# Derivation of Gravitational Constant from Cavendish Experiment

by Ron Kurtus (20 February 2015)

By examining the relationships between the various factors in the Cavendish Experiment, you can derive the equation for the Universal Gravitational Constant, G.

The experiment uses a torsion balance device to measure the movement of smaller lead balls toward the larger balls. The gravitational force attracting the balls provides a torque on the moment arm and twists the wire holding the balance. Light is reflected off a mirror to measure the angle the balance turns and its oscillation rate.

The derivation of the equation for G is in two parts. The first part shows the relationship between G and the angle, final distance, bar length, masses of the balls, and torsion coefficient at equilibrium point. The second part of the derivation defines torsion coefficient in terms of oscillation period and moment of inertia.

Combining those parts yields the equation: G = 2π2LθRe2/T2M.

Questions you may have include:

• What are the relationships at the equilibrium point?
• What is the formula for the torsion coefficient?
• What is the completed result?

This lesson will answer those questions. Useful tool: Units Conversion

## Relationships at 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 as a torque. Since there are two moment arms of L/2 each on the bar, the torque on the wire is:

τ = FL

where

• τ (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

• τT 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 2π radians = 360°.

Thus, the torque is proportional to the angle turned.

### Equilibrium point

At the equilibrium point, τ = τT 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 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 = 2π√(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 = 2π√(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

This completes the derivation.

## Summary

You can derive the equation for G showing the relationship with the angle, final distance, bar length, masses of the balls, and torsion coefficient at equilibrium point. Then find the torsion coefficient in terms of oscillation period and moment of inertia.

Combining those parts yields the equation: G = 2π2LθRe2/T2M.

## Resources and references

Ron Kurtus' Credentials

### Websites

Cavendish Experiment - Harvard University Natural Science Lecture Demonstrations

The Cavendish Experiment - Good illustrations of experiment from Leyden Science

Cavendish experiment - Wikipedia

Gravitation Resources

### Books

Top-rated books on Gravity Top-rated books on Gravitation Do you have any questions, comments, or opinions on this subject? If so, send an email with your feedback. I will try to get back to you as soon as possible.

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