18 Jan 09
Originally posted by KazetNagorraWhat do you mean by 'no G'? You mean G would be 1 or something in some other units system? What units would that be? I guess it would have to be 6.7 E+11 times bigger, eh to be equal to 1 if that's what you mean.
It's just a proportionality constant that arises because of the use of SI units. You can transform to a different set of units and get no G.
18 Jan 09
Originally posted by sonhouseYes, it could be 1 in some other unit system. It doesn't really matter how you would define those units as long as you understand it's an artifact of SI units.
What do you mean by 'no G'? You mean G would be 1 or something in some other units system? What units would that be? I guess it would have to be 6.7 E+11 times bigger, eh to be equal to 1 if that's what you mean.
19 Jan 09
Originally posted by sonhouseThe constant is derived experimentally, first done by I believe Cavendish. It was a torsion pendulum set-up, you can probably find it on the internet.
This constant comes up in so many gravity field and lens formula's, just wondered if anyone knew how it was derived and what is it anyway?
I know it's 6.67 E-11 and all that but what does that mean?
6.67e-11 indicates that it is very weak.
19 Jan 09
Originally posted by RamnedHey Ramned, whatever happened to your physics question thread? It was fun.
The constant is derived experimentally, first done by I believe Cavendish. It was a torsion pendulum set-up, you can probably find it on the internet.
6.67e-11 indicates that it is very weak.
19 Jan 09
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Originally posted by PBE6It's hidden in the Posers and Puzzles somewhere, I left the site for awhile. Though the other day when I was tutoring I ran into a good one - in fact, it relates to gravitation.
Hey Ramned, whatever happened to your physics question thread? It was fun.
"Show how, guided by Kepler's law of periods, Newton could deduce that the force holding the moon in its orbit (assumed circular) depends on the inverse square of the moon's distance from the center of the Earth."
20 Jan 09
Originally posted by RamnedWould that be in terms of earth radii? Calling the earth's radii about 4000 miles and the moon 240,000 miles out, 60 radii? 1/60^2 or about 1/3600 of surface pull, so the acceleration of gravity from earth on the moon would be about 9.8/3600 M/S^2 or 2.7E-3 M/S^2 at the moon. I don't have time to go further, am at work, midnight shift, am tending a lot of testing equipment, we produce laser modulators for fiber optic networks, 40 Ghz and up (and down🙂 So a one Kg weight on earth would be 0.36 grams at the distance of the moon if there were an elevator that high🙂
It's hidden in the Posers and Puzzles somewhere, I left the site for awhile. Though the other day when I was tutoring I ran into a good one - in fact, it relates to gravitation.
"Show how, guided by Kepler's law of periods, Newton could deduce that the force holding the moon in its orbit (assumed circular) depends on the inverse square of the moon's distance from the center of the Earth."