# Gravitational Constant Measurement ?

1. Oct 15, 2012

### Debo Industry

How did we measure the gravitational constant G = 2.034 x 10 ^ 17 cm ^ 2 and the Earth average density (specific gravity) p = 3.45 g / cm ^ 2 s ^ 2.

2. Oct 15, 2012

### PhysicoRaj

The gravitational constant was measured by Henry Cavendish, using a setup, like this:

When the lead spheres are moved the gravitational force acting on the other set of spheres induces a force on the wire which twists, and the restoring torque can be equated to the gravitational force as follows:
torque = force * length
τθ=GMm/r2]*l
(where l is the length of bar, τ is the torque per unit angle of twist, θ is the angle of twist caused by the force).
So,
G=τθr2/Mml
θ, angle of twist can be measured by many methods.( in the figure it is measured by laser, and hence u see a mirror).

Last edited: Oct 15, 2012
3. Oct 15, 2012

### Debo Industry

PhysicoRaj,
After working out the gravitation formula and obtaining gravitational constant G there appeared a chance to check up the correctness of finding by the Cavendish the force of gravitation between the first ball with the mass of 1.0 g and the second ball with the mass of 1.0 g, located at the distance of Rball-ball = 1.0 cm, and equal to 6,6742 x 10 ^ - 8 gсm / s ^ 2. Due to the fact that Cavendish carrying out his calculations in grains used the units of weight with the dimension of mass, the mass of each ball was found first of all.
The mass of the ball Мball was found as the relation of the ball Рball = 1.0 gcm / s ^ 2 weight to the Earth gravity acceleration gear = 980.665 cm / s ^ 2 by the formula: Mball = 1 / 980.665 = 1.0197 x 10 ^ - 3g. The gravity acceleration of the ball gball was found as the product of the ball mass Мball by the gravity acceleration of 1g of the body g1-m = 2.5645 x 10 ^ - 22 cm / gs ^ 2 by the formula:

gball = 1.0197 x 10 ^ - 3 x 2.5645 x 10 ^ - 22 = 2.615 x 10 ^ - 25 cm / s ^ 2.

The force of gravitation between the first mass standard Мball = 1.0197 x 10 ^ - 3 g and the second mass standard Мball = 1.0197 x 10 ^ - 3 g, located at the distance Rball-ball = 1.0 сm:

Fball-ball = 2.034 x 10 ^ 17 x (1.0197 x 10 ^ - 3 x 2.615 x 10 ^ - 25 + 1.0197 x 10 ^ - 3 x 2.615 x 10 ^ - 25) / 1.0 ^ 2 = 1,0847 x 10 ^ - 10 gcm / s ^ 2.

Thus, the force of gravitation between two balls in the experiment of Sir H. Cavendish was Fball-ball = 1,0847 x 10 ^ - 10 gсm / s ^ 2 , and not 6,6742 x 10 ^ - 8 gcm / s ^ 2 that is appeared to be 615.3 less

http://tsiganok.cc.ua/files/laws_of_motion_of_bodies_draft_220911.pdf (p.114).

4. Oct 18, 2012

### PhysicoRaj

The force between the balls is equal to the gravitational constant if the masses taken are equal to unity and are placed at unit distance apart. Since 1 g is itself unit mass, the need to convert 'weight' of 1 g to 'mass' by dividing by 980.6 cm s^-2 is the compulsion of -
F=G M1 g2+M2 g1/r^2

if you use
F=G M1*M2/r^2
Then you need not do that.
to get 6.67x10^-8 gcm/s^2 , unit 'mass' is necessary but not unit 'weight'.