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Comment by alkonaut | Hacker News Reader

> Cavendish painstakingly recorded those oscillations to measure the gravitational force of the larger spheres on the smaller ones, and from that he could infer Earth’s density.

I can't wrap my head around this sentence. How is the force between two objects, none of which is the earth, related to any property of the earth (its mass or density)? Wouldn't Cavendish's experiment have worked - even better - in zero g and given the same result? There is something more to his setup beyond two big weights pulling on each other that I'm probably missing?

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Force of gravity for spherical objects of constant density is calculated from 2 masses + 1 distance + a constant.

Before the experiment you can measure the mass of both objects. In the experiment you measure the force and distance to calculate the constant.

The weight either object gives you the force between that object and earth (adjusting for atmospheric buoyancy). Altitude at your location + size and shape of earth gives distance between object and center of earth, you just learned the constant. So you know 4 out of five variables in an equation and can thus calculate the mass of the earth.

Technically that excludes the weight of the atmosphere above your altitude, but you can get that from the air pressure. Similarly the density of the earth isn’t constant but it is very close to symmetrical so you can get a reasonable estimate.

By ascertaining an approxiamte value of G , perhaps? After that, you know M_earth, and already knowing Earth’s geometry, one arrives at average density rho.

Yup, that's exactly it:

- get the gravitational constant with these two known masses

- then can deduct the mass of the unknown Earth by its interaction with other masses (say the "g" gravitational acceleration value)

- then from the mass and the otherwise measured size of Earth the density pops out

More details in good ol' Wikipedia: https://en.wikipedia.org/wiki/Cavendish_experiment#Derivatio...

It's also a notoriously difficult experiment to perform. When I did it at university, the value of G I got was out by an order of magnitude - and that was considered a good result!

Laying out the math (assuming earth is an homogeneous sphere) just in case it's not clear:

F_gravitational = G m1 m2 /r^2

g = G Mass_earth / r_earth^2

Mass_earth = r_earth^2 * g/G

Density_earth = r_earth^2 * g/G / V_earth

Density_earth = 3*g / (4*Pi*G*r_earth)

Prior to Cavendish we already new g and r_earth, just missing G.

Oh so the earth density is merely the motivation for the experiment? I read it as the earth mass actually being used somewhere in the formulas within the setup itself which was what confused me.

He uses his experiment to calculate G based only on the test masses and spring and then the _result_ of the calculation was just used as a final step to calculate the mass of the earth, and then from that the density?

Several years back when I first learned about the Cavendish experiment, I was indeed surprised that he was able to measure anything related to gravity without using a planetary-scale mass as one of experimental masses.

Clever experiments to measure tiny values are probably my favorite kind of science. Millikan found the charge of an electron by looking for a common integer multiple of charge in tiny oil droplets. Or LIGO measures gravitational waves by watching the interference pattern generated over time by two light beams taking different paths.

Y’know, I went and looked at the current Wikipedia page for the Cavendish experiment. There’s apparently more nuance than my simple outline offers, particularly as G wasn’t treated as an isolated quantity till decades after his experiment.

Linking directly to section anchor: https://en.wikipedia.org/wiki/Cavendish_experiment#Reformula...

There is less description than I would like, but I am happy to have been made aware of the discrepancy at all.

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