# Rotation of planets

1. Sep 14, 2010

### Parbat

why should the planets rotate?

2. Sep 14, 2010

### D H

Staff Emeritus
Because they were rotating when they formed.

Edit
By rotate do you mean orbit (the Earth orbits the Sun in a year) or revolve (the Earth revolves about its axis in a day)?

Last edited: Sep 14, 2010
3. Sep 14, 2010

### NeoDevin

Because they have angular momentum.

4. Sep 14, 2010

### Parbat

I mean rotating about their own axis.

5. Sep 14, 2010

### zhermes

As NeoDevin said: 'because they have angular momentum'

The planets formed out of a roughly continuous disk of gas around the sun; as the planets form (complex process overall), the probability of having near-zero angular momentum is extremely small. This is amplified by the many large-scale collisions that happen in the later formation-history of planets.

6. Sep 16, 2010

### Kometkaj

If you were to calculate on a dusty gascloud collapsing, you would decipher it into "multi-particles motion" eg. 1000 elements heading for the same gravity-center.
1. As they close in on the center, density rises and chances of collisions increase.
2. The gascloud also orbits something else: planetary formations usually orbits a protostar. This means that the parts of the collapsing dustcloud that are closest to the protostar have a higher orbital velocity than the farthest (This is actually the "Coreolis force").
3. So when they collide, the dustgrains in the outer cloud are gaining speed, as they are drawn inwords, while the inner part of the cloud are drawn outwords, loosing velocity
4. This creates the rotation when they accidentially collides.

So the rotation is not only a question of momentum-probability due to random collisions, but is actually an inevidable consequence of the diffenerences in orbital momentums in the inner and outer part of the collapsing cloud.

Last edited: Sep 16, 2010
7. Sep 16, 2010

### Kometkaj

Btw:
This is not only the case for planetary formation, but for everything that orbits something.
This is also why stars rotate, and why they rotate in the same orientation of the galactic plane.

Ever wondered why we can see so many exoplanetary transits, when the chance of seeing an eclipse should be small?
It's because they all rotate in roughly the same orientation, for the same reasons as my above answer

I hope i helped you, rather than confused you :O)

Henrik

Last edited: Sep 16, 2010
8. Sep 16, 2010

### Andre

There is a hypothesis currently that assumes that our solar system is the remnant of a super nova. Not sure if that agrees with a dusty gas cloud collapsing.

9. Sep 16, 2010

### Kometkaj

Hi Andre.

Well actually it is a well established theory, that there was a type II supernova prior to the creation of the Sun. This is quite normal for starcreations; When heavy stars that usually only lives <1mio years are born, they explode and tricker a secondary wave of starformations in the surrounding interstellar clouds.

The new hypothesis is that there also was a type Ia supernova somewhere around the creation. This theory is based on findings of some characteristic isotopes in meteorites.
Se http://news.uchicago.edu/news.php?asset_id=2087

This will not effect smaller scale collapses, such as individual star- and planetary creations.

The interesting thing about this theory is in my opinion, that it makes Earth special, and could thus give us something to look for, when we are trying to find extra-terrestial life.

/Henrik

10. Sep 16, 2010

### D H

Staff Emeritus
Not true. The plane of the ecliptic is inclined by about 63 degrees with respect to the galactic plane. Other star systems, who knows? It's pretty much random.

As I mentioned above, the orientation of a star system is pretty much random. Suppose some other Sun-like star truly has an Earth-like planet in an Earth-like orbit. Since the orientation of one star system with respect to another is pretty much random, the chance of seeing that planet transit its star is only about 1/200. This low probability is the driving reason why the Kepler mission is looking at more than 100,000 stars. They want to ensure that a null result (failing to find any Earth-like planets in Earth-like orbits about Sun-like stars) is truly significant.

11. Sep 16, 2010

### Kometkaj

I must admit, that i have never seen any observational documentation for the claim
I have seen several sources that refer to a common inclination of the stellar orbits with respect to the galactic plane.
But whether it is a 20/80 relationship or a more significant number, I am not sure ..?

Anyway; If the orientation was completely random, a 1:200 succes-rate for planetary eclipses wouldn't even begin to describe it ;O)

/Henrik

12. Sep 16, 2010

### Rebound

Another minor point of interest:
These precursor stars are absolutely necessary for the formation of the elements aside from hydrogen, helium, and lithium so without them, we would not exist.

13. Sep 16, 2010

### D H

Staff Emeritus
Yeah, and they're all wrong. The internet is full of nuts. You can find lots of sites that say that the ecliptic is only inclined at an angle of 5 degrees with respect to the ecliptic. I don't comprehend why people bother publishing information that is obviously false. That's the internet for you.

Have you done the math to prove this assertion? The Kepler mission people did. The 0.5% figure assumes zero correlation between the orbital plane of the exoplanets about some other star and the Earth's equatorial plane. That is one of the reasons it has to look at so many stars.

14. Sep 16, 2010

### Barakn

It certainly would be difficult to look for Earth-like planets, but there are a handful of factors that influence the probability of whether a planet gets found via the transit method. Of primary importance is the semi-major axis of the orbit. Imagine the very hypothetical scenario of a (doomed) planet whose orbital radius is the same as the radius of the star it orbits, i.e. they are touching. In this case, the only point of view from which we would NOT see transits would be when looking at this system along the axis of revolution, a very low probability. If you look at the statistics of planets found via the transit method, you'll find the vast majority have a semi-major axis less than 1/20 of an AU. http://exoplanet.eu/catalog-transit.php

Obviously the larger a planet is, the more likely that some portion of it will eclipse the star if it happens to pass around front. Also the bigger the planet the more light of the star it will block, and thus the easier it will be to pick out of the statistical noise. A quick look at the stats reveals the peak of the radius distribution of planets found via transits is just over one Jupiter radius. This effect is less strong than for orbital radius.

One might also initially think one would be more likely to catch a planet passing in front of a star with a bigger radius - not only is the stellar disk bigger but the planet is less likely to be passing along the stellar limb and more towards the stellar center, where the star is brighter. But also the bigger the star the less of a percentage of its surface area is blocked by the planet, making it harder to spot in statistical noise. But then again the larger a star is the more likely it is more luminous (easily said for main sequence stars), and the more luminous (at a given distance) the cleaner the data. I haven't had time to peruse the properties of the stars around which planets have been seen transiting, but if I get a chance I will.

15. Sep 16, 2010

### Kometkaj

Yes; Assuming that we can track and monitor an entire orbital revolution, then the apparent diameter of the planet from the stars perspective, should cover 360'/200 = 1,8degrees, if the probability for an alignment that allows us to see an eclipse should be 1:200 or 0,5%

Most of the exoplanets found so far have been closely orbiting Jupiters, but still; Whitout having looked into it, 1,8degrees sounds bigger or closer than the average of the observed exoplanets i have read about.

There might be a bias in my perception, since i have not done any quantitative statstics, but... Am i mistaken here?

/Henirk

16. Sep 16, 2010

### Dr Lots-o'watts

Besides all the formation mechanisms, note that practically anything that flies will rotate (unless it's designed not to), it's just natural. Non-rotation is a special, rare case of of zero angular speed. A ball that is thrown or that has just collided will also rotate.

17. Sep 16, 2010

### D H

Staff Emeritus
Transits are not really eclipses. The planet is a tiny speck that blocks a tiny amount of the light coming from the star. For example, an Earth-like planet in an Earth-like orbit about a Sun-like star will reduce the amount of light coming from the star by 80 or so parts per million during a transit.

To see a transit the transit has to occur. Imagine a planet that is orbiting a star such that the orbital angular momentum vector is pointing straight at our solar system. We will never that planet transit its star. The planet never passes in front of its parent star from our perspective. Now imagine that you are a god and you can grab onto the orbital angular momentum vector and tilt the planet's orbital plane downward. As you tilt the orbit more and more, the orbit will at some point clip the top of the star (from our perspective). Tilt it a just bit more and the planet will pass in front of the star, but only for a short while. Kepler can't see these short duration transits that only block a small amount of the star's output. The transit duration will increase as you tilt the orbital plane ever more until you reach the point where we are seeing the orbit edge-on. For Kepler to reliably see an Earth-like planet in an Earth-like orbit the orbit has to be close to edge-on.

18. Sep 17, 2010

### Barakn

I'm not sure why you think the apparent diameter of the planet from the star's perspective matters, as the planet and star are viewed from the perspective of Earth. Assuming the Earth is infinitely far away (not a bad approximation for geometrical optics), and that the planet is much smaller than the star (allowing yet more judicious use of the Small Angle Approximation), a planet will be seen to transit the star when it's orbital inclination i varies from $$\pi$$/2 radians to arccos (R/d), where R is the stellar radius and d is the planet's orbital radius at the time of transit. Since d>>R, i is usually only slightly smaller than $$\pi$$/2 radians. Since, as noted by other posters, i varies randomly, the fraction of planets that transit is ($$\pi$$/2 - arccos (R/d))/($$\pi$$/2) = 1 -2 arccos(R/d)/$$\pi$$ for any particular values of R and d held constant, but coming up with a grand answer would entail knowing the parameter space of R and d.

Since the planetary diameter seems to matter to you, I came up with a formula including r, the planetary radius. The inclinations resulting in transits vary from $$\pi$$/2 radians to arccos ((R + r)/d). Using the Maclaurin expansion for arccos, arcos x = $$\pi$$/2 -x - x3/6 - 3 x5/40 ... (radians)
i = arccos ((R+r)/d) ~= $$\pi$$/2 - (R+r)/d - (R3+3R2r + 3Rr2/ (6 d3)) ...
Since typically d>>R>>r, this simplifies to i ~= $$\pi$$/2 - R/d.

Once again I note how important d is. I also mentioned that bigger R increases the probability of a transit, but as R gets bigger one the percentage of the light blocked by planet of radius r gets smaller, making it harder to see.

Some problems with Latex, am getting an approximately equals sign where a pi should be and can't fix it.

19. Sep 17, 2010

### Barakn

When the Moon is far away enough from the Earth and comes between it and the Sun, the Moon doesn't quite cover the Sun's surface and appears surrounded by an ring of light. This is known as an annular eclipse, and is quite similar to a transit, which would also feature something smaller passing in front of and not completely obscuring a star's surface. Furthermore, if you look up the definition of the word eclipse, you'll see that the word eclipse is used when any celestial body is obscured by another celestial body or is obscured by another's shadow. In other words, a transit is an eclipse.

20. Sep 17, 2010

### D H

Staff Emeritus
I would not call a housefly eight miles or so distant passing in front of the Moon an eclipse. Nor would astronomers. The generic term for occultations, transits, and eclipses is syzygies (singular: syzygy). Astronomers reserve the word eclipse for those syzygies in which the obscuring body and the obscured body have nearly equal angular sizes. Yes, there is a gray area between eclipses and transits, or eclipses and occultations. A housefly at eight miles passing in front of the Moon is not anywhere close to that gray area. It is transiting, not eclipsing, the Moon. The amount of moonlight obscured by a housefly eight miles away transiting the Moon is what the Kepler satellite is trying to see, BTW.