Ellipsoidal nature of the Earth

[natski]

Flattening of a planet

Hi all,

Does anyone know how to calculate the flattening of a planet, say a gaseous one for simplicity, simply from it's radius and rotational period?

Natski

 

[Redbelly98]

I think you could calculate the gravitational potential due to two terms, gravity itself and the centrifugal force due to the rotation. Set that potential equal to a constant to find a possible surface shape.

Grav. potential:
PE_grav = - G M1 M2 / r
where
M1 is the planet's mass and M2 is a "test mass"
r is the distance from the planet's center to the surface.

Centrifugal force:
M2 w^2 r cos(latitude)
where
M2 is the same test mass as before
w is the rotation rate in rad/sec

From the cent. force, we get the Centifugal potential is
PE_cent = -(1/2) M2 w^2 r^2 cos(latitude)^2

Set the total potential equal to a constant, and solve for r as a function of latitude:

PE_grav + PE_cent = constant.

Some hints:

1. To find the constant, at least approximately, neglect PE_cent and plug in the known mass and average r in the PE_grav expression.

2. Looks like you'll get a cubic equation to solve. Ouch! This approximation might be helpful:
|PE_cent| << |PE_grav|
Not that you'd neglect PE_cent. But you might be able to do some sort of series expansion, and drop higher-order terms.

 

[stevebd1]

Equatorial bulge (or flattening) can be calculated using the following-

$$\textit{f}=\frac{a-b}{a}=1-b:a \approx\frac{3\pi}{2GT^{2}\rho}$$

where a is the long radius (equatorial), b is the short radius (polar) G is the gravitational constant, T is the rotation period (in seconds) and $$\rho$$ is density

http://en.wikipedia.org/wiki/Equatorial_bulge

regards
Steve

 

[natski]

Thanks for that Steve, it is a shame Wikipedia has no reference for that equation...

 

[Redbelly98]

I was able to derive the Wikipedia equation using the method I described in msg #2 of this thread.

I can post it if anybody is interested, but it'll be ascii math as I'm not yet proficient in Latex.

 

[stevebd1]

I'd previously had a look at how to reformat the equations so that they equalled the equatorial radius which I posted over in the maths forum-

https://www.physicsforums.com/showthread.php?t=231579

I got it to a level where it was a relatively straight forward trial and error process, then it turned into a cubic equation.

Redbelly, feel free to post your method though I suggest you try latex, it's a lot better than using type set and quicker to understand.

Steve

 

[Redbelly98]

I'm about halfway through a Latex version of my derivation, but have to log off for the night and will pick it up again tomorrow.

By the way, I inadvertently posted an incomplete version that appeared here briefly; I've deleted that post until the complete derivation is ready.

 

[natski]

Sounds good, are you assuming some kind of uniform density sphere and using the equations of hydrodynamics?

 

[Redbelly98]

The density need not be uniform, it can vary with distance from the planet's center.

I use a a reference frame rotating with the planet, so that the forces on a test mass m are the gravitational force and a centrifugal force. I find the potential energy (of the test mass) due to these forces, which is a function of:
r, the distance from the planet's center
The latitude.
The planet's rotation rate and mass.

Setting the potential equal to a constant defines an equipotential surface, which must be the case for the planet's surface (if it's not, then material would flow or fall "downhill", in order to make it so).

By solving the potential=constant equation we get the center-to-surface distance as a function of latitude, and can use that to determine the ratio of polar/equatorial radii.

As I'm using Latex for the first time, for me it's slow going typing in the symbols.

edit added:
Time for me to head into work soon, tonight I should have time to finish it up.

 

[Redbelly98]

Okay, here is my derivation by request

This is Part 1 (out of 2). Here I'll derive an exact relation between radius and latitude. In part 2 I'll make approximations to get a solution for r.

In a reference frame rotating with the planet, the forces on a test mass m are the gravitational force and a centrifugal force:

$$F_{grav} = -\frac{GMm}{r^2}$$, directed inward toward the planet's center

$$F_{cent} = m\omega^2 r \cos(\theta)$$, directed outward from the planet's axis

where
$$M$$ is the planet's mass
$$r$$ is the distance from the planet's center to a surface point
$$\omega$$ is the planet's rotation rate (rad/sec)
$$\theta$$ is the latitude at a surface point (0 at the equator, 90 degrees at the poles)

The potential energy of the test mass is then

$$U = -\int\vec{F}\cdot\vec{dr}$$

$$= -\frac{GMm}{r} - \frac{1}{2} m \omega^2 r^2 \cos^2(\theta)$$

Note we picked up an extra cosine factor in the centrifugal term, because force and radial vector make an angle to one another:

Work = force x displacement x $$\cos(\theta)$$

The planet's surface conforms to an equipotential surface. In particular, the potential energy at a pole equals the potential energy anywhere else on the surface:

$$-\frac{GMm}{r_{pole}} = -\frac{GMm}{r} - \frac{1}{2} m \omega^2 r^2\cos^2(\theta)$$

Multiply both sides of this equation by the factor

$$-\: \frac{r}{GMm}$$ :

$$\frac{r}{r_{pole}}= 1 + \frac{r^3 \omega^2}{GM}\cos^2(\theta)$$

or

$$\frac{r}{r_{pole}}= 1 + (\frac{r}{r_{pole}})^3 \cdot\frac{r_{pole}^3\omega^2}{GM}\cos^2(\theta)$$

This is an exact equation, using dimensionless variables and parameters:

$$\frac{r}{r_{pole}}$$ ,

$$\frac{r_{pole}^3\omega^2}{GM}$$ ,

and

$$\theta$$

While cubic equations are analytically solvable, the formula involved is quite cumbersome. In part 2 I'll use the approximation $$r \approx r_{pole}$$ to get a solution for r.

 

[Redbelly98]

Will have to continue this tomorrow (Friday). I'm getting error messages when I try to preview my Latex code and am getting too tired to figure out what the problem is. Frustrating.

 

[Redbelly98]

I dropped a factor of 1/2 earlier, the relation between r and latitude should read

$$\frac{r}{r_{pole}}= 1 + (\frac{r}{r_{pole}})^3 \cdot \frac{1}{2} \left[\frac{r_{pole}^3\omega^2}{GM}\right] \cdot \cos^2(\theta)$$

Part 2 of derivation

We're after solutions for r in the above equation. We assume that r does not vary too much over the planet's surface, in other words

$$\frac{r}{r_{pole}} = 1 + \delta$$

where

$$\delta \ll 1$$

Our equation becomes, to 1st order in $$\delta$$ :

$$(1+\delta) \approx 1 + (1+3\delta) \cdot \frac{1}{2} \left[\frac{r_{pole}^3\omega^2}{GM}\right] \cdot \cos^2(\theta)$$

or

$$\delta \approx (1+3\delta) \cdot \frac{1}{2} \left[\frac{r_{pole}^3\omega^2}{GM}\right] \cdot \cos^2(\theta)$$

Evidently the term in square brackets is of order $$\delta$$ , which becomes of order $$\delta^2$$ upon multiplication by $$3\delta$$ . As we are neglecting terms of order $$\delta^2$$ and higher, we may drop the $$3\delta$$ from the equation so that

$$\delta \approx \frac{1}{2} \left[\frac{r_{pole}^3\omega^2}{GM}\right] \cdot \cos^2(\theta)$$

and

$$\frac{r}{r_{pole}} = 1 + \delta \approx 1 + \frac{1}{2} \left[\frac{r_{pole}^3\omega^2}{GM}\right] \cdot \cos^2(\theta)$$

.

Part 2a. Comparison with the Wikipedia result.

Wikipedia gives the ratio of polar to equatorial radii as

$$1-\frac{r_{pole}}{r_{equator}} \approx \frac{3\pi}{2GT^{2}\rho}$$

or

$$\frac{r_{pole}}{r_{equator}} \approx 1 - \frac{3\pi}{2GT^{2}\rho}$$

where $$\rho$$ is the planet's density and T is the rotation period.

Our equation gives, using $$\theta = 0$$ degrees for the equatorial radius,

$$\frac{r_{equator}}{r_{pole}} \approx 1 + \frac{1}{2} \left[\frac{r_{pole}^3\omega^2}{GM}\right]$$

We make substitutions

$$\omega = 2 \pi / T \text{ and } \frac{M}{r_{pole}^3} \approx \frac{4\pi\rho}{3}$$
(The expression involving density is valid in the approximation $$r \approx r_{pole}$$ everywhere on the surface.)

and get

$$\frac{r_{equator}}{r_{pole}} \approx 1 +\frac{3\pi}{2GT^{2}\rho}$$

So that

$$\frac{r_{pole}}{r_{equator}} \approx \frac{1}{1 +\frac{3\pi}{2GT^{2}\rho}}$$

or

$$\frac{r_{pole}}{r_{equator}} \approx 1 - \frac{3\pi}{2GT^{2}\rho}$$ ,

in agreement with the result given in Wikipedia.