Weights at different altitudes

If gold were always sold by weight, could you make money buying gold at one altitude
above the ground and selling it at a different altitude? Where would you buy – at high or
at low altitude?

Answers and Replies

ZapperZ
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If gold were always sold by weight, could you make money buying gold at one altitude
above the ground and selling it at a different altitude? Where would you buy – at high or
at low altitude?

You are assuming that it isn't sold by "mass", and that each of the weighing instrument at different location is not calibrated. I highly suspect that it isn't true for something as valuable as gold.

Zz.

You are assuming that it isn't sold by "mass", and that each of the weighing instrument at different location is not calibrated. I highly suspect that it isn't true for something as valuable as gold.

Zz.

Balance scales are direct measuring tools for the mass of relatively small objects like gold bars.

D H
Staff Emeritus
Science Advisor
If gold were always sold by weight ...
In the US, gold *is* sold by weight. Then again, in the US, weight is legally a synonym for mass.

You are assuming that it isn't sold by "mass", and that each of the weighing instrument at different location is not calibrated. I highly suspect that it isn't true for something as valuable as gold.

Zz.

you are being too practical, this is a hypothetical question in which of course I am assuming that gold isn't sold by "mass", hence the "if" in my first sentence.

D H
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It's not sold that way precisely because of the problems you mentioned. Nothing of any substantial value is sold that way, and that includes a one pound can of peas.

ZapperZ
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you are being too practical, this is a hypothetical question in which of course I am assuming that gold isn't sold by "mass", hence the "if" in my first sentence.

But it makes your question moot!

You should have made your question more generic (just like you did with the topic of this thread before it was changed). Ask something like "if we use weighing scales to measure mass, then what happens if the value of g changes?", which is a question that we had addressed already in a few threads in this forum.

Zz.

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That might make a non-profitable difference for an electronic scale. But for a weighted balance scale, the counterweights would just as much less as the gold would.

Also, the weight of gold is proportional to the reciprocal of the square of the distance from the center of the earth: 4000 miles. So if you climb 10000 ft (2 miles), the weight would decrease by 4002^2/4000^2 = 1.001. Since gold is $1600 per ounce, you'd make$1.6 per ounce, assuming you don't lose that in gas, and you can find a difference of 10,000 ft. For 1,000 ft, \$0.16 per ounce. You might be luckier doing it above an iron deposit that has a bit more local gravity so you don't have to spend so much gas driving up that hill. Climbing takes lots of power and fuel.

arildno
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Can we measure DENSITY irrespective of measuring weight?
Yup, we can!
Take with a control block (equal in size) with the gold you want to buy on your mountain trip, along with a free hanging spring. Let the conrol block's density be d, "good gold"'s density D. Let local g be denoted as g*, spring constant k. Letting elongation of spring when a control block is hanging from it be called "l", that of gold L.

Thus, on equilibrium, we will have:

kl=dVg*
kL=DVg*, that is:

l/L=d/D

meaning that you can determine the density of the purported gold block without caring a damn about differences in g* at different localities.

I'm sure someone told me something similar; that a bar of gold would be heavier if you weighed it at the north pole rather than at the Equator (which would have more of an effect than the altitude thing I suspect), so the mass of the gold is sometimes measured with an instrument whose operation does not depend on the gravitational field. It was a long time ago but it was based on the principle of measuring inertia using some sort of horizontal spring. As long as you held it perfectly level, the variations in the Earth's gravitational field strength wouldn't have any influence on the measurement.