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We saw in the last lecture that once something starts to grow,

Â it begins to grow even faster.

Â And you might think, okay, what's ever going to stop it?

Â Why doesn't it take over all of the material in the entire universe?

Â Well, first, that would be a lot of material to take over, but at least why

Â doesn't it take over all the material in the disc that it's forming out of?

Â Well, let's think about why this would be.

Â If there was a star in the middle of this disk.

Â I'm now doing a top down view of it.

Â And the disk is here around the star, and we're talking about,

Â let's say something that was starting to form right here by all the collisions that

Â are happening inside of it.

Â Eventually, it's true that the body that starts to grow there,

Â the one largest body will continue to get larger and larger and larger,

Â until it has basically cleared out all the material in one annulus of the disk.

Â And it will continue to clear out the annulus until it gets to

Â a stage where the cleared out annulus is so

Â big that the gravitational influence of this singular body

Â is smaller than the gravitational influence of the star.

Â That means that this stuff on this side of the annulus

Â is no longer gravitationally attracted to that body.

Â Do you remember when we did that idea of gravitational focusing, we considered

Â a body, we considered something coming in, we considered it coming like this?

Â And this thing was off at infinity, infinitely far away.

Â That part's not really true,

Â all the things are actually in orbit around the star.

Â And the further you are away from this body,

Â the more the influence of the star will be.

Â So lets figure out how far this is.

Â Once we figure out how far this is,

Â we'll know how big of a range this body is able to feed from.

Â And then if we can figure out how much material is in this range,

Â we can figure out how big this object gets to be.

Â So how big of a range does this object feed through?

Â Well you could think of doing it in a couple ways,

Â you could say there's a star here, there's this growing planetesimal here.

Â And you want to figure out whether a particle here,

Â at distance r from the planetesimal, is more influenced by the planetesimal and

Â can have its gravitational pull pulled into there.

Â Or if it's more influenced by the star over here.

Â You can do something like simply calculate the force from the star

Â to this little particle, and this force from this planetesimal to the particle.

Â But you wouldn't get the right answer if you did that, the reason is because

Â this is an orbit around, this is not just two things sitting there nailed down and

Â you're waiting to see which has a stronger force on there.

Â The more accurate way to do it is to take a particle right here and

Â ask yourself the question, if this particle were in orbit around the star,

Â what would its orbital period be?

Â If this particle were more in orbit around the planetesimal,

Â what would its orbital period be?

Â If its orbital period is smaller around the planet than it is around the star,

Â then you can see that it's being influenced by the planet more.

Â It doesn't actually mean that it's orbiting around the planet

Â necessarily, but

Â it does that mean that that gravitational influence of the planet is strong.

Â And that's a good indication that it will eventually come under the gravitational

Â influence of that planet, and in this case planetesimal, and

Â in this case it will be gobbled up.

Â And that would be part of the feeding zone of that planet, planetesimal.

Â If instead you're over here and the orbital period around here is quite long,

Â this is quite short, or they're comparable.

Â Then you're really more just in orbit around the star, you never will be really

Â influenced by that, influenced enough to cross its orbit.

Â Okay, let's calculate the period of something that's sitting here in orbit

Â around here, or something that's sitting here in orbit around here.

Â We need one more distance, we need the distance between these two.

Â We'll call that a.

Â We need the mass of the star.

Â We need the mass of the planetesimal, little m.

Â And we need the mass, we don't actually need it, but we'll use it for

Â min at the mass.

Â The little particle we'll call mass dot, where dot means little particle.

Â How do we figure out the orbital period of one of these things?

Â Well, the easiest way is to remember that the force on this object is

Â force is GMm dot over r.

Â We're going to assume that the distance is really essentially a,

Â because this r will be quite small, over r squared.

Â So that's the force due to the sun, that force is counterbalanced by

Â the centrifugal force as this object is going around the sun.

Â The centrifugal force, as you remember from high school physics,

Â is m.v squared over r.

Â So that r is an a.

Â So we have a velocity, but now we wanted to know the period.

Â What is the period?

Â Well, the period is how long it takes all the way to go around,

Â which is simply the circumference of this circle, which is 2 pi a.

Â Over the velocity, which is v and that's equal to the period.

Â Okay, so let's calculate the period of this particle that

Â would be in orbit around this star.

Â And we get that period equals 2 pi a over v

Â now is going to be the square root of GM.

Â The M dots cancel out, I said they didn't actually matter,

Â divided by a or, it's equal to 4 pie squared a cubed over GM.

Â If we do the same thing for the particle in orbit around the planetesimal we would

Â get something very similar.

Â We would get that that period Is 4 pie squared r cube,

Â because that's the distance from here to here, divided by G little m now.

Â Now let's set those to equal and see what we get.

Â Well, these 4 pi squares are going to cancel out,

Â the G's are going to cancel out.

Â We can square both sides so the square roots go away, and

Â what we really end up with is that a cubed over big M equals r cubed over little m.

Â Or simply that r is a times the cube root of little m over big M.

Â This sort of makes sense, it's proportional to the distance away.

Â The further away the object is from the sun,

Â the more of an influence it can have compared to the sun.

Â And it is proportional to the cubed root now of the mass

Â of the planetesimal divided by the mass of the star.

Â The bigger the planetesimal, the bigger its influence.

Â The bigger the star, the smaller its influence.

Â Okay, we can now figure out how much material this thing gets.

Â If the star's in the middle, and this is here,

Â there's an annulus in this feeding zone that this guy gets to eat up, and

Â that is the annulus that has a length of 2r.

Â It has a circumference of a, and so

Â we can approximate the area of this annulus as 2 pi a.

Â That's the circumference here, times 2r, where r is this r.

Â And then, all we need to know is how much material per unit area this has.

Â I'm going to give that, call that a sigma.

Â We'll talk about sigma in a minute.

Â Sigma has the units of kilograms per meter squared.

Â That means if I'm looking down at the disk,

Â I'd like to know regardless of how thick that disk is in height.

Â I'd like to know how much material per meter squared is there.

Â Because I don't care how high it is, how low it is.

Â All that material is going to go into this object.

Â This surface density, it's called,

Â is an important parameter in thinking about planetary systems.

Â We'll talk about it in just a second.

Â So the mass is equal to, as I said, this.

Â But r is equal to this.

Â So we could say that the mass, the isolation mass we're going to call it now,

Â because that's how much mass you can get before you become isolated from the disk.

Â The isolation mass is 2 pi a, times 2r,

Â which is a times cubed root of m isolation.

Â That's your isolation mass, mass of the star times this sigma.

Â Notice we have isolation mass over here, isolation mass over here.

Â We're going to cube both sides, and we're going to then start to ignore factors like

Â 2pi and 2 and all those cubed because we really just care about the variables.

Â And also because we're doing astronomy, we can ignore that stuff.

Â So cube both sides.

Â We're going to get mass isolation cubed.

Â We're going to divide it by one factor of mass isolation.

Â So we're going to get isolation mass squared.

Â And that's going to be equal to, that's a squared, so that's a of the sixth.

Â M cubed is just 1 over m, or m to the 1/3 cubed is 1 over m, I should have said.

Â And that sigma is just sigma cubed.

Â So we can solve directly for that isolation mass,

Â which is a pretty amazing thing.

Â We can get that it's equal to, or I should say proportional to a cubed m to the -1/2,

Â 1 over the square of m, times sigma to the 3/2.

Â This is a pretty cool result.

Â This tells you that there is a certain-sized, certain-massed object

Â that will isolate itself at every distance away from the star.

Â Notice that it depends, the mass of it depends on the distance cubed.

Â So close to the star, those will be quite small, far away they can be quite large.

Â It depends on the mass of the star which, of course, is not changing.

Â And it depends on this surface density of the disk.

Â Not surprising, a bigger disk, a higher surface density will allow you to

Â accumulate more mass before you've isolated yourself.

Â If we'd like to know what was going on in the solar system,

Â the one thing we need to know then is what is that value?

Â What was the surface density of the initial disk that

Â the planets formed out of?

Â It's hard to know the answer to that question exactly,

Â but it's actually easy-ish to make an estimate of it.

Â How do you make an estimate?

Â Well, you do this thought process,

Â take the planets that exist now, squash them flat.

Â Spread them out into an annulus around the sun, each one of them and make sure

Â that annulus spreads from one planet to the next planet to the next planet.

Â If you do that,

Â you will approximate the disk of material that went into the planets.

Â This is called the minimum mass solar nebula.

Â And if you do that, you get a plot that looks something like this.

Â Okay, this is a complicated-looking plot, or sort of a bizarre-looking plot, and

Â what do you see?

Â Well, a couple of things that jump out at you right away.

Â What is this going on in through here?

Â This is the asteroid belt.

Â In the asteroid belt, if you flatten all the asteroids and spread them out into

Â the annulus, there's much less mass in the asteroid belt than in

Â the region before it, out to Mars, and the region after it where there's a huge jump.

Â We'll talk about this a lot, but

Â this is a consequence of the fact that a lot of material was lost here.

Â Same thing happens out here beyond Neptune,

Â Neptune is this last thing that was smashed right here.

Â And that's a consequence of the fact that a lot of material was lost at the edge of

Â the solar system in the Kuiper belt.

Â Again, we'll talk in detail about that later.

Â The other thing that you probably noticed though is there's this huge jump

Â between here and here in the total mass of the planets.

Â You knew that. This is Mercury, Venus, Earth, Mars here

Â and suddenly jumping to Jupiter, which has so much extra mass up through there.

Â And what is all that extra mass?

Â Well, we know the answer.

Â It's that all that extra hydrogen and

Â helium that Jupiter has that we don't have.

Â And you can see what happens is that right here,

Â the giant planets begin to capture that hydrogen and helium.

Â There would have been hydrogen and helium in the initial disc right here too, but

Â it all got blown away.

Â And in fact, what you can do is augment the masses of the planets by the amount

Â of hydrogen and helium they would have had, had they captured all that.

Â You get something like a pretty nice, straight line going up through here.

Â And this, including the extra hydrogen and

Â helium, is what we like to call the minimum mass solar nebula.

Â Now, we're talking, though, about the formation of solid planetesimals.

Â So we need to ignore the hydrogen and helium for now, and

Â just think about if we were for example at the Earth.

Â Well, at the Earth we know what this number is for spreading out all

Â the material, and it's something like 5 grams per centimeter squared.

Â Again, it's a surface density, changing units on UI sub kilograms per

Â meter squared, I'm giving you grams per centimeter squared.

Â I'm doing that on purpose because astronomers tend to use different units

Â all over the place, and it's pretty good to be able to keep them all in your head.

Â So about 5 grams per centimeter squared, right here.

Â And you might think, as if you continued on through Jupiter, and

Â you only talked about the solids, you would continue like that.

Â But you would be wrong.

Â At the location of Jupiter, even the solid material takes a big jump.

Â The big jump goes something like this in solid material.

Â 12:35

There's extra solid material at the orbit of Jupiter.

Â What's that extra solid material at the orbit of Jupiter?

Â It's ice.

Â At the Earth's location, that ice is in the form of water which is in

Â the form of vapor, it doesn't become part of the solid.

Â At Jupiter's location, all of that water, all of that oxygen that

Â is available turns into water which turns into ice,

Â which might as well be a rock at these locations.

Â So there's a huge jump in surface density out there.

Â This is called the waterline, and

Â it's an important component of where you can form things like Jupiter.

Â If there had been no water in the disk, and

Â the masses were small as you'll see in a little bit, Jupiter could not have formed.

Â The addition of water, and that sudden jump in water out here,

Â is something that very likely led to the formation of Jupiter.

Â And very naturally leads to the formation of Jupiter right outside there,

Â where that jump occurs.

Â The question was, what is that isolation mass?

Â If you take these numbers, like 5 grams per centimeter squared,

Â and calculate the isolation mass at the location of the Earth.

Â You get something like 0.1 Earth masses.

Â It's interesting, that is not the mass of the Earth,

Â that is a small fraction of the mass of the Earth.

Â So planetesimals, proto-planets, these words are often used together.

Â Proto-planets become isolated around the distance of the Earth, but

Â they're not the size of the Earth.

Â We'll worry later about how the Earth itself forms if you're isolating things

Â that are smaller than the Earth.

Â But, the interesting thing is, at Jupiter's location two things happen.

Â One, you have a jump in surface density.

Â And two,

Â you remember that that isolation mass was proportional to the cube of the distance.

Â We're now 5 AU away, so we're 125 times larger right then,

Â and you end up with isolation masses of something like 10 Earth masses.

Â 10 Earth masses?

Â That's a huge amount, so that means something that was starting to form just

Â by particles sticking together, and then grabbing more and

Â more because it was slowing things down, and grabbing them.

Â Gravitational focus was happening, dynamical friction was going on,

Â it continued to grow larger and larger, that the solid body.

Â The one largest solid body around there could grow something as

Â large as 10 Earth masses.

Â 10 Earth masses is an interesting number because 10 Earth masses, if you recall,

Â is what we think is approximately the mass of Jupiter's core.

Â