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(I asked this questions in Mathematics section but there were no answers.)

  • I'm working on Cavendish's original analysis of his experiment. Cavendish does not use equations or standard units. For instance, he states the weight of the lead weights by taking the weight of a sphere of water of 1 foot in diameter as unit. He says, the weight weighs 10.64 times the weight of a water sphere of 1 foot diameter.

  • To calculate the weight of the earth he uses the definition of density, as, density = weight/volume (Cavendish does not use mass). Then he uses the relation Force :: weight x inverse square of distance to calculate the forces.

  • This far I understand. But, Cavendish states his final result by skipping several steps. Luckily, in the book that I'm using the editor, included a footnote explaining the missing steps.

  • My question is, in the footnote, he says, "the volume of the earth in spherical units is (41,800,000)^3..." But, 41,800,000 feet is the diameter of the earth. I thought (radius)^3 would be the volume. Can you explain why he is using (diameter)^3 for volume and not (radius)^3?

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Can you explain why he is using (diameter)^3 for volume and not (radius)^3?

Because, as per your first paragraph, Cavendish's spherical volume unit is a sphere 1 foot in diameter, not a sphere 1 foot in radius. In other words, Cavendish's spherical volume unit is $\frac {\pi} 6$ cubic feet, not $\frac {4 \pi} 3$ cubic feet.

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What does it matter if you use the diameter ($d$) or the radius ($r$)?

$$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(\frac{d}{2}\right)^3 = \frac{1}{6}\pi d^3.$$

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  • $\begingroup$ @gandalf61's answer clarified it for me. What was confusing was that $radius^2$ and $diameter^2$ are not same numerically. But he is using a 1 foot diameter sphere as the unit, so he does not need to use $\pi$ $\endgroup$
    – zeynel
    Jan 5, 2022 at 11:44

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