How To Find The Universal Law Of Gravitational Constant? - Henry Cavendish (Tamil)
Newton's Law Of Gravitational is introduced by Sir Issac Newton but he did not give the value of gravitational constant. So these first research papers of newton but give only valid point is to theorem statement to make corrected with respect to Newton's First Law Of Motion, Newton's Second Law Of Motion, and Newton's Third Law Of Motion is explained. But that time so many researchers said to be hypothetically correct statements.
So Many Years Pass the Henry Cavendish is given the value of Gravitational Constant or Universal Law Of Gravitational or Cavendish Gravitational Constant.
Henry Cavendish is conducting the experiment to make the model with respect to Newton's Law Of Gravitational.
Relationships at the equilibrium point
The first part of the derivation is to find the angle at the equilibrium point, where the gravitational force pulling the lead balls together equals the opposing torque from the twisted wire.
Force-torque relationship
When you apply a force on a torsion bar, the twisting of the wire is measured torque. Since there are two-moment arms of L/2 each on the bar, the torque on the wire is:
Ï„ = FL
where
- Ï„ (small Greek letter tau) is the torque at the pivot point of the torsion bar in newton-meters (N-m)
- F is the applied gravitational force on the small balls in newtons (N)
- L is the total length of the bar in meters (m)
Torque resistance relationship
A fiber or wire resists being twisted, similar to Hooke's Law for springs. The torque required to twist a wire a certain angle is related to the torsion coefficient of the wire and the angle it is twisted. Likewise, a twisted wire will result in a torque:
τT = κθ
where
- Ï„T is the torque resulting from a twisted wire in N-m
- κ (small Greek letter kappa) is the torsion coefficient in newton-meters/radian
- θ (large Greek letter theta) is the angle from the rest position to the equilibrium point measured in radians
Note: A radian is a unit of angular measurement where 1 radian = 57.3° and 2Ï€ radians = 360°.
Thus, the torque is proportional to the angle turned.
Equilibrium point
At the equilibrium point, Ï„ = Ï„T and:
FL = κθ
F = κθ/L
Find G
Newton's Universal Gravitation Equation at the equilibrium point is:
F = GMm/Re2
where
- F is the force of attraction between the balls in newtons (N)
- G is the Universal Gravitational Constant in N-m2/kg2 or m3/kg-s2
- M is the mass of the larger ball in kg
- m is the mass of the smaller ball in kg
- Re is the separation between the centers of the balls at the equilibrium point in meters
Substitute F = κθ/L:
κθ/L = GMm/Re2
G = κθRe2/LMm
Although θ, Re, L, M, and m can be measured, κ is still an unknown, depending on the wore used.
Find torsion coefficient
When the balance bar is initially released and the moving balls approach the larger balls, the inertia of the smaller balls causes them to overshoot the equilibrium angle. The torsion coefficient must be calculated by measuring the resonant oscillation period of the wire.
Oscillation period
This results in the torsion balance oscillating back-and-forth at its natural resonant oscillation period:
T = 2Ï€√(I/κ)
where
- T is the oscillation period in seconds
- π (small Greek letter pi) is 3.14...
- I is the moment of inertia of the torsion bar in kg-m2
- κ is the torsion coefficient in newton-meters/radian.
Note: Since the balls are heavy lead, the mass of the bar is considered negligible and not a factor in the inertia.
Solve for torsion coefficient
Square T = 2Ï€√(I/κ) and solve for torsion coefficient:
T2 = 4π2I/κ
κ = 4π2I/T2
Moment of inertia
The moment of inertia of the smaller balls is:
I = mL2/2
Substitute inertia in the torsion coefficient equation:
κ = 4π2mL2/2T2
κ = 2π2mL2/T2
Completing the derivation
Substitute κ = 2π2mL2/T2 into G = κθRe2/LMm:
G = 2π2mL2θRe2/LT2Mm
Simplify the equation:
G = 2π2LθRe2/T2M
At last Model reading data given in the Formula Of Gravitational Constant
G = 6.67 X 10^-11 Nm^2/Kg^2
This is the value of finding by Henry Cavendish
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