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# Derivation of Gravitational Constant from Cavendish Experiment

by Ron Kurtus

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.

(See Cavendish Experiment for more information.)

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 = 2π ^{2}LθR_{e}^{2}/T^{2}M**.

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:

τ_{T}= κθ

where

**τ**is the torque resulting from a twisted wire in N-m_{T}**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: Aradianis a unit of angular measurement where1 radian = 57.3°and2π radians = 360°.

Thus, the torque is proportional to the angle turned.

### Equilibrium point

At the equilibrium point, **τ = τ _{T}** and:

FL=κθ

F = κθ/L

### Find G

Newton's *Universal Gravitation Equation* at the equilibrium point is:

F = GMm/R_{e}^{2}

where

**F**is the force of attraction between the balls in newtons (N)**G**is the Universal Gravitational Constant in in N-m^{2}/kg^{2}or m^{3}/kg-s^{2}**M**is the mass of the larger ball in kg**m**is the mass of the smaller ball in kg**R**is the separation between the centers of the balls at the equilibrium point in meters_{e}

Substitute **F = κθ/L**:

κθ/L = GMm/R_{e}^{2}

G = κθR_{e}^{2}/LMm

Although **θ**, **R _{e}**,

**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-m^{2}**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:

T^{2}= 4π^{2}I/κ

κ = 4π^{2}I/T^{2}

### Moment of inertia

The moment of inertia of the smaller balls is:

I = mL^{2}/2

Substitute inertia in the torsion coefficient equation:

κ = 4π^{2}mL^{2}/2T^{2}

κ = 2π^{2}mL^{2}/T^{2}

## Completing the derivation

Substitute **κ = 2π ^{2}mL^{2}/T^{2 }**into

**G = κθR**:

_{e}^{2}/LMm

G = 2π^{2}mL^{2}θR_{e}^{2}/LT^{2}Mm

Simplify the equation:

G = 2π^{2}LθR_{e}^{2}/T^{2}M

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 = 2π ^{2}LθR_{e}^{2}/T^{2}M**.

Perform your tasks one step at a time

## Resources and references

### Websites

**Cavendish Experiment** - Harvard University Natural Science Lecture Demonstrations

**The Cavendish Experiment** - Good illustrations of experiment from Leyden Science

**Cavendish experiment** - Wikipedia

### Books

(Notice: The *School for Champions* may earn commissions from book purchases)

**Top-rated books on Gravitation**

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## Derivation of Gravitational Constant from Cavendish Experiment