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Big Bang and Alternatives
19 years 3 months ago #14265
by Cindy
Replied by Cindy on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">In a universe of infinite extent (most of us here take this as a given) there will be equal amounts of mass in all directions so the net gravitational force on a particular galaxy is zero. Except for the occasional chance collision with another galaxy.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
We know that in the univer, there are zones in which density of galaxies is very high. Now we consider galaxies in such a zone. Are these galaxies getting closer under the effect of gvavity ?
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">But most of us here lean toward Meta Model's concept of a limited range for the force of gravity.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Can you tell me where the concept comes from ? Other words, What makes you think the range for the force of gravity is limited ? And, within the concept, what is the fomula of the force of gravity ?
We know that in the univer, there are zones in which density of galaxies is very high. Now we consider galaxies in such a zone. Are these galaxies getting closer under the effect of gvavity ?
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">But most of us here lean toward Meta Model's concept of a limited range for the force of gravity.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Can you tell me where the concept comes from ? Other words, What makes you think the range for the force of gravity is limited ? And, within the concept, what is the fomula of the force of gravity ?
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- Larry Burford
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19 years 3 months ago #14223
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
You will need to do some homework to come up to speed on the latest thinking about gravity. In addition to a dozen or so gravity related articles available via the page tabs mentioned before there are two books you need to read: Dark Matter, <i>New Comets and Missing Planets</i> by Tom Van Flandern and <i>Pushing Gravity</i> edited by Matthew Edwards (containing an article by TVF). Both are available via this Website's store or from places like Amazon.com. They are also available at many libraries.
But the really short version goes something like this:
===
The universe is filled with a lot of very small particles flying around in all directions. The vast majority of these particles pass through normal matter w/o interaction, but a few will either bounce or stick. These interactions result in a slight attenuation of particles comming from the direction of any nearby mass and that attenuation creates a net force toward the nearby mass.
Also, these particles can travel about 5,000 lightyears on average before they hit another similar particle and scatter. This is the source of the range limit for gravitational force. Once you get more than about 5 to 10 thousand lightyears away from a mass the attenuation of particle flux from the direction of that mass is filled in by scattered background particles and the net force drops toward zero.
Galaxies would thus have to be very close to one another for them to feel each other's gravity.
At distances of less than a few thousand lightyears the formulas of Newton (or Einstein if you need to consider relativistic effects) are used for computing trajectories. As distance from a mass approaches the average scattering distance for the particles that cause gravitational force these equations would be modified by a distance-dependant-exponential-decay term.
Meta Model is not a new mathematical description of gravity. (The math we already have is more than adequate.) It is a new physical explanation of what gravity is and how it works.
Regards,
LB
But the really short version goes something like this:
===
The universe is filled with a lot of very small particles flying around in all directions. The vast majority of these particles pass through normal matter w/o interaction, but a few will either bounce or stick. These interactions result in a slight attenuation of particles comming from the direction of any nearby mass and that attenuation creates a net force toward the nearby mass.
Also, these particles can travel about 5,000 lightyears on average before they hit another similar particle and scatter. This is the source of the range limit for gravitational force. Once you get more than about 5 to 10 thousand lightyears away from a mass the attenuation of particle flux from the direction of that mass is filled in by scattered background particles and the net force drops toward zero.
Galaxies would thus have to be very close to one another for them to feel each other's gravity.
At distances of less than a few thousand lightyears the formulas of Newton (or Einstein if you need to consider relativistic effects) are used for computing trajectories. As distance from a mass approaches the average scattering distance for the particles that cause gravitational force these equations would be modified by a distance-dependant-exponential-decay term.
Meta Model is not a new mathematical description of gravity. (The math we already have is more than adequate.) It is a new physical explanation of what gravity is and how it works.
Regards,
LB
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19 years 3 months ago #14267
by Michiel
Replied by Michiel on topic Reply from Michiel
Assuming gravity has a finite propagation speed can have consequences. Here's an example.
When two equal masses are in circular orbit their orbital speed is given by:
vo = sqr( m * G / (2 * r) ) [m/s]
m is the mass of each body
G is the gravitational constant
r is the distance between the bodies
Let vg [m/s] be the speed of gravity then the time it takes for gravitation to travel from one body to the other is:
Tg = r / vg
During that time each body will move a certain distance:
d = Tg * vo = r * vo / vg [m]
The force caused by gravity will now have a different direction as compared to gravity without propagation delay. The new force component is in the direction of vo, the orbital speed. The orbit will slowly get wider.
The ratio d / r is a measure for the 'error angle' but only if it's really small.
d / r = vo / vg = sqr( m * G / (2 * r) ) / vg [m/m]
We can see that the angle gets smaller with higher values of vg , no surprise there.
When r increases the angle gets smaller too. But higher m will give a bigger error angle.
Or you could say: all systems with a certain m / r will have the same error angle.
___
What this means in practice depends on vg , I think. The effect may not be that big after all.
When two equal masses are in circular orbit their orbital speed is given by:
vo = sqr( m * G / (2 * r) ) [m/s]
m is the mass of each body
G is the gravitational constant
r is the distance between the bodies
Let vg [m/s] be the speed of gravity then the time it takes for gravitation to travel from one body to the other is:
Tg = r / vg
During that time each body will move a certain distance:
d = Tg * vo = r * vo / vg [m]
The force caused by gravity will now have a different direction as compared to gravity without propagation delay. The new force component is in the direction of vo, the orbital speed. The orbit will slowly get wider.
The ratio d / r is a measure for the 'error angle' but only if it's really small.
d / r = vo / vg = sqr( m * G / (2 * r) ) / vg [m/m]
We can see that the angle gets smaller with higher values of vg , no surprise there.
When r increases the angle gets smaller too. But higher m will give a bigger error angle.
Or you could say: all systems with a certain m / r will have the same error angle.
___
What this means in practice depends on vg , I think. The effect may not be that big after all.
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19 years 3 months ago #14226
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
If you set vg equal to c, the speed of light, and run your calculation you will find that all orbits controlled by gravity are unstable. In just few tens of thousands of years or so all of the planets in our solar system would be boosted to escape velocity.
If vg is set to several million times c, the time required for detectable orbital changes in our solar systm rises to several billion years. But there are some other objects we have observed that would still have unstable orbits unless vg is at least 20 billion c.
LB
If vg is set to several million times c, the time required for detectable orbital changes in our solar systm rises to several billion years. But there are some other objects we have observed that would still have unstable orbits unless vg is at least 20 billion c.
LB
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19 years 3 months ago #14228
by Michiel
Replied by Michiel on topic Reply from Michiel
I agree that vg = c is right out of the question (although I don't think planets would reach escape velocity).
And even Isaac Newton couldn't accept gravity being an instantanious action at a distance - still we don't have an upper limit.
At least we do seem to have a lower limit.
Larry, do you know if this lower limit is low enough to have a significant effect on two orbiting galaxies?
And even Isaac Newton couldn't accept gravity being an instantanious action at a distance - still we don't have an upper limit.
At least we do seem to have a lower limit.
Larry, do you know if this lower limit is low enough to have a significant effect on two orbiting galaxies?
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19 years 3 months ago #11171
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
I'm not convinced that galaxies can orbit anything. The range of gravitational force is limited and that limit is small compared to the size of a galaxy. When two galaxies collide, however, or have a near miss, the parts of each one come close enough to the parts of the other for gravitational interactions to occur.
LB
LB
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