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

it begins to grow even faster.

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

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

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

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

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

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

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

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

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

are happening inside of it.

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

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

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

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

a stage where the cleared out annulus is so

big that the gravitational influence of this singular body

is smaller than the gravitational influence of the star.

That means that this stuff on this side of the annulus

is no longer gravitationally attracted to that body.

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

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

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

That part's not really true,

all the things are actually in orbit around the star.

And the further you are away from this body,

the more the influence of the star will be.

So lets figure out how far this is.

Once we figure out how far this is,

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

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

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

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

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

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

And you want to figure out whether a particle here,

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

can have its gravitational pull pulled into there.

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

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

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

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

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

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

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

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

what would its orbital period be?

If this particle were more in orbit around the planetesimal,

what would its orbital period be?

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

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

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

necessarily, but

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

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

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

in this case it will be gobbled up.

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

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

this is quite short, or they're comparable.

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

influenced by that, influenced enough to cross its orbit.

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

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

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

We'll call that a.

We need the mass of the star.

We need the mass of the planetesimal, little m.

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

min at the mass.

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

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

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

force is GMm dot over r.

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

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

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

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

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

is m.v squared over r.

So that r is an a.

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

What is the period?

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

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

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

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

would be in orbit around this star.

And we get that period equals 2 pi a over v

now is going to be the square root of GM.

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

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

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

get something very similar.

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

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

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

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

the G's are going to cancel out.

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

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

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

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

The further away the object is from the sun,

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

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

of the planetesimal divided by the mass of the star.

The bigger the planetesimal, the bigger its influence.

The bigger the star, the smaller its influence.

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

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

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

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

It has a circumference of a, and so

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

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

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

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

We'll talk about sigma in a minute.

Sigma has the units of kilograms per meter squared.

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

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

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

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

All that material is going to go into this object.

This surface density, it's called,

is an important parameter in thinking about planetary systems.

We'll talk about it in just a second.

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

But r is equal to this.

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

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

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

which is a times cubed root of m isolation.

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

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

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

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

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

So cube both sides.

We're going to get mass isolation cubed.

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

So we're going to get isolation mass squared.

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

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

And that sigma is just sigma cubed.

So we can solve directly for that isolation mass,

which is a pretty amazing thing.

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

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

This is a pretty cool result.

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

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

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

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

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

And it depends on this surface density of the disk.

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

accumulate more mass before you've isolated yourself.

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

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

What was the surface density of the initial disk that

the planets formed out of?

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

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

How do you make an estimate?

Well, you do this thought process,

take the planets that exist now, squash them flat.

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

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

If you do that,

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

This is called the minimum mass solar nebula.

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

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

what do you see?

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

What is this going on in through here?

This is the asteroid belt.

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

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

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

We'll talk about this a lot, but

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

Same thing happens out here beyond Neptune,

Neptune is this last thing that was smashed right here.

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

the solar system in the Kuiper belt.

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

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

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

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

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

And what is all that extra mass?

Well, we know the answer.

It's that all that extra hydrogen and

helium that Jupiter has that we don't have.

And you can see what happens is that right here,

the giant planets begin to capture that hydrogen and helium.

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

it all got blown away.

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

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

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

And this, including the extra hydrogen and

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

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

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

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

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

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

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

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

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

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

So about 5 grams per centimeter squared, right here.

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

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

But you would be wrong.

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

The big jump goes something like this in solid material.