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Originally posted by MegaMind
The OP has derived a formula for the attractive force between two masses in a special case (one orbiting another) from a set of knowns (T period, M mass and r radius). From this formula the OP declares that (G), the gravitational constant, is not necessary to calculate the attractive force (F) using his formula, indeed it is not in his special case with these knowns. Then the OP extrapolates that (G) is not needed at all.
But OP can you, as others have called for, show us how your formula or another derivation can give us the force (F) between two stationary objects without the use of (G)?
You cannot. Newton's law is fundamental. Your formula is not.edit on 8-10-2010 by MegaMind because: (no reason given)
The value of G has been refined over the years as more accurate measurements have been made. That is much different than saying it's non-constant.
Originally posted by mnemeth1
G has proven to be NON-CONSTANT in nature, no matter what method of measurement has been used.
Originally posted by nataylor
The value of G has been refined over the years as more accurate measurements have been made. That is much different than saying it's non-constant.
Originally posted by mnemeth1
G has proven to be NON-CONSTANT in nature, no matter what method of measurement has been used.
Originally posted by MegaMind
reply to post by mnemeth1
Yes and Newtonian physics breaks down in predicting the extreme. But that doesn't change the fact that the OP cannot even come close to calculating the force of stationary objects without G. You would admit that Newton's formulas give very, very close approximations wouldn't you?edit on 8-10-2010 by MegaMind because: (no reason given)
Originally posted by mnemeth1
Logic would dictate that there is no attraction, other than of the electro-magnetic type, which still ultimately relies on movement.
Since we can't measure the attractive force between stationary objects with any accuracy other than arriving at the conclusion the attraction is zero, I don't see any reason why it should not be declared as zero given his findings.
Measurements are averaged so as to average out the uncertainty in the measurements. It is difficult to measure because gravity is so weak. That they have it narrowed down to a value with an uncertainty of less than 0.0014% is pretty darn good.
Originally posted by mnemeth1If it was constant, its measurements would not need to averaged.
Originally posted by MegaMind
Originally posted by mnemeth1
Logic would dictate that there is no attraction, other than of the electro-magnetic type, which still ultimately relies on movement.
Since we can't measure the attractive force between stationary objects with any accuracy other than arriving at the conclusion the attraction is zero, I don't see any reason why it should not be declared as zero given his findings.
I did a version of the Calvendish Experiment in my physics course. The attraction is clearly observable and neither mass is moving relative to another. The force is definitely not zero.
Accuracy of measurement is surely relative. Its accurate to a significant digit.
If G = 0, then F = 0 and Gravity does not exist. Yet clearly I observed the two lead balls being attracted to one another. How do you suppose we send satellites into orbit when our calculations are using G = 0 and F = 0? Yeah maybe we should just throw out the last 200 years of science.edit on 8-10-2010 by MegaMind because: (no reason given)
Given the electromagnetic force is about 10^39 times stronger than gravity, it's pretty safe to say it wasn't electromagnetic.
Originally posted by mnemeth1
The question is what type of attraction did you observe.
Electromagnetic or "gravity"
Originally posted by mnemeth1
The gravitational force between two stationary objects is virtually unmeasurable in the real world.