read and react to my last post prior to this one.

 

[derekmcd]

Yes and no. As I have stated before. The side opposite the moon is due to centrifugal force and the bulge facing the moon is due to the moon's pull of gravity. The center of mass wobble can not create a bulge on both sides simultaneously. <br /><br />Edit: I added "and no"... as I don't believe both high tides are due to "sloshing" <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>

 

[bonzelite]

<font color="yellow"><br />Yes and no. As I have stated before. The side opposite the moon is due to centrifugal force</font><br /><br />centrifugal force of....(your answer) <br />

 

[derekmcd]

Exactly my answer. As I asked before... How do you account for the simultaneous high tide closest to the moon if it not related to mass as you have stated before. <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>

 

[bonzelite]

i will address that as soon as you complete my fill in the blank.<br /><br />centrifugal force of... (your answer)

 

[derekmcd]

There is no answer. Centrifugal force is what it is. The water wants to continue on it's natural path... hence the bulge.<br /> <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>

 

[bonzelite]

you answered this way:<br /><font color="yellow"><br />Yes and no. As I have stated before. The side opposite the moon is due to centrifugal force </font><br /><br />and i'd like you to just clarify what you did agree upon. i have been very clear so far in my explaining. i'd like to remain on the same page and on track with that. please complete, the bulge on the side opposite the moon is due to:<br /><br />centrifugal force of....(your answer)

 

[derekmcd]

There is no 'of'. Centrifugal force is an effect. The water is bulging due to centrifugal force, period. How about this. The centripital force OF the earth and moon orbiting each other creates a centrifugal force upon the water thus creating a bulge. <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>

 

[bonzelite]

nice. ok. <br /><br /><font color="yellow">The centripital force OF the earth and moon orbiting each other creates a centrifugal force upon the water thus creating a bulge.</font><br /><br />agreed. due to the barycenter "wobbling" that is our lopsided merry-go-round. you accept this, yes? <br /><br />now. you accept that the bulges we observe on opposite sides of the world; the opposite-sided high tides, are in exact alignment with the moon?

 

[derekmcd]

<font color="orange">agreed. due to the barycenter "wobbling" that is our lopsided merry-go-round. you accept this, yes?</font><br /><br />As I have stated before... yes.<br /><br /><font color="orange">you accept that the bulges we observe on opposite sides of the world; the opposite-sided high tides, are in exact alignment with the moon?</font><br /><br />Well... there is tidal lag, so it is not exact. That's another thread. For the sake of this thread and my contention that gravity is also a direct result of mass, then, yes. <img src="/images/icons/smile.gif" /><br /> <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>

 

[bonzelite]

ok. nothing is exact. <br /><br />we can continue this tomorrow. i'm too tired. see you later <img src="/images/icons/wink.gif" />

 

[siarad]

That's a long answer to me but as I said there are 1, 2 or 4 tides produced by the Moon. There's also the 0 tide too, the Mediterranean has neither Sun or Moon tides & the one locked to the Sun thus occurring at the <i>same time</i> each day. <br />You seem to be equating inertia with gravity, spinning or speed increases inertia but not gravity.<br />Do you have a link whereby these gravitational effects have been <i>measured.</i><br />I still don't understand how gravity being an acceleration, produced as you say, doesn't come from an asymptotic system.<br />It is true the <i>effect</i> of gravity isn't simply due to mass but also density, squeezing the Earth to 0.5cm? produces a black hole able to capture light without increasing it's gravity. I'm not a scientist so that figure may not be correct but an example.

 

[why06]

What Are You talking about!!!! Are you saying all matter is one charge? If that's the case than how do you explain a static shock. any way charge only deals with a lack or excess of electrons!! <div class="Discussion_UserSignature"> <div>________________________________________ <br /></div><div><ul><li><font color="#008000"><em>your move...</em></font></li></ul></div> </div>