I'm trying to compute the value of G from Cavendish's own observations. I get $G_{Cav}=5.27501×10^{−10}$ which is 8 times bigger than the accepted value of $G_{True}=6.67430×10^{−11}$. Do you see anything wrong with my computations below?

I'm using this formula (from Wikipedia)

$$ G = \frac{2\pi^2 L r^2 \theta}{M T^2} $$

G = Gravitational constant

L = Length of torsion balance (the distance between the centers of balls)

r = The distance of attraction (between weights and balls)

$\theta$ = Deflection of the arm from its rest position due to gravitational attraction

M = Mass of attracting lead weight

T = Natural period of oscillation of the balance

I take $\theta$ and N from the 4th experiment, (Page 520 in Cavendish's paper), the rest are constants,

L = 1.862 m

r = 0.2248 m

$r^2$ = 0.05053 $m^2$

$\theta$ = 0.00806788 radians

M = 158.04 kg

T = 421 s

$T^2$ = 177241 $s^2$

Substituting in the numbers,

$$ G_{Cav} = \frac{2 \times \pi^2 \times 1.836 \times 0.05053504 \times 0.00806788}{158.04 \times 177241} = 5.27501\times 10^{-10} $$

This is eight times bigger than the accepted value,

$$ \frac{G_{Cav}}{G_{Tru}} = 7.90 $$

There are more details here

  • $\begingroup$ What values does Cavendish quote for G? $\endgroup$
    – AChem
    May 21, 2022 at 18:23
  • $\begingroup$ Cavedish did not compute G. G was defined about a century after his death. This answer hsm.stackexchange.com/a/14308/16575 refers to a nice paper about the subject. $\endgroup$
    – zeynel
    May 21, 2022 at 20:21
  • $\begingroup$ Cross-posted to Physics. $\endgroup$
    – rob
    Jun 1, 2022 at 21:16
  • $\begingroup$ There are answers at Physics Stack Exchange. $\endgroup$
    – Farcher
    Jun 6, 2022 at 7:23


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