My answer is more of a summary of insight that others have presented on a number of questions on Physics.SE.
It is true that all astronomical bodies larger than a certain mass will take on a nearly spherical shape. The logical process to arrive at this conclusion involves several steps. I will try to enumerate these with the smallest number of non-trivial steps.
- Material strength matters less and less as the scale of your system increases. A material yield strength (or other definable limit) has units of pressure (like MPa). When material structure is acting to hold the shape of something against an external force, it offers some some anisotropy in the stress tensor. As we increase the size of a self-gravitating system, the scale on which this anisotropic component can matter becomes less and less. Basically on larger scales, even rocky bodies behave more and more like a liquid than a solid. This does not preclude the existence of complex solid structures and mountains on the surface even though the vast majority the vast majority of the planet behaves mostly like a liquid (see Earth).
- When the material strength matters very little, then a single self-gravitating body will either: attain a shape that is consistent with hydrostatic equilibrium, or it will break up into pieces. In the presence of angular momentum, the hydrostatic equilibrium shape is non-spherical.
- Most astronomical bodies (but not all) have sufficiently small angular momentum such that they are sphere-like. If the angular momentum is very high, it will break up into pieces, although practically it probably never forms in such a way to begin with. There are relatively few bodies spinning below the break-apart threshold but fast enough to be highly non-spherical for reasons that I do not fully understand. The Kepler space telescope is offering new insights into the variety of planets including those with super-fast rotation and will likely shed new light on the subject.
I have a hard time understanding what the Star Wars pictures are even trying to depict, but I will focus on Lola Sayu since I think I can make out what the picture is showing. The depiction is akin to an apple with a bite taken out of it.
(Image license CC-BY-SA-3.0, Wikimedia Commons)
Specifically, here are the various reasons such a shape is unphysical for a planet:
- The core of the planet is molten, therefore it behaves as a liquid, therefore it assumes a hydrostatic equilibrium shape, and the above shape is not included, QED. Now, it would be over-generalizing without some qualifiers. The inner of most rocky planets is molten because of heat left over since its creation and internal heat production. We can potentially think of a sci-fi scenario where both of these will not be present (just set it for a trillion years in the future). We defer to the next reason.
- Even with the entire volume being fully solid, there is a separate, distinct, reason that mountains can not be higher than what the material properties will permit, and a planet with 1/4th of the matter blown away as in the case of Lola Sayu, the edges of that crater appear like a mountain to gravity. Limited material strength cannot, against the gravity force, uphold shapes so severely deformed from the hydrostatic condition.
As a final note, the pieces blow off from the planet in the picture are either in orbit, or they will be cleared away within a fairly small amount of time. Most objects will probably not remain in orbit since they are ejected from the surface, and it will more than likely return to the surface at some point in the future, per:
(Image license CC-BY-SA-3.0, Wikimedia Commons)
Now, obviously you can't tell if something is in orbit from the picture (since a still picture doesn't show movement), but it leaves plenty of unanswered questions. Where did that mass of the planet go?! Perhaps it blew away faster than escape velocity. Either way, I'm pretty sure none of the concerns mentioned here were given consideration in the creation of the artwork.
How long time does it take before three planets achieve the same relative position?
The answer is never, except for the case when their orbital periods can be expressed with low integers, like the 4:2:1 resonance of Io, Europa and Ganymede
However, what you are asking about is when they are going to be in almost the same position again, a quazi-period.
To find those periods, we are pretty much only left by brute forcing as our method. A nice little detail about the case with three planets is that the inner planet is always aligned with one of the other ones at the closest three-planet alignments. That allows us to calculate accurate solutions. In the cases of four or even five planets I simply give up.
To check all the possibilities, we can use a program. Here is an example of a function in JavaScript returning a list of quazi-periods and alignment error:
sameLine = function (period1,period2,period3,limit){
results = [];
newMargin = 1;
synodic1 = 1/(1/period1 - 1/period2);
anomaly1 = (synodic1/period1) % 1;
synodic2 = 1/(1/period1 - 1/period3);
anomaly2 = (synodic2/period1) % 1;
alert(synodic1+","+synodic2);
for (i = synodic2; i < limit; i+=synodic2){
numb1 = i/synodic1 - (i/synodic1) % 1;
numb2 = i/synodic2 - (i/synodic2) % 1;
err1 = Math.abs((numb1 * anomaly1 - numb1 * synodic1/period3) % 1);
err2 = Math.abs((numb2 * anomaly2 - numb2 * synodic2/period2) % 1);
if (err1 > 1 - err1){
err1 = 1 - err1;
};
if (err2 > 1 - err2){
err2 = 1 - err2;
};
if ((err1 < newMargin) && (numb1 > 0)){
results.push([numb1 * synodic1,err1]);
newMargin = err1;
};
if ((err2 < newMargin) && (numb2 > 0)){
results.push([numb2 * synodic2,err2]);
newMargin = err2;
};
};
return results;
};
For Jupiter, Saturn and Uranus, I get the following output:
Time,,,error
13.81170069444156,,,0.30449020900657225
39.71676854387252,,,0.12441143762575813
41.43510208332468,,,0.08652937298028318
139.00868990355383,,,0.06455996830984656
138.1170069444156,,,0.04490209006572288
179.55210902774027,,,0.041627282914560304
317.6691159721559,,,0.0032748071511630172
3991.581500693611,,,0.002329597100612091
4309.250616665767,,,0.0009452100505313865
The first of this periods is of no use, as the error in alignment is almost a third of an orbit. Note that the one you found (that is really impressive you did,actually) gives an error in the alignment of less than a percent. We have to look at periods more than a thousand years long to find any better alignment.
Be sure to feed this function with accurate orbital periods.
Best Answer
First, Mercury "aligns" with the ecliptic plane only twice in its "year", when it comes from above to below and vice versa.
Luckily for our calculations, Pluto is not a planet any longer, because it would completely rain on our parade with its 248 Earth years of orbital period and another two points within it that it crosses the plane again. Getting Pluto and Mercury aligned alone would take millennia.
Now, what do we count as "aligned"? This is a very vague term because it doesn't state any tolerances. If you mean discs of the planets overlapping, just forget it, their own minor deviations from the ecliptic plane will suffice that it will never ever happen. Let us assume a tolerance of one earth day of their movement. This is fairly generous, in case of Mercury it's over 4% tolerance of its total orbit radius, which considering their size on the sky is quite a lot - in case of all planets the distance traveled over one earth day far exceeds their diameter. So, we're not taking a total alignment, just one night where they are closest to each other, a pretty loose approximation.
Now, we pick the day the rest of the planets are on the plane as Mercury, so let us simply take the 2 in 88 days of its orbital period and continue dividing by orbital periods of other planets.
1 in (44 * 225 * 365 * 687 * 4332 * 10759 * 30799 * 60190) days. That is one day in $5.8 \cdot10^{23}$ years. The age of the universe is $1.375 \cdot 10^{10}$ years.
It means planets would align for one day in 42 trillion times the age of the universe.
I think it's a good enough approximation to say it is not possible, period.
Feel free to divide by 365, if you don't want aligned with the Sun but only with Earth. (one constraint removed.) It really doesn't change the conclusion.