# of members ? - Page 4

Suicide Junkie · #31 June 9th, 2003, 03:13 PM

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 ]

Perrin · #32 June 9th, 2003, 05:12 PM

Wait I thought that there was only Fyron. At that the rest of us are just sub personalities or figments of his imagination.

Fyron · #33 June 9th, 2003, 07:52 PM

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.

Jack Simth · #34 June 9th, 2003, 07:56 PM

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 ]

Suicide Junkie · #35 June 9th, 2003, 10:12 PM

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 · #36 June 9th, 2003, 11:08 PM

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.

Atrocities · #37 June 10th, 2003, 02:05 PM

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)

The universe made simple.

mottlee · #38 June 10th, 2003, 03:16 PM

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

Greybeard · #39 June 10th, 2003, 03:25 PM

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 · #40 June 10th, 2003, 03:39 PM

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 ]