- Thank you received: 0
gravity effects
21 years 11 months ago #4050
by Jim
Replied by Jim on topic Reply from
There are no cliffs here-you asked me where I want this to go and I told you. My views and yours are quite different and it is not a problem for me. The model as you have so far described is good as are the several other models I'm aware of. This one assumes acceleration is a primary factor I think you say and that is ok by me. I look forward to the day when force, energy and the tidal effect are factored in. It is good to learn that tides are a gravity effect and that must mean energy is part of the overall modeling. So once I have the other part understood I'll be ready for that.
Please Log in or Create an account to join the conversation.
21 years 11 months ago #4051
by Jim
Replied by Jim on topic Reply from
A correction- tides are a non-gravitional effect that force orbital problems to occur.
Please Log in or Create an account to join the conversation.
- tvanflandern
- Offline
- Platinum Member
Less
More
- Thank you received: 0
21 years 11 months ago #4618
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>I look forward to the day when force, energy and the tidal effect are factored in.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Force is not going to appear in gravitation for macroscopic bodies. Force occurs only at a sub-sub-quantum level for individual gravitons operating on individual matter ingredients. See the book <i>Pushing Gravity</i> for a complete model of how gravity works. Macroscopic bodies simply accelerate. "Forces" for macroscopic bodies would imply that inertia existed for these, which it does not. In gravitation, all bodies, large and small, accelerate at the same rate in a given gravitational field. No bodies resist acceleration, so there is no inertia. {An article in the current MRB explains this.]
Energy is an effect of gravity, but not a cause. A body falling in a gravitational field is changing potential energy into kinetic energy; and vice versa for bodies rising in the field.
Tidal effects are non-gravitational. They come in longitudinal, radial, and latitudinal flavors. These operate differently for solid, liquid, and gaseous bodies. In all cases, friction must be operating for tides to occur. There is a nice discussion of theory and applications in my book, <i>Dark Matter, Missing Planets and New Comets</i>. But since these effects have so little to do with gravity, why do you care? About the only local application you are likely to run into is the tidal evolution of the Moon's orbit. But that is only a few centimeters each year -- significant only over millions or billions of years.
I stress again that these are advanced concepts for someone still struggling to understand Newtonian gravitation. You need a firm foundation before attempting to build on it. -|Tom|-
Force is not going to appear in gravitation for macroscopic bodies. Force occurs only at a sub-sub-quantum level for individual gravitons operating on individual matter ingredients. See the book <i>Pushing Gravity</i> for a complete model of how gravity works. Macroscopic bodies simply accelerate. "Forces" for macroscopic bodies would imply that inertia existed for these, which it does not. In gravitation, all bodies, large and small, accelerate at the same rate in a given gravitational field. No bodies resist acceleration, so there is no inertia. {An article in the current MRB explains this.]
Energy is an effect of gravity, but not a cause. A body falling in a gravitational field is changing potential energy into kinetic energy; and vice versa for bodies rising in the field.
Tidal effects are non-gravitational. They come in longitudinal, radial, and latitudinal flavors. These operate differently for solid, liquid, and gaseous bodies. In all cases, friction must be operating for tides to occur. There is a nice discussion of theory and applications in my book, <i>Dark Matter, Missing Planets and New Comets</i>. But since these effects have so little to do with gravity, why do you care? About the only local application you are likely to run into is the tidal evolution of the Moon's orbit. But that is only a few centimeters each year -- significant only over millions or billions of years.
I stress again that these are advanced concepts for someone still struggling to understand Newtonian gravitation. You need a firm foundation before attempting to build on it. -|Tom|-
Please Log in or Create an account to join the conversation.
21 years 11 months ago #4053
by Jim
Replied by Jim on topic Reply from
So, the model requires no energy? And nothing resists acceleration? These are the two things that make the model work? I'm struggling to understand a model-you have no lock on understanding real events.
Please Log in or Create an account to join the conversation.
- tvanflandern
- Offline
- Platinum Member
Less
More
- Thank you received: 0
21 years 11 months ago #4427
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>So, the model requires no energy? And nothing resists acceleration?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
For gravitation, both of those statements are true at a macroscopic level. All you need is the acceleration formula. All of orbital dynamics flow from that one formula until you get to relativity effects, which are very, very small.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>These are the two things that make the model work?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Only the acceleration formula "makes the model work". The absence of energy in the formula and the absence of inertia are aids to understanding, not to operation. People who introduce those concepts into gravitation are often led into paradoxes. I'm trying to simplify things so you can see the essence of the standard model (not "my model").
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>I'm struggling to understand a model-you have no lock on understanding real events.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
I never claimed to have any "lock". But any teacher expects his/her students to understand the existing standard model or models before critiquing them and branching out. Is that reasonable enough for you? -|Tom|-
For gravitation, both of those statements are true at a macroscopic level. All you need is the acceleration formula. All of orbital dynamics flow from that one formula until you get to relativity effects, which are very, very small.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>These are the two things that make the model work?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Only the acceleration formula "makes the model work". The absence of energy in the formula and the absence of inertia are aids to understanding, not to operation. People who introduce those concepts into gravitation are often led into paradoxes. I'm trying to simplify things so you can see the essence of the standard model (not "my model").
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>I'm struggling to understand a model-you have no lock on understanding real events.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
I never claimed to have any "lock". But any teacher expects his/her students to understand the existing standard model or models before critiquing them and branching out. Is that reasonable enough for you? -|Tom|-
Please Log in or Create an account to join the conversation.
21 years 11 months ago #4750
by Jim
Replied by Jim on topic Reply from
This is very reasonable as long as we are chatting about MODELS-the standard model now is no better or worse to me than the flat world model of a 1,000 yaers ago. You have posted a few items that are new to me and I want to have a good understanding these details. Even so, I would rather have data that has been measured since as I said about data posted by NASA is generated from models and not at all measured-even so, it is very good data I'm not questioning that point.
Please Log in or Create an account to join the conversation.
Time to create page: 0.453 seconds