OK, dumb question: When astronomers describe the new planets that have been discovered outside the solar system, they use the term "gas giant." This term is especially vexing because it implies these are Jupiter-like planets: huge bodies composed of lighter elements that are "blown" into the farther reaches of the solar system by the parent star. Yet some of these planets orbit their suns very closely, which has astronomers scrambling to alter their theories about solar system formation. My question is this: Why must these planets be "gas giants"? Why can't they simply be very large terrestrial-style planets, or very dense planets? I mean, gravity doesn't care if it's a big planet with low density or a little planet with high density, right?
I don't think this is a dumb question at all. There is no visual evidence of the planet itself. Its presence is inferred by the famous "wobble measurement" on its parent sun. Forgive me if I wing the following, without rigorously checking the maths, but here's what I imagine is the process: 1. Measure the period of the wobble, and measure the amplitude of the wobble. 2. Come up with an estimate of the sun's mass, by using the database of stars and the measured redshift / distance estimate. I think this is possible. 3. We can now deduce (I THINK) the mass and orbital radius of the unseen planet. The above implies that planetary density has nothing to do with it (OK, to a first order, in case Boris or Plato are spying on me Please Register or Log in to view the hidden image!). In other words, I think you're right on.
You are right of course in assuming the planet's volume haven't been observed but there are some reasons that plead for gas giants. For one if a planet would have a mass of 10 times the one of Jupiter (which is the case for some of the wobbles) and it has a volume of the earth then it's density must be 10000 times that of water. It is impossible for ordinary matter to have such a density (for example lead or gold has a density of 15-20 times that of water). What we are talking about here is degenerated matter like we find in white dwarves. The only way (we know) a white dwarf can evolve is as a result of star evolution and 10 jupiter masses is not enough for star ignition. So unless some other exotic mechanism is at work we are safe to assume the wobble are gas giants with normal densities. ------------------ we are midgets standing on the backs of giants, Plato
OK, how about a terrestrial-style planet with a 10th the volume of Jupiter? Wouldn't that produce the gravitational effects while avoiding the density problems?
I assume with a terrestrial planet you mean a planet with the same density of Earth ? This means about 5 times the density of Jupiter (earth is more massive). So then you say : V_p = 1/10 V_J (with V_p the volume of the planet and V_J the volume of Jupiter) and D_p = 5 D_J (D_p density planet and D_J density Jupiter) So the mass of the planet is M_p = V_p D_p = 1/2 M_J. So you only have half of the Jupiter mass, while the planets are up to 20 times more massive. So if you want a 10M_J planet with earthly density you need a planet with a volume two times the one of Jupiter. Maybe such an object exists, as long as we don't have direct visual images of the planets (and this means telescopes with mirrors of 100 meters or more) we can't be sure. ------------------ we are midgets standing on the backs of giants, Plato
The reason they believe the planets are made of lighter elements is (probably, my guess) because they know the abundance of heavy elements in the parent star. The 'metal' content of a star can be easily measured by looking at its spectrum. Sun is a rather unusual star because its metal content is very high compared to most other stars out there. Hence, it is not surprising to find so many rocky bodies in the inner solar system. However, if a star is metal-poor, one would be very surprised to find large metallic objects orbiting close to the star, with masses on the order of 10 Jupiters! (By the way, in astronomical jargon 'metal' refers to anything heavier than Hydrogen.) ------------------ I am; therefore I think.
I am confused how astronomers can calculate both orbital distance & mass don't both variables affect he stars wobble? wouldn't a planet with a mass 2.5 times of jupiters at say 2 AU effect the same wobble as a planet with a mass 5 times jupiters at 4 AU [This message has been edited by god (edited June 29, 1999).]
That is exactly right, Boris. Besides 99 % of all matter is in the form of the extreamly light elements hydrogen and helium so it would be highly unlikely that one planet in a solarsystem suddenly has this abundance of iron. That is why the inner planets in our solar system are so small in comparison to the outher gas giants, they are a 'good' representation of the distribution of heavy elements on the kosmological scale. About the wobble, what the astronomers mean by the wobble is actually the period of a revolution of a planet around it's star. Remember the third law of Kepler : T^2 = k R^3 with T the period, k a constant depending on the mass of the star (this starmass is an estimation based on spectral type and its velocity around the centre of the galaxy) and the gravitational constant and R the mean radius of the orbit. So you see the mass of the planet has nothing to do with its distance from its star. ------------------ we are midgets standing on the backs of giants, Plato [This message has been edited by Plato (edited June 30, 1999).]
A planet with 2.5 jupiter's mass at 2 AU and a planet with 5 jupiter's mass at 4 AU would produce same MAGNITUDE wobble, but the DURATION of each wobble would be longer for the longer orbit. Wobble duration, i.e. orbital period, does not depend on the planet's mass, but only on the star's mass and the planet's distance from it.
Hmmm ... I'm a little confused by this idea of auditing the elements of a star via spectroscopy ... I mean, wouldn't only hydrogen, helium, and maybe carbon show up on any spectrograph? How do the heavier elements show up? Where are they?
Del: Anyone who ever had even an elementary introduction to atomic structure wouldn't be asking such questions. So, I assume you aren't very familiar with the subject. Hence forgive the, perhaps, somewhat too detailed explanation. Atoms are composed of a nucleus 'orbited' by a cloud of electrons. For light elements like hydrogen or helium, the nucleus is very small (one proton-neutron pair for H, two for He -- ignoring isotopes for the purposes of this discussion). For heavy elements, there are dozens of protons and even more neutrons in the nucleus. Because the nuclei are made of protons (charge +1) and neutrons (charge 0), they are overall positively charged, and the amount of their charge is equal to the number of protons a nucleus contains. The number of protons in a nucleus also defines the element itself: one proton -> H, two protons -> He, three protons -> Li, etc. Normally, elements are neutral -- which means that an atom has no overall charge. To make this happen, we must cancel out the positive charge of the nucleus with the corresponding number of electrons (charge -1 each). Hence, an element would have as many electrons orbiting its nucleus as it has protons in the nucleus. Thus, H has one electron, He has 2, Li has 3, etc. Due to quantum mechanics, electrons can't orbit the nucleus just anywhere. They are confined to narrow regions of space, called 'orbitals'. The electrons also possess something called a 'spin'. Due to Pauli exclusion principle, no two electrons with the same spin can inhabit the same orbital. So, each 'full' orbital would contain an electron spinning 'up', and an electron spinning 'down'. The orbitals form a spacial 'hierarchy', i.e. they are arranged in 'levels'. Closest to the nucleus, you have a level 1 'S' orbital which can only contain two electrons. Above this S1 orbital, we have level 2, which contains its own S orbital, and 3 P orbitals; above level 2 we have level 3 with S, P, and D orbitals, and so on. The further away from the nucleus an orbital is, the higher its energy. Electrons naturally want to occupy the lowest-energy orbital available, to be as close as possible to the nucleus. However, when an atom receives an energy influx, such as from an incident photon, the electrons in its outer shells can be kicked up to higher-energy orbitals. They stay in those high-energy orbitals for a while, but then transition back to the lower-energy orbital, at the same time giving off the excess energy in the form of a photon. Now, the energy differentials between various orbitals are all different. And, the energy given off by an electron transitioning to a lower-energy orbital is directly represented in the frequency of the photon that is created. Hence, by just looking at the frequency of different photons emitted by excited atoms, and using some black-body emission physics, we can figure out exactly what element the atom belongs to, and what the atom's 'temperature' is. In addition to emitting at wavelengths specific to their element, atoms also absorb light at very specific wavelenghts. This is how scientists identify elements in stars -- by looking at the 'absorption spectrum' of a star. The spectrum of a star in general is just a plot of intensity vs. frequency -- i.e. y represents how many photons of frequency x we managed to detect in time delta t. Because various elements absorb light at specific frequencies, and re-emit the energy at other frequencies, emission spectra in general tend to have 'absorption lines' in them -- where for certain frequencies the light intensity is much less than for other frequencies. Again, for each element the set of absorption lines, and the frequency spacing between them, is unique. So, by simply performing absorption spectrum analysis of a star, it can be determined what elements, and in what abundance, are present in the star (and especially in the star's outer layers). In reality, the process is somewhat complicated by the fact that atoms are usually ionized in stars -- i.e. they lack at least one electron from their outer shells -- which somewhat affects their absorption spectra. Nevertheless, ions as well as intact atoms have unique absorption signatures, so direct measurements of element abundance in stars are indeed practical. I hope I didn't confuse you too much, but the main point is that spectroscopic analysis and absorption spectra apply to light elements just as well as to heavy ones -- there's no fundamental distinction between the two classes, because all elements and their ions have unique spectral behavior. ------------------ I am; therefore I think.
Boris, Wow! Thanks very much for the explanation (and you're right -- I've had no such training). Very interesting.
