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June 9th, 2003, 07:39 AM
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Re: # of members ?
Quote:
Originally posted by Kamog:
Is the gravitational constant really constant?
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To the best of my knoweledge, within a given system of units (e.g. metric: meters, seconds, kilograms; "English": feet, seconds, slugs) the gravitational constant is constant. However, we have only had the tools to measure it for a fairly short period of time, and only on Earth.
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June 9th, 2003, 03:13 PM
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Shrapnel Fanatic
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Re: # of members ?
And around earth, and all over the solar system, and all over the galaxy...
If the gravitational constant was different over at alpha centauri, for example, then the stars would be orbiting each other at the "wrong" speed.
PS: And, variables can also be functions in certain cases.
[ June 09, 2003, 14:15: Message edited by: Suicide Junkie ]
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June 9th, 2003, 05:12 PM
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Sergeant
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Re: # of members ?
Wait I thought that there was only Fyron. At that the rest of us are just sub personalities or figments of his imagination.
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June 9th, 2003, 07:52 PM
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Re: # of members ?
No silly, we are all Puke.
And the gravitational constant is indeed constant everywhere. We may not have the exact value down to the nanometer (and probably never will), but that doesn't really matter. 
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June 9th, 2003, 07:56 PM
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Re: # of members ?
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.
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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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June 9th, 2003, 10:12 PM
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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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June 9th, 2003, 11:08 PM
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Re: # of members ?
Quote:
Originally posted by Suicide Junkie:
Star masses can also be found by looking at colour, age, size, etc.
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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.
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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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