San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 The Shape of a Rotating Fluid Mass

A static blob of fluid, small or large, will assume the shape of a sphere, if for no other reason than surface tension. The question of how that shape will change when the fluid is rotating has been investigated by some of the finest minds in physics from Isaac Newton to S. Chandrasekhar. The immediate application of the analysis is for the explanation of the shapes of stars and planets. Casual observation indicates that these rotating bodies are very nearly spherical. In Newton's time it was not known whether the deviation from sphericity was oblate, like a pumpkin, or prolate like a watermelon. Newton correctly deduced oblateness but Cassini and his followers argued for prolateness.

## Newton's Analysis

Newton focused on the question of the shape of the Earth. Here is his analysis. Let a be the radius of the Earth perpendicular to the axis of rotation and b the radius along the axis of rotation. The eccentricity or ellipticity ε of the shape is defined as:

#### ε = (a-b)/a = 1 − (b/a)

Newton considered a hypothetical construction involving holes drilled to the center of the Earth from the Equator and from a Pole and those holes filled with a fluid. At the center the weights of the two columns would have to be equal. But the weight of a section of the equatorial column would be diminished by the centrifugal force due to the rotation. Therefore this column would have to be longer than the polar column. The extent of the dimunition of the weight of the fluid in the equatorial column would depend upon the ratio of the centrifugal acceleration to the gravitational acceleration at the surface. Let Ω be the rate of rotation. Then at the equator the centrifugal acceleration is Ω²a. The gravitational acceleration at the equator is GM/a², where G is the G is the gravitational constant and M is the mass of the Earth. This ratio of these two is denoted as m where

#### m = (Ω²a)/(GM/a²) = Ω²a³/(GM)

To keep matter simple assume the cross section areas of the holes are equal to unity. Then the masses of the fluids in the holes are proportional to a and b. Let those masses be denoted as γa and γb. Let the accelerations due to gravity at the equator and the pole be denoted as ge and gp, respectively. The weight of the polar column of fluid is then γbgp. The weight of the equatorial column is diminished by the factor m so the weight of the equatorial column is γage(1-m). Since the weights of the columns have to be equal

#### age(1-m) = bgpand thus 1−m = (b/a)(gp/ge)

From the definition of ε it follows that (b/a) is equal to (1-ε). Newton had deduced that for a homogeneous oblate sphere

#### gp/ge = 1 + (1/5)ε + O(ε²)

Therefore to the first order of approximation; i.e,, ignoring terms order ε² or higher

#### 1−m = (1-ε)(1+(1/5)ε) = 1 − (4/5)ε and hence m = (4/5)ε or, equivalently ε = (5/4)m

In Newton's time it was known that m=1/290. Thus ε=1/230. The value of ε is actually about 1/305, so Newton's estimate was off by (only) about 32 percent.

## A Refinement of Newton's Analysis

Newton's analysis can be refined. What is needed is the pressures within the columns rather than the weights of the fluid. The hydrostatic equation is

#### dp/dr = −F

where r is the distance from the center of the Earth, p is the pressure and F is the downward force per unit area on a element of the column. The gravitation force per unit mass at a distance r from the center is that due to the mass of the sphere of radius r; i.e.,

#### Fp = Gρ(4/3)r³/r² = (4/3)Gρr

where ρ is the density of Earth.

For the equatorial column

#### Fe = (4/3)Gρr − Ω²r

In both cases the force is proportional to r.

The hydrostatic equation then is of the form

#### dp/dr = −fr and hence integration from the surface R down to r yields p(R)−p(r) = −½f(R²−r²)

This means that at r=0

#### p(0) = p(R) + ½fR²

The pressure at the surface can be taken to be essentially zero compared to the pressures below the surface.

For the polar column fp=Gρ and for the equatorial column fe=Gρ−Ω².

Thus pe(0)=pp(0) means that

#### (Gρ−Ω²)a² = Gρb² and therefore (b/a)² = [1 − Ω²/Gρ] or, equivalently (1−ε)² = 1 − Ω²/Gρ

The angular rate of rotation Ω for the Earth is 2π radians per day or 7.277×10-5 radians per second so Ω² is 5.2882×10-9 per sec². The gravitational constant G is 6.6743×10-11 m³/(kg*s²). The density of the Earth is 5520 kg/m³. Thus Gρ is 3.6842×10-7. The ratio of Ω² to Gρ is then 1.4353×10-2, a dimensionless number. The ellipticity is therefore approximately one half of this number; i.e., 7.1765×10-3 or 1/139.3. This is farther from the actual value than Newton's estimate.

## Colin Maclaurin's Analysis

Maclaurin established in 1742 more general formulas for the acceleration due to gravity at the equator and a pole that did not assume the deviation from sphericity is necessarily small. These formulas were:

#### ge = 2πGρa(1-e²)½[sin-1(e) − e(1-e²)½]/e³ ge = 4πGρa(1-e²)½[e − sin-1(e)(1-e²)½]/e³

The Newton condition of equal weights of the columns leads to the equation

#### ge−aΩ² = gp(1-e²)½

This condition and the formulas for ge and gp lead to the the following equation for the eccentricity of the spheroid:

#### Ω²/(πGρ) = [2(1-e²)½(3-2e²)sin-1(e)/e³ − 6(1-e²)/e²]

According to Chandrasekhar, who gives the above equation, below a certain level of the LHS there are two solutions for each value of the LHS. One solution is for a near spherical shape (small value of e) and the other for a highly flatten shape (large value of e). This was first noted by Thomas Simpson. However the graph of the RHS of the above equation shown below reveals no instance of two solutions for a value of the LHS. ## Carl Gustav Jacob Jacobi's Analysis

(To be continued.)