# why there is no accuracy of the measured value of $G$?

With the advancement of Modern Technology still there is no accuracy of the measured value of $G$ Gravitational Constant, why!?

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①Gravitation is too small.②The prototype limits the accuracy of mass measurement.③Gravitation of unwanted sources (equipment, lab, you) cannot be screened. – Siyuan Ren Sep 4 '12 at 0:32
What does "no accuracy" mean? – Ben Crowell Aug 14 '13 at 4:25

With all due respect, I think both of these answers are way out of date. Read item 4 here http://www.npl.washington.edu/eotwash/experiments. Systematic errors associated with the torsion fibers commonly used made the bias much larger than the standard error. The U Wash group developed a feedback torsion balance which virtually eliminated anelasticity. It's still a hard measurement, but at least we are beyond the embarrassment of several reputable measurement whose 95% CI's didn't overlap. Having a torsion balance in which feedback keeps the fiber from moving is pretty sweet. I have no connection with the EotWash group.

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It's very hard to measure the magnitude of the gravitational force between objects of well-known mass. For the mass to be well-known, as a multiple of the kilogram prototype (which is how we still define the unit of mass) they have to be rather small objects. But the gravity between the small objects is too weak. It can be measured but it has been impossible to measure the force at a better accuracy than three or four significant figures.

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alfered,

This is a good question with a potentially long answer, although there are good sources that discuss some of the difficulties and recent solutions in measuring G. A good one is the thesis of Joshua Schwarz which gives a good overview in the first six pages and covers how to perform gravity measurements in free fall. One obstacle in traditional torsion type approaches is having an accurate estimate of the anelasticity of the fiber used in the torsion experiment. As one can imagine, one must understand the physics of the material itself in order to estimate the actual force that is being applied. Hope this helps.

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