Paul, you realy dont understand MS at all... Im affraid we have got to master some fundimentals first....... I attached my diagram here again so no one has to go looking for it.... Points - 1. Inertia is not a force. Inertia is a resistance to the grav feild of distance stars, thus inertia is just a reaction to grav feilds thus inetia is the reation to the force of grav....grav is inertia and accelerations in MS ........I dont know of anyone who holds to GTR in MS who would even argue this fact....?! 2. Your two descriptions external of the graphic end with '?' I never know with you -- are these questions or statements????? im asking a question whos answer is both obvious and rediculous...it is the implied conclusion of a given argument that is obviously inconsistant.. .......also note: in the last bottom explaintion the significance of all that is the earths accelerations are not constant... 3. Your statements headed "MS states ..." Come on Allen! If you pull the other one it plays jingle bells! First -- the Sun pulls us down -- I can live with that. Second -- distant stars pull us up -- horse feathers. it most certainly does other wise there would be nothing to keep it from falling downward by the suns grav force in MS ..?!Third -- distant stars create inertia -- horse feathers. Inertia is a resistance to the grav feild of distance stars...it most certainly is in MS!? Inertia is a property of mass only wrt chagnes in inertial/grav feilds.....-- it is the resistance to a change in velocity. wrt all the grav/inerita feilds in the universe.... The role of the distant stars -- are you saying that MS states that in a two body universe one body cannot orbit another body? yes you conclusion is right on the money.....now go find someone in MS who would debate that...you wont! however note: im just using what MS does claim wrt to what we observe so as to show its problmes. Not what i can imagine. Yes your 2 body universe has that very problem......... and MS has no solution to it except that there is no 2 body universes to worry about so they get to "skate" by on that one..... coz it is only a imaginary experiment with no way to falsify it..........which realy is not science is it...ummmmm ..........I'm trying to focus you all on: 1 Attention on the term "free fall" itself and what that means exactly wrt a orbit vrs a stright line path...... 2. Even taking the postion that ...."not until the "extended object" effect begins to dominate"...... how do we expalin the fact that the tides are obviously accelerated differently wrt the earth...what is the extended object here the earth or the oceans both bulge? And if we are at a point where the extended object dominates as eximplified by the tides then we are back to where we started from....a detectable accelertion in free fall...because our orbit is at the majic spot where some extended object has begun to dominate......and thus the question beggs if we are at that majic spot then why can't we detect the earth's other changes wrt its inertial feilds..../ accelerations....??? Look at the diagram aggian if we are at the majic spot to detect the variation in V1 as shown in the tides then why not changes in V2/..it is the exact same force?.... 3. Notice the inertial argument in the diagram for a arc verse a circut/orbit............ ----- Original Message ---- From: Paul Deema <paul_deema@xxxxxxxxxxx> To: geocentrism@xxxxxxxxxxxxx Sent: Wednesday, May 7, 2008 5:06:19 PM Subject: [geocentrism] Re: Tides Allen D I see an object in free fall as being a body in motion in a straight line, in an orbit of any possible shape, or an undefined path through the universe being deflected randomly by gravity fields but having no ability to change its own speed or direction. Do you argue against that? Now, to being able to detect the change in velocity of a body in a non circular orbit ie an orbit where the velocity does actually change -- let's choose elliptical with the maximum sustainable eccentricity. I don't know how you define that but I want a very long, very skinny ellipse. A body in this orbit will have the maximum delta v. Now let us put a very very very long skinny space ship in that orbit, longitudinally aligned tangential to the orbit at the periapsis, having a prograde rotation in the plane of the orbit and having a sidereal period of one of its orbit's years. I'll admit that the acceleration measured at the front of the vehicle will be different from the acceleration at the rear of the vehicle EXCEPT when its longitudinal axis is tangential to its orbit at periapsis and apoapsis. I'm unable to determine at what point in the orbit the greatest difference will be measured. However, if the accelerometer is placed at the centre of gravity of the very very very long skinny spaceship -- or indeed if the vessel is of any shape right through to and including a sphere, the indicated acceleration will always be zero. Do you argue against that? Note that there is an admission contained here. On this basis, you should be able to determine what the relative positions of the red sphere and the green sphere will be as they pass perihelion. On this basis also, with your proficiency at mathematics, you should be able to calculate the expected changes in acceleration of the Earth in its slightly elliptical orbit about the Sun, though I suspect that compensating for the acceleration due to Earth's rotation will keep you busy, as will compensating for that rotation being inclined to the orbit. Now to the contents of this post. Actually the only points that I think I follow are in your illustration. Points - 1. Inertia is not a force. 2. Your two descriptions external of the graphic end with '?' I never know with you -- are these questions or statements????? 3. Your statements headed "MS states ..." Come on Allen! If you pull the other one it plays jingle bells! First -- the Sun pulls us down -- I can live with that. Second -- distant stars pull us up -- horse feathers. Third -- distant stars create inertia -- horse feathers. Inertia is a property of mass -- it is the resistance to a change in velocity. The role of the distant stars -- are you saying that MS states that in a two body universe one body cannot orbit another body? Paul D ----- Original Message ---- From: Allen Daves <allendaves@xxxxxxxxxxxxxx> To: geocentrism@xxxxxxxxxxxxx Sent: Wednesday, 7 May, 2008 9:43:18 PM Subject: [geocentrism] Re: Tides ok ....see attached diagram...... A. HOw can i be pulled away differently then the earth in a "free fall"?...I thought grav in f a "freefall "pulls centers of mass at the same rate?.....if that is true about a person on the planet then why is it not true for the accelerometer.....well could say the accelerometer is at the center of mass...ok....so then what we are saying is all we have to do to detect our acceleration is move the mass or person in our accelerometer/ planet offset wrt the center of totlal mass by puting it on the sun side (side of the direction of our "free fall") ?.......ummmmm that is not very sporting of the sun if it's gravity is pulling all centers of mass at the same rate wrt each other....where is the additional energy comming from.....ummmm....perhaps we are saying the further we fall the larger the deviation between any two centers of masses are because of the rate is determined by the square of the distance............. in a orbit the square of that distance changes by how much?.......oh the tides thats right.....so can we or can we not detect the acceleration of the earth in "freefall"...? B. If there was no sun would we weigh anything at all?..we would say yes....because the earth has it's own grav feild..ok... so how does the sun negate the earth's grav in free fall?..even more to the point if it can why does it not do that when we are on the sun side during the day?. if it accelerates the center of mass of the oceans then why not the center of my mass ? ....why not?... We are said to "freefall" toward the sun even now! ( on the sun side of the earth) But we dont appear to fly off toward the sun....why not, if we are "freefalling" and the sun's grav feild would negate the pull of the earth's grav feild?.if it does not negate/overcome the pull of the earth's feild then how would we fall away toward the sun?!......If the pull is equal from the sun and the earth then how can we weigh anything....if the earths pull is greater at our location then the sun's then how do the tides get accelerated differently then we do by the sun?...why does our weight not fluctuate with the tides with the same magnitude of force that accelerates all that sea water as the sun passes over us.... C. on the night side of that earth, there is the difference between a person on that earth "freefalling" to the center of the sun's center of mass and a person on the elevator "freefalling" to the earth's center of mass ....? is the earth's grav over power the suns...then how come there are tides on the night side as well?....what is accelerating them in a new moon where the sun and moon are on the same side wrt the earth.... Summary: The earth still has a gravity that attracts mass with or without the sun.....There is no way to negate the earth's pull on person by claiming a "freefalling" toward the sun even if we say the earths is stronger we still have a detectable acceleration wrt the sun in free fall the tides!?.....or more precisely the whole idea of "free falling" to the sun is not really "freefalling" in the same way that a apple falling toward the earth. The falling apple is not the same as a satellite in orbit...using the term "freefall" is just a free and loose use of terminology/cop out for the fact that those two circumstances are not even remotely the same and yet they are explained away the exact same way....? Neville Jones <njones@xxxxxxxxx> wrote: -----Original Message----- From: allendaves@xxxxxxxxxxxxxx Sent: Wed, 7 May 2008 08:14:47 -0700 (PDT) ... I will ask a question to everyone.....if i am on a planet of one earth mass that is falling toward the sun will i have any weight and how much will i weigh and why?..... Your weight is a force and is therefore given by Newton's F=ma, where m is your mass (in kg for example) and a is the acceleration (m/s/s) produced by this force. There will be a force acting on you directed towards the centre of the World, where the acceleration resulting from this force at the World's surface is g m/s/s. And there will be a force acting on you and directed towards the centre of mass of the Sun, where the acceleration due to this at the World's surface is g' m/s/s. These will be acting in opposite senses. Your weight, W, will therefore be W = mg - mg' = m(g - g') N and will act towards the centre of the World until such time (since the World is in 'free fall' towards the Sun) as g' = g, when you will become weightless. Then, as g' > g, you will be pulled off the World and fall into the Sun before the World does. I.e., your weight would be constantly changing. The above assumes that you are positioned on the same side of the World as the Sun and that you are not an 'extended' object, at least not in relation to the strength (size) of the gravitational field of either the World or the Sun. Neville. ________________________________ Receive Notifications of Incoming Messages Easily monitor multiple email accounts & access them with a click. 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