That page didn't have a formula, in either cartesian or polar coordinates, for the shape of the object. Lots of formulas, but I didn't see anything I could use to create a 3d mesh and print one of these things out on my printer.
It's the general ellipsoid formula: x*2/a*2 + y*2/b*2 + z*2/c*2 = 1, where a, b and c are all unequal. The interesting part is actually that this shape could be made of a liquid, held by gravitation and maintain this asymmetrical shape. Normally, one would imagine it would be ellipsoid of revolution, where two of the axes are equal.
Relevant to this is the J2 Perturbation [1], commonly used when accurately modeling the Earth as an oblate spheroid (like in the article image, but less drastic) rather than a perfect sphere. This has resultant effects on orbits, such as the "gravity wells" in the GEO belt at 105degW and 75degE. There are higher-order perturbations [2] as you closer approximate the Earth's actual shape, such as J3, J4, etc.
Hmm, that can't be true; it's of uniform density, and the thing it's rotating about has to be its center of mass.
What's odd about it (to me) is the optimal solution isn't symmetric (cylindrically symmetric). It's an intuition trap that you'd expect symmetric solutions. If the Wikipedia history is to be believed, Lagrange fell for this wrong assumption, and there was a 45-year gap before anyone noticed the subtle wrongness of it.
It has a lower symmetry than cylindrical, but it still has axes of symmetry and planes of symmetry and a center of symmetry.
The equilibrium condition only forces the symmetry condition that the points of intersection between any straight line that passes through the center of rotation with the surface of the body must be equidistant to the center of rotation.
This symmetry condition is satisfied by the symmetry group of the rectangular parallelepiped, which is also the symmetry group of the triaxial ellipsoid.
The ellipsoid can rotate around its minor axis (more stable), around its major axis (also stable) or around its medium axis (unstable, the direction of the axis will “flip”). The three axes go through the center.
That page didn't have a formula, in either cartesian or polar coordinates, for the shape of the object. Lots of formulas, but I didn't see anything I could use to create a 3d mesh and print one of these things out on my printer.
It's the general ellipsoid formula: x*2/a*2 + y*2/b*2 + z*2/c*2 = 1, where a, b and c are all unequal. The interesting part is actually that this shape could be made of a liquid, held by gravitation and maintain this asymmetrical shape. Normally, one would imagine it would be ellipsoid of revolution, where two of the axes are equal.
https://en.wikipedia.org/wiki/Ellipsoid
To make a plot in software such as JMol, you really need parametric equations, which are also given in the above page.
Relevant to this is the J2 Perturbation [1], commonly used when accurately modeling the Earth as an oblate spheroid (like in the article image, but less drastic) rather than a perfect sphere. This has resultant effects on orbits, such as the "gravity wells" in the GEO belt at 105degW and 75degE. There are higher-order perturbations [2] as you closer approximate the Earth's actual shape, such as J3, J4, etc.
[1] https://ai-solutions.com/_freeflyeruniversityguide/j2_pertur...
[2] https://oer.pressbooks.pub/lynnanegeorge/chapter/chapter-10-...
This is surprising, but less surprising when you realise it is rotating about one of the foci, not the centre.
Hmm, that can't be true; it's of uniform density, and the thing it's rotating about has to be its center of mass.
What's odd about it (to me) is the optimal solution isn't symmetric (cylindrically symmetric). It's an intuition trap that you'd expect symmetric solutions. If the Wikipedia history is to be believed, Lagrange fell for this wrong assumption, and there was a 45-year gap before anyone noticed the subtle wrongness of it.
It has a lower symmetry than cylindrical, but it still has axes of symmetry and planes of symmetry and a center of symmetry.
The equilibrium condition only forces the symmetry condition that the points of intersection between any straight line that passes through the center of rotation with the surface of the body must be equidistant to the center of rotation.
This symmetry condition is satisfied by the symmetry group of the rectangular parallelepiped, which is also the symmetry group of the triaxial ellipsoid.
Remember the spontaneously flipping spinning hammer on the ISS?
...rotation for sufficiently non-spherical(-cow) objects in free fall is kinda weird.
The ellipsoid can rotate around its minor axis (more stable), around its major axis (also stable) or around its medium axis (unstable, the direction of the axis will “flip”). The three axes go through the center.