Computing Newton's gravitational constant from Cavendish's data

Question

[I asked this at History of Science and Mathematics but there was no answers. I'm asking here too, after all the Newton's gravitational constant is fundamental in physics.]

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

I'm using this formula (from Wikipedia)

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

where

• $G =$ 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 T (Cavendish's 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,

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

What's wrong with these calculations?

Cavendish uses the formula $D= \frac{N^2}{10683\times B}$ to compute density $D$ of the earth. When the weights are moved from + to - and from - to +, the arm of the pendulum moves twice as much then when the weights are moved from middle position to + or - positions. He does not say it explicitly but for experiments with double values ($2B$) Cavendish uses the formula, $D= \frac{N^2}{10683\times (B/2)}$.

I computed $G$ by halving $B$ but I still did not get the correct value. But when I halved $2B$ values and double the period I got the correct value.

And, this is how I computed the radian value of B:

On page 509, Cavendish gives the distance of the ivory scale from the center of motion: "But the ivory scale at the end of the arm is 38.3 inches from the center of motion." On the same page he says that each division of the ivory scale is 1/20 of an inch, that is, 0.050 inch. By the radian rule, $\theta = \frac{l}{r} = \frac{0.050}{38.3}= 0.0013054$.

For this experiment $B=6.18$ So, I multiply 6.18 by 0.00130 to get 0.008067 radians as the angle of deflection.

There are more details here

Answer 1

A factor of two is relatively easy to account for in that in Wikipedia formula $G = \frac{2\pi^2 L r^2 \theta}{M T^2}$ the angle $\theta$ is the angle of deflection from not having any large masses present to having the two large masses present.

The value for the movement of the arm from Cavendish's table of results that you used from experiment 4 is $6.18/2$ as the large masses were moved from one side $(+)$ to the other $(-)$.
Note the value above $(3.1)$ where the deflection is halved because it is a no large masses present $(m)$ to large masses present deflection $(+)$.
This will make your calculated value of $G$ smaller by a factor of two.

The other factor of four is to do with the period of the oscillation which had a meaning to Cavendish which differs from that now commonly accepted.

The modern definition of a period is the time taken for one complete oscillation, eg the time taken from one extreme to the other extreme and then back to the stating extreme. For Cavendish it was the time taken from one extreme to the next extreme and so is half the period as now defined.

In going from a maximum deflection of $15$ to a maximum deflection on the other side of $22.4$ [A] the beam passed through the "mid point of vibration" at an estimated time $\rm 10h\, 20'\, 31''$ [A'] and then when moving in the other direction with the beam going from a deflection of $22.4$ to $15.1$ [B] passed through the "mid point of vibration" at an estimated time of $\rm 10h\, 27'\, 31''$ [B'].
The time interval between the two crossing points is quoted as $7'\,0''$ and thus the (modern) period $T = 840\,\rm s$.

Since the period is squared a factor of four is involved.

So there is a reduction by a factor of two in the numerator and an increase by a factor of four in the denominator giving a total reduction by a factor of eight. So the calculated value of a value for $G$, without applying any of Cavendish's corrections, now becomes $6.59\times 10^{-11} \,\rm m^3kg^{-1}s^{-2}$.

Answer 2

As you know, the Cavendish did not try to measure the $G$, but he was trying to measure the $\rho_{\rm earth}$. So, he performed many experiments and you can see the result of these experiments on page 520 of Cavendish Paper.

A single experimental result cannot determine the value of $G$ (or any measured value). In general, we perform many measurements and then take the average.

Hence, when you take the average value of the $\rho_{\rm earth}$ measurements, you'll obtain (it seems Cavendish made an error while taking the mean value. See this)

$$\rm\rho_{earth} = 5.448~g\,cm^{-3} \equiv 5.448\times 10^{3}~kg\,m^{-3}$$

Later, when you use this equation (which is given on the Wikipedia page)

$$G = \frac{3g}{4\pi R_{\rm earth}\rho_{\rm earth}}$$

you'll obtain

$$G = 6.738 \times 10^{-11}{\rm m^3kg^{-1}s^{-2}}$$

So to answer your question, it's not possible to obtain $G_{\rm True}$ by doing calculations based on a single experiment. You have to take the average of the results. What you can do this, Calculate $G_i$ for each $T_i$ and $\theta_i$ (I guess these are the variables) and then calculate the mean

$$\bar{G} = \frac{1}{n}\sum_i^n G_i$$

and see if $\bar{G}$ is close to $G_{\rm True}$ or not.