gravity effects

Then Gm=rv2 must be the right road. Is this true? And if so how is this rule equally applied to large and small bodies when the moon is only moving the Earth according to its mass ratio of 81:1? The equivalance of m and v2 is evident since the other factors are exactly the same quantities.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Jim]: Then Gm=rv2 must be the right road. Is this true?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

No, that is another red herring. It is true under special circumstances only, such as circular orbits in the 2-body problem. But that is an "advanced topic", and should be forgotten for now until we get the basics right.

The only law that matters for 99% of all gravitational phenomena is acceleration = GM/r^2, where M is the mass of the source of gravity and r is the distance from source mass to target body.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>And if so how is this rule equally applied to large and small bodies when the moon is only moving the Earth according to its mass ratio of 81:1?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

When the acceleration law is applied to source mass Earth and target body Moon, we get the Moon's normal orbit around Earth. When it is applied to source mass Moon and target body Earth, we get the Earth's much smaller (by a factor of 81) orbital motion around the Moon. Earth keeps trying to fall toward the Moon, just as the Moon tries to fall toward the Earth. The difference is that the Earth doesn't get very far before the Moon moves on and starts pulling the other way.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>The equivalance of m and v2 is evident since the other factors are exactly the same quantities.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

GM/r = v^2 only for circular orbits, and only when there are just two bodies involved. Forget these complicated advanced matters until we get the basic motions straight. Nothing in this so-called "virial theorem" will help you understand the basics. -|Tom|-

OK I'll do this with a=Gm/r2. a for the moon is 1/81 if Earth a=1. The r2 is the same for both bodies. Gm=ar2. a=v2/r for circular orbits if that is ok with you? so v2/r=Gm/r2 to v2=Gm/r to Gm=rv2. It looks like the same idea to me. (I hope my math is right)

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>OK I'll do this with a=Gm/r2. a for the moon is 1/81 if Earth a=1. The r2 is the same for both bodies. Gm=ar2.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Okay so far.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>a=v2/r for circular orbits if that is ok with you?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

The relation is correct for 2-body circular orbits. But I advise against going there. Stick with formulas that tell you something about the motion. It is too easy to get into paradoxes when you switch from motion equations to energy equations, which is where you appear to be headed.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>so v2/r=Gm/r2 to v2=Gm/r to Gm=rv2. It looks like the same idea to me. (I hope my math is right)<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Yes, the math is right, and so is the relationship, with the previous caveats still in effect. Your second-last form is an energy equation, and no longer contains information about how any single body moves. Instead, it connects the kinetic and potential energies for one body in a circular orbit. The last formula tells you that the product rv^2 is constant for all the planets. But it would be a different constant for the Moon and Earth satellites because they are orbiting a different source mass.

So where do you plan to go with this formula? -|Tom|-

I want to resolve the energy effect of mass in orbital motion it that makes sense. You lost me on the last post when you divided the math relationships into two types-Energy and Motion. I have never run into that before and will need to ponder it some. I guess tidal effects are one or the other and energy is needed to perform either type. Back to the topic-I just did a bit of math with acceleration and mass since the mass of the sun moon and Earth are known and the acceleration is proportional to mass. Now, I know you are not going to agree but I get a very large difference in the force of these three bodies have on each other. And I am now more sure the moon does not move Earth and Earth does not move the sun. Also tides are for sure a gravity effect that is currently modeled wrongly. I say this because you asked where I want to get to with this link not to belabor my point. I'm looking for truth and find some once in a while.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>I want to resolve the energy effect of mass in orbital motion if that makes sense.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

The sentence is not self-explanatory.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>You lost me on the last post when you divided the math relationships into two types-Energy and Motion. I have never run into that before and will need to ponder it some. I guess tidal effects are one or the other and energy is needed to perform either type.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Tidal effects are a type of "non-gravitational force", even though gravity is involved. But we can learn little or nothing about the nature of gravity from tides because they disrupt normal orbital motion.

And I don't know what your interest in energy is. It is an effect of gravity, not a cause.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Back to the topic-I just did a bit of math with acceleration and mass since the mass of the sun moon and Earth are known and the acceleration is proportional to mass. Now, I know you are not going to agree but I get a very large difference in the force of these three bodies have on each other.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

We agreed to leave forces out of the discussion because they are mathematical fictions. Stick with accelerations -- that is what drives orbital motion. There certainly is a large difference in the accelerations produced by Sun, Earth, and Moon on one another. Why wouldn't I agree? Only the accelerations of Earth and Moon produced by the Sun are nearly equal. All other accelerations are very different. The acceleration formula is easy to apply and easily gives those results.

Examples:
If acceleration of Sun caused by Earth is 1,
then acceleration of Earth caused by Moon is 1900,
and acceleration of Moon caused by Earth is 150,000,
and acceleration of Earth or Moon caused by Sun is 330,000.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>And I am now more sure the moon does not move Earth and Earth does not move the sun.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Regarding logical consistency, you just jumped off a cliff. <img src=icon_smile_sad.gif border=0 align=middle> If each body applies an acceleration to each other body, that means all the bodies move by definition of the word "acceleration". How did our discussion derail so badly, out of the blue?

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Also tides are for sure a gravity effect that is currently modeled wrongly.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

From your descriptions, I have to doubt that you know what a "tidal effect" really is. But let's get gravity straight first, and leave more complicated matters for chapter two. -|Tom|-