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[Suicide Junkie]

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 ]

[Jack Simth]

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.

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.

[Me Loonn]

Sigh
As usual, this thread faced the same fate as all the others ...

[Atrocities]

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)

The universe made simple.

[mottlee]

Man did this one get DEEP! (way over my head)

[Greybeard]

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!"

Greybeard

[Suicide Junkie]

Gravity is not the only force out there, so if gravity is off, and mass is off to compensate, you'll also have to adjust just the speed of light (E=MC^2), the electromagnetic, weak and strong nuclear forces, etc because the mass of nucleons has all changed.

Is that really what you meant to imply?
I am not a physicist, but I suspect you'd be hard pressed to find a stable universe with different constants and still have anything close to the same observations made.

[ June 10, 2003, 14:46: Message edited by: Suicide Junkie ]