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Old June 9th, 2003, 07:56 PM
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Quote:
Originally posted by Suicide Junkie:
If the gravitational constant was different over at alpha centauri, for example, then the stars would be orbiting each other at the "wrong" speed.
But to tell how fast they "should" be orbiting each other, you need to know how massive they are, which is calculated from G and how fast they are orbiting each other. Using G to calculate G in such a manner is circular reasoning; it isn't reliable. It is probably the same everywhere, but we can't be certain until we send people over there to take local measurements.

[ June 09, 2003, 18:57: Message edited by: Jack Simth ]
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Old June 9th, 2003, 10:12 PM
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Default Re: # of members ?

Star masses can also be found by looking at colour, age, size, etc.

And having a third body in the system to observe helps a lot.

G can be calculated from the masses, distance between, and the observed acceleration.

[ June 09, 2003, 21:14: Message edited by: Suicide Junkie ]
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Old June 9th, 2003, 11:08 PM
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Quote:
Originally posted by Suicide Junkie:
Star masses can also be found by looking at colour, age, size, etc.
Getting the mass from the color, age, size, etc. implicitly uses G, as the plasma physics that produce such results include gravitational effects from the mass of the star. Again, G is used to calculate G, and as such is circular logic and is not reliable. Mind you, G is probably constant throughout the universe - but there isn't any good way to be certain of that until we get out there and measure things up close. Until then, G is constant makes for a good working theory, but it can't be proven.
Quote:
Originally posted by Suicide Junkie:

And having a third body in the system to observe helps a lot.

G can be calculated from the masses, distance between, and the observed acceleration.
For that, the masses have to be known. Getting the masses uses G, although sometimes it is implicit rather than explicit. Again, G is used to obtain G, which is circular reasoning; not reliable. Mind you, G is probably constant throughout the universe - but there isn't any good way to be certain of that until we get out there and measure things up close.

Without knowing both the masses and G, some simple numerical manipulation on the gravitational formulas can tell you that the distance and acceleration alone won't help:

F = G(M*m)/(d^2)
F' = G'(M'*m')/(d'^2)
A = F/m = (G*M)/(d^2)
A' = F'/m' = (G'*M')/(d'^2)
If A = A' and d = d', then
(G'*M')/(d^2) = (G*M)/(d^2)
-> (G'*M') = (G*M)

Example: Suppose G' = 2G:
-> 2G*M' = G*M -> 2M' = M -> M' = M/2

Then M' = M/2 results in the same acceleration for the same distance. The number of bodies won't make a difference for this aspect of things.
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Old June 11th, 2003, 01:13 AM
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Default Re: # of members ?

Sigh
As usual, this thread faced the same fate as all the others ...
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Old June 10th, 2003, 02:05 PM
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Default Re: # of members ?

Quote:
F = G(M*m)/(d^2)
F' = G'(M'*m')/(d'^2)
A = F/m = (G*M)/(d^2)
A' = F'/m' = (G'*M')/(d'^2)
If A = A' and d = d', then
(G'*M')/(d^2) = (G*M)/(d^2)
-> (G'*M') = (G*M)
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Old June 10th, 2003, 03:16 PM
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Default Re: # of members ?

Man did this one get DEEP! (way over my head)
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Old June 10th, 2003, 03:25 PM
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Well, I consider my reading the Forum as active. I try to read it every day. However, I only post occassionally, usually because someone has already made the point I was going to make and I don't want to just say, "yea, me too!"

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