San José State University
Department of Economics

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 The Acquisition of Rotation by a Contracting Feature in a Revolving System

When a feature such as a tropical cyclone in the atmosphere of a revolving Earth or a proto-planet in a revolving planetary disk of the solar system contracts it acquires rotation in the same direction as as the revolving system. For expanding features the direction of rotation acquired would be in the opposite direction from that of the revolving system.

This mechanism for the acquisition of rotation is similar to what happens when an ice skater goes into a spin with her arms outstretched. When she draws in her arms she spins faster. For a more precise analogy consider someone sitting in a swivel chair on a rotating platform facing the direction of rotation. If the sitter has his arms initially outstretched and draws them in he and the chair will begin to rotate with respect to the platform.

## Preliminary Analysis

Let V be the tangential velocity in a revolving system at a distance R from the center of the revolving system. This tangential velocity will be a function of R, say V=F(R). For a disk turning as a unit F(R)=ΩR where Ω is the angular velocity of the disk. A Keplerian planetary disk is a ring of particles around a central body such as the Sun or a planet in which centrifugal force exactly balances the gravitational attraction. For a Keplerian disk F(R)=γ/R½ where γ is the square root of the product of the gravitational constant and the mass of the central body.

Consider a body of radius r at a distance R from the center of the revolving system that bring in material from (R-ΔR) and from (R+ΔR). The material from (R-ΔR) will lose velocity while preserving angular momentum with respect to the center of the revolving system; that from (R+ΔR) will gain velocity. The differential in velocities at (R-r) and (R+r) will produce rotation of the feature.

It is convenient to work with the angular momentum per unit mass function L(R)=VR=F(R)R. The matter which moved from (R+ΔR) has the same L at (R+r) it had at (R+ΔR) and likewise for the matter that moved from (R-ΔR). Therefore the difference in velocities at (R+r) and (R-r) is

#### ΔV = L((R+ΔR)/(R+r) − L(R-ΔR)/(R-r)

Since ΔR<<R the values of L(R+ΔR)/(R+r) and L(R-ΔR)/(R-r) can be closely approximated by

Thus

#### ΔV = 2(L'(R)/R)ΔR − 2(L(R)/R²)r

The rotation of the feature is given by

#### ω = (ΔV/2)/r

Thus the relationship for the acquisition of rotation is

## The Case of a Uniformly Revolving Disk

For a disk revolving as a unit F(R)=ΩR and hence L(R)=ΩR². Therefore L'(R)=2ΩR and thus

#### ω = 2Ω(ΔR/r) − Ω or, equivalently ω/Ω = 2(ΔR/r) − 1

For a hurricane the winds take approximately three hours to circle the eye. This means that ω/Ω is about 24/3=8 so (ΔR/r)=4.5. Thus for a hurricane of diameter of 50 km the value of ΔR is about 112.5 km. However the Earth surface is not a flat plane so a further refinement of the analysis may be required.

## The Case of a Planetary Disk

For a Keplerian disk F(R)=γ/R½ and thus L(R)=γR½ and L'(R)=½γ/R½. Therefore

#### ω = ½(γ/R3/2)(Δ/r) − γ/R3/2which reduces to ω = (γ/R3/2)(½(ΔR/r)−1)

In the Keplerian disk, (γ/R3/2) is equal to the angular rate of revolution Ω. Therefore

#### ω/Ω = ½(ΔR/r)−1

If τ represents the period of rotation and Τ the period of revolution then

#### Τ/τ = ½(ΔR/r)−1.

For the Earth Τ/τ=365.25 and thus ΔR/r = 732.5. This means that ΔR =(3960)(732.5)= 2,900,700 miles, a not unreasonable figure of 3.1 percent of the orbit radius of 93.5 million miles. This would mean that the material of the Earth was collected over a band of width equal to about 6 percent of its orbit radius. However this contraction is not the only means by which a planet could acquire spin. For another mechanism see Planetary Sweep.

## Generalization

Consider the case in which the tangential velocity function is V(R)=αRβ. This includes the uniformly rotating disk and the Keplerian disk as special cases. For this general case

Thus

#### ω = (β+1)αRβ−1(ΔR/r) − αRβ−1which reduces to ω = αRβ−1[(β+1)(ΔR/r) − −1]

However the angular velocity of revolution is

#### Ω = V/R = αRβ−1and therefore ω = Ω[(β+1)(ΔR/r) − 1] or, equivalently ω/Ω = (β+1)(ΔR/r) − 1

In terms of the period of revolution Τ and the period of rotation τ the above relation is

#### Τ/τ = (β+1)(ΔR/r) − 1

This relationship was derived by methods which presumed that r<<ΔR. The ratio (ΔR/r) is a contraction ratio. The contraction would involve material from r to ΔR so the relevant contraction is from some average of ΔR and r down to r.