# Computing Newton's gravitational constant from Cavendish's data

[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.]

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

• I want to help but I did not understand your question clearly..Can you elaborate more..? May 26, 2022 at 14:20
• @seVenVo1d Thanks. I computed G from Cavendish's observations and I got a value which is 8 times greater than the accepted value of G. I was expecting to find the accepted value of G. Is there a mistake in my calculations? And why do I get a different value? May 26, 2022 at 17:30
• Your calculations seems correct. Have you checked the numerical values ? They might be incorrect. May 26, 2022 at 20:28
• Note: you don’t have to (and shouldn’t) clutter your question with metadata like “edited on [date].” People who are interested in the evolution of your question can see the edit history by clicking the “edited on” link at the bottom. Try to make your question read like a single well-formed document.
– rob
Jun 1, 2022 at 19:35
• @PM2Ring Yes, I agree. There is a confusion about the period. I get the right answer when I halve the motion of the arm and double the period. I halve the motion of the arm (deflection) because when he moves the weights from one near position to the other near position the arm moves as 2B. But I cannot figure out why I need to double the period. As you say, his period may be what we would call half period today. Jun 3, 2022 at 10:33

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 $$6.59\times 10^{-11} \,\rm m^3kg^{-1}s^{-2}$$.

• Great answer, thanks. I tried to fix a minor typo "stating extreme" should be "starting extreme" but it won't let me. Also you call the motion of the arm "deflection": "In going from a maximum deflection of 15 to a maximum deflection on the other side of 22.4". Of course, this is correct but, I was calling "deflection" the motion of the mid point ("point of rest") due to the attraction of the weights, which in this case is 6.18 divisions. Maybe this can be clarified. Thanks again. Jun 6, 2022 at 7:36
• For the angle $\theta$ the value on Cavendish's table is given as $2B$ divisions, that's why I halve it. As you said the motion is from + => - and that makes 2B. Jun 6, 2022 at 7:40

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.

• Yes, I will calculate the mean value. But now I'm wondering if Cavendish's period is half-period, as mentioned here large.stanford.edu/courses/2007/ph210/chang1 «The time of vibration was determined by choosing a fixed point and measuring the time between successive returns to that point, divided by the number of vibrations during that interval. (It seems to me that what was actually being measured was a half-period.)» We measure the period the same way, right? I don't need to multiply Cavendish's period by 2? May 28, 2022 at 11:15
• Cavendish's calculation of period is on paga 475 of his article. May 28, 2022 at 11:17
• I am not sure. But we know that $G \propto 1/T^2$. So if you multiply the period by $2$, then there will be $1/4$ difference. So after doing your calculations multiply the result by $1/4$ and see if there's a difference or not. May 28, 2022 at 11:57
• I don't think $G \propto 1/T^2$ is correct. If it were we would get a different value for $G$ by changing the period of the pendulum. In fact, the ratio $B/T^2$ is constant. As an example, the ratio $B/T^2$ for the first 4 experiments are like this: 3.48E-5, 3.30E-5, 3.26E-5, 3.31E-5 May 28, 2022 at 18:47
• I computed the average values, there was only a small change: Average B = 5.63 Average B_Rad = 0.00734986945 Average N = 424 and G from average values is: 4.804E-10 G_Average/G_True is: 7.12 May 28, 2022 at 18:56