Since there are variations of $g$ depending on location on Earth's surface, why not use a reproducible lab experiment using a vertical axis centrifugal balance, and say that one kg is defined by centrifugal force of $x$ atoms of silicon which center of mass is rotating at $y$ rpm, at a distance $R$ from the center?


closed as unclear what you're asking by sammy gerbil, stafusa, Kyle Kanos, Asher, Bill N Nov 14 '17 at 2:47

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  • $\begingroup$ See en.wikipedia.org/wiki/Proposed_redefinition_of_SI_base_units $\endgroup$ – JamalS Nov 10 '17 at 15:41
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    $\begingroup$ @JamalS That's irrelevant. The question is based on incorrect assumptions for both the existing and the proposed definitions. As to why a question this ill-researched is getting upvotes, I'm stumped. $\endgroup$ – Emilio Pisanty Nov 10 '17 at 15:44
  • $\begingroup$ Not clear what you are trying to define, mass or force, or what your exact procedure is. The kilogram of mass historically comes from the mass of one liter of water at $4^oC$ (maximum density), but water is not ideal for precise standards. $\endgroup$ – safesphere Nov 10 '17 at 17:33
  • $\begingroup$ The kilogram presently is defined as the mass (not the weight) of a certain metal artifact that is kept by the International Bureau of Weights and Measures. In future, it will be redefined as the mass of a certain number of silicon atoms, and it will be possible (if you can afford it) for you to make your own, extremely accurate, 1kg silicon artifact by following the instructions that presently are being tested by the Avogadro Project. en.wikipedia.org/wiki/Kilogram $\endgroup$ – Solomon Slow Nov 10 '17 at 18:06
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    $\begingroup$ @jameslarge That's a misunderstanding of how the Avogadro-project route to the new kilogram works; the kilogram won't be defined as the mass of a certain number of silicon atoms. Instead, the atomic mass of silicon is measured with respect to $\hbar$ using atomic recoil spectroscopy (i.e. the kilogram is the mass unit in which $\hbar$ has a fixed value, and you measure $\hbar/m(^{28}\rm Si)$ empirically), and you scale that up to the kilogram scale with an object (the shiny black sphere) with a precisely-measured number of silicon atoms. $\endgroup$ – Emilio Pisanty Nov 10 '17 at 18:51

Simple: the kilogram is not defined using Earth's gravity.

The kilogram is a measure of mass. Earths' gravity is only relevant when computing weight, which is a completely different concept.

There is a proposed redefinition of the kilogram in the works (which will get rid of the comparison to the IPK or any artifacts, and make the definition reproducible on any lab without the need for any reference materials) but neither the proposed new definition nor the existing one depend on Earth's gravity. (Or, when they do, this is measured and controlled for to remove that dependence.)


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