# Computing Gravitational constant G from Cavendish's data

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

• What values does Cavendish quote for G? May 21, 2022 at 18:23
• 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. May 21, 2022 at 20:21
• – rob
Jun 1, 2022 at 21:16
• There are answers at Physics Stack Exchange. Jun 6, 2022 at 7:23