Bailyn: Welcome back to more of Astronomy 160.
The syllabus, if you haven’t got it yet,
is on the classes server. We have extra copies but I have
to say I don’t know where they are, and in any case,
everything that you need to know is on the classes server
and there’s actually more than is on the printed syllabus there
anywhere. Let me say a couple things;
we’re going to stick to the rule that there should be no
science majors taking this class.
And from the looks of the enrollment, there’s going to be
no problem for juniors and seniors.
I think we’ve got plenty of sections to accommodate everyone
in here, provided you’re not a science major.
But the way it’s going to work in terms of signing up for the
class, I think, as I understand it,
is as follows: You need a section,
sections are mandatory for this course.
So, you have to go through the online section sign up process.
That’ll start on Monday but it’ll only start for freshman
and sophomores on Monday. Juniors and seniors will be
able to sign up for the course and sign for sections on
Tuesday, and so please do that. Sections will start–sections
are going to be on Mondays, so they obviously won’t start
next week they’ll start the following week.
I have no doubt looking around the room that we’re going to be
able to accommodate juniors and seniors if they want to take the
class. All right, other things,
yes, the first problem set is assigned as of today.
When I get back to my office I’ll stick that on the classes
server as well, and you can pick it up from
there. It’s due one week from today.
The way I like to do this, I like to have problem sets due
at the start of class because it bugs me to look out and watch
people copying each other’s problem sets.
So, you have to hand it in at the beginning of class otherwise
it’s late. On the classes server,
there’s a whole little thing about lateness policy which you
ought to read. And also something–let me make
a couple general remarks about problem sets.
The purpose of problem sets is different from the purpose of a
take-home test. This is not something where the
major purpose is to evaluate how much you know.
The point of problem sets is to get you to actually do those
particular problems because that’s how one learns stuff in
this kind of subject matter. It’s kind of like doing the
reading; if you don’t do it,
you’re just going to have trouble.
And so, the reason it’s a substantial fraction of your
grade is just to make sure that you do it.
So, this has implications for working in groups.
Working in groups is strongly encouraged.
Please do that. It’s a very good thing,
make friends with the other people in the class,
work together, talk to each other,
start early, and make use of each other’s
intelligence. However, at the same time,
we want to make sure that you actually do things.
So, we want the work you hand in to be your own.
This sounds contradictory. On the one hand,
I’m tell you to work in groups; on the other hand,
I’m telling you, you have to hand in your own
work. What does that mean?
So, as is written down on the classes server,
the way we try and work this is when you are actually writing
down the thing you’re going to hand in,
do that alone. So, if you work in a group,
figure out what you’re doing, get everything all set up,
then split up and write down the answer, because the process
of actually writing the thing down yourself will make sure
that you’ve actually understood what you were doing.
We’ll check on this. We’ll look for identical
wording or identical multiplication mistakes and so
forth, and so we’ll keep a little bit of an eye on it.
I do encourage you to work in groups.
There’s other ways of getting help if you need it.
The primary source of help are the sections,
but of course, those won’t be meeting this
week so not before the first problem set is due.
Office hours; I’ll be in Starbucks Monday
from 9:30 to 11:00; please come and say hello,
and ask any questions you might have.
The teaching fellows are going to have office hours on
Wednesdays; conveniently located the day
before the problem sets are due you’ll notice.
And let me strongly recommend, don’t send us email with
questions about it. Send things into the forum on
the classes server. That way–and look first on the
forum to see if somebody has sent your questions.
This works really well. We’ll keep an eye on that all
the time. We’ll be answering it–not all
the time, not probably after around 8:00 at night on
Wednesday. If you have a question at 2:00
in the morning the day before something is due you’re kind of
on your own. Yes?
Student: Is that Monday 9:30 to 11:00 in the morning or
the evening? Professor Charles
Bailyn: Excellent question, thank you so much.
Morning, I’m afraid, sorry. Yeah, I’m not as young as I
used to be. Let me say, if you do want to
meet at some ghastly hour of the evening or something,
send me email. We can find a time,
same with the teaching assistants.
Send stuff into the classes server at all times,
we just don’t guarantee to look at it after 8:00 the night
before it’s due. Start the problem sets early,
that way if you have a question you can answer it.
Not only that, if you even just look at the
problem set, your brain will work on it in your subconscious
and you won’t freak out quite so much when you get past midnight
the night before it’s due. Let’s see, in general,
I’m going to try and make it so that everything you need to know
to do a particular problem set has been covered in class by the
day the problem set is handed out.
That would be today for this one.
In the very first week this week I’m not quite going to make
it and so there’s one problem, it’s noted on the problem set,
there’s one problem that you’ll need stuff that I’ll talk about
on Tuesday. There are going to be a bunch
of help sheets for various topics that are going to also be
linked to the classes server. By all means,
take a look at the problem set and start thinking about it as
soon as you can. Okay. Let me remind you what we had
started to talk about. The class is organized into
three fairly specific topics. The first of which is
extrasolar planets. Planets around stars other than
the Sun. Exoplanets, so-called.
That’s our topic. One of the things that I
pointed out last time is that, surprisingly enough,
very–until ten years ago none of these were known.
And it’s only a very recent development that there’s any
actual evidence that these things exist.
So, one question you might ask is why are these things so hard
to find? The science fiction folks seem
to have no trouble; they just sort of go around in
their spaceships and find these things all over the place.
And to consider that question let me show you a picture.
Here’s a picture of a star. This is the star Sirius.
It’s the brightest. It’s a blowup obviously of a
photograph plus a little Photoshopped [computer program
to edit photos] arrow.
That’s not a celestial object [laughter]. So, this is a blowup of a
photograph of the star Sirius. Sirius is the brightest star in
the sky to the–easily visible with the naked eye.
In fact, as I say, the brightest star.
It’s one of the closest stars. It’s a little bit brighter than
the Sun intrinsically, but it’s ten light years away
or so. And let me comment on a couple
features of this picture. First of all,
you can see the star covers a fair amount of area on this
blown up picture, but that has nothing to do with
the actual size or shape of the star itself.
It’s not like it’s a big round ball with spikes coming off it.
That’s not what’s happening. In fact, if you had perfect
optics and perfect vision, this star would be a point
basically so small you couldn’t resolve it right in the middle
here. That’s the physical extent of
the star at this scale. It’s something like this
[pointing to photo], and you wouldn’t be able to see
it at all. The reason it’s extended is for
two reasons; first of all,
the atmosphere distorts–this was a ground-based picture,
the atmosphere distorts how you see the star,
and in particular, it’s like looking up from the
bottom of a swimming pool. There’s distortion in the
atmosphere and it makes the stars seem to jump around.
And so if you take a long photographic exposure of
something the star has been jumping around during the
exposure and it smears out the image.
Plus also, the optics of the telescope aren’t perfect,
and these spiky things here are due to the optics of the
telescope. This sort of–there are all
these spikes and that has to do with how one of the mirrors of
the telescope is held up. So, the combination of optics
and atmosphere sort of spreads out the light by a whole bunch.
This is always the case even when you’re doing observations
from space. Now, the little arrow here is
pointing at a little bump off in the side, but you might not have
thought anything more of than you did about these spikes.
But in fact, that’s not a property of
optics; there’s actually something
there. That’s an object in orbit
around Sirius. And you might think that this
is the way you would go about finding planets orbiting other
stars. It would look kind of like this.
Here’s a great big star; there’s a planet.
And if you watch this thing over a course of many years,
you actually see the position of this point move around.
It’s because it’s in orbit. It’s in about a forty-year
orbit. Here’s the thing.
This isn’t a planet. It’s another star.
This is another star somewhat fainter than this [pointing to
the photo of Sirius]. A planet would be 10,000 times
fainter than this thing, but probably in about the same
relative position to this. Now, picture all this light,
all this mess from this star, and try to find something right
here that’s 10,000 times fainter than this.
That’s the problem. The problem is that planets are
relatively faint things. That, in itself isn’t
catastrophic. We have big telescopes;
we can see pretty faint things. It’s just they are faint things
that are really near to things that are bright,
and so basically, the problem with finding
planets around other stars is the same as trying to find stars
in the sky during the daytime. Why can’t you see the stars
during the daytime? Because the background of the
sky is much, much brighter than the stars are.
Similarly, if you had a little planet here you wouldn’t be able
to distinguish it from the atmospheric distortions and the
optical distortions caused by the telescope that takes this
picture. So, the problem is you’ve got
stuff that’s faint near things that are bright,
and the bright things have all kinds of distortions on them
anyway, and so you wouldn’t be able to figure out what actually
is a planet. So, just to summarize this,
the problem is not that the planets are too faint,
although they are quite faint. The problem is that the planets
are too close to the star. That prompts–we’ve got to
quantify this statement I think and so the question arises,
“Well how close is it?” That leads into a discussion,
which we already started last time, of planetary orbits. If you want to answer that
question in any quantitative sort of way.
You will recall that in the last class I wrote down an
important equation of–regarding planetary orbits,
that looks like this: a^(3)=P^(2)M,
where this [M] is the mass of the objects in
orbit around each other in units of solar masses.
This [P^(2)] is the orbital period in units
of years. And this [a^(3)]
is–orbits you will recall are ellipses, and this is the
so-called semi-major axis of the ellipse in units called
Astronomical Units, which is the distance from the
Earth to the Sun. Or the semi-major axis of the
Earth-Sun orbit, more properly. So, that’s in a particular set
of units. Before going on,
let me make a couple general remarks about equations.
First thing to say about equations is there’s this idea
that equations, mathematics,
is some terribly different language, not at all.
This is an English sentence. The sentence reads A cubed is
equal to P squared times M, period.
If you wrote this down in a textbook you’d actually see they
put the period in there. It’s got a verb,
is equal to, it’s got a subject,
it’s got an object. All this is is a different way
of writing stuff down. So, it’s not a different
language; it’s just a different writing
system, that’s all. So, don’t freak out,
that’s the first thing. It’s just ordinary English.
The second thing to say is that this is a physics equation and
not a math equation, and that’s actually a subtler
point. You will recall back in
eighth-grade algebra when they started writing down X is equal
to whatever X was equal to. And they said,
you know, the amazingly cool thing about algebra is that X
could be anything. That’s why it’s so powerful,
and so you can solve whole classes of problems all at once
just by letting X equal something unknown.
That’s math; physics–it isn’t true.
It is not the case that this equation is true regardless of
what a is, or regardless of what P
is. This is only true for very
specific meanings of those quantities.
And the amazing thing about the Universe is it turns out that
math equations apply in very specific physical situations.
This didn’t have to be true. Either that you could express
physical things in mathematical equations, and that math and
physics would have anything to do with each other.
And it is different to have a physics equation than to have a
math equation because it has to be so specific.
If you substitute something that isn’t the semi-major axis
in for a, then that equation isn’t true
anymore. It seems kind of obvious,
I know, but worth keeping in mind.
The other thing is that physics equations tend to have units
associated with them. Not always, but most of the
time. And so that’s why it’s so
important to say, well, this is true not only if
M stands for mass but also if mass is expressed in a
particular set of units, in this case solar masses.
You’ll recall from last time that this is actually a specific
version of a slightly more general equation which looks
like this: a^(3)=P^(2)GM/4π^(2),
where G is a constant that depends on what units you’re
using. And last time I derived this
equation from that. But another way of saying it is
if you choose your units to be solar masses,
years, and astronomical units; then G is equal to 4π^(2) and
so it cancels out. That’s another way of thinking
about why this specific version of the equation works out.
All right. So, once you’ve got your
physics equation, what are you supposed to do
with it? Well, the first thing you
should do is take a careful look at it and figure out what kinds
of problems you can solve. This, again,
is fairly straight-forward, obvious stuff.
If you’ve got a^(3)=P^(2)M,
there are obviously three different problems you can
solve. If you know a and
M you can compute P, right?
If you know a and P you can compute
M. And if you know P and
M you can compute a.
So, you know two of these things, you find out the third.
This problem–an example of this problem we did last time.
We had–we solved for the orbital period of Jupiter by
knowing how far away from the Sun it is and what the mass of
the Sun is. So, like Jupiter last time.
So, let’s do another one of these examples just quickly
here. What is the mass of the Sun? So, that’s obviously this
version of the problem [if you know a and M you
can compute P]. And so, what we’re going to do
is we’re going to take a and P from the Earth’s
orbit. Use Earth’s orbit to determine
the mass of the Sun. And this is,
in fact, how one goes about determining the mass of the Sun.
So, let’s use this in the units we have, a=1
Astronomical Unit for the Earth. P is equal to 1 year for
the Sun. 1^(3)=1^(2) x M.
So, M=1 in units of solar masses.
That’s not especially surprising, right? Not particularly illuminating,
right? That the mass of the Sun comes
out to be one in units of solar masses.
If we use a different set of units this is a more interesting
calculation. So, let’s use a more standard
set of units, this so called mks
units, this is meters, kilograms for mass,
and seconds for time. All right, so now,
turns out an Astronomical Unit is expressed in meters is 150
billion meters, approximately.
It looks something like this [150,000,000,000].
And what’s a year expressed in seconds?
Well, let’s see a year is 365.24 days, a day is 24 hours,
an hour is 60 minutes, and a minute is 60 seconds,
so that’s a year expressed in seconds.
And at this point we better pause and get our large numbers
straight. You are all probably familiar I
imagine with scientific notation.
This becomes very important, otherwise the numbers,
as you can clearly see, get completely out of hand.
And because it’s so important I’m going to tire you by writing
down some facts about scientific notation.
There’s a help sheet on this if you have any trouble but recall
how this works. You’re supposed to express
numbers as in the following form: N times 10^(m),
this [N] is a decimal with one digit out
in front of the dot. This [m]
is supposed to be an integer, a whole number either positive
or negative. And the reason that this is
easy to deal with is because of one of the great limitations of
the human brain. We only deal well with really
small numbers, basically one-digit numbers.
If I take a few coins here and I sort of throw them up you
immediately know that there are four of them,
you don’t have to count one, two, three, four to satisfy
yourself that there are four coins on this table because we
have a concept of fourness, all of us in our minds,
and it’s clear that this pattern satisfies that concept.
If I threw two coins or six coins, or seven coins you’ll
probably be okay. If I threw thirteen,
you’d start to have real trouble.
If there were 271 of them, there’s no way you could guess
just by looking at them how many there were.
You’d have to count them one by one.
So, the human brain really only deals well with one-digit
numbers and you can tell this because of the names we give to
numbers. When you start getting big
numbers we call them millions or billions, or trillions and they
all sound alike, that’s because we can’t tell
the difference between a million and a trillion,
it’s just big. So, anytime you get up into big
numbers we start to completely lose our grip on what these
numbers mean, and you have to have elaborate
metaphors; a trillion is a million
millions. And if you fill up the Moon
with ping-pong balls there are–I don’t know.
You have to do that kind of mental exercise to get a grip on
it. The nice thing about scientific
notation is that it turns big numbers into small numbers,
right? N is a number between one and
nine, m is an integer, it’s one of those numbers four,
five, six that we can get our minds wrapped around.
Recall what this means, right, 10^(m) is 1 with m zeros
after it. So, that’s the point of
scientific notation, it’s to take big numbers and
allow us to deal with them. I think this is a really
valuable thing and I kind of wish that budget discussions in
the political context would take place in scientific notation
because people are always making mischief from the fact that all
big numbers sound alike. You have congress critters
saying things like, “oh terrible disaster,” there’s
a million dollars of waste in the Pentagon budget.
The Pentagon budget is $500 billion dollars.
One million dollars of waste–that’s me dropping a
nickel behind the couch on my budget.
So that’s–nobody would accuse me of being wastefulness if I
didn’t move the couch and get that nickel out,
all right. It’s because a million sounds
big, and so does 500 billion. We have no real sense of how
big these things are relative to each other.
But we do have a sense of the difference between 6 and 11,
which is the difference in scientific notation between one
million dollars and $100 billion dollars,
so that’s how this works. So, arithmetic rules.
Again, there’s a help sheet on this if you’re rusty.
If you have N times 10^(m) and you multiply it by another
scientific notation number, A times 10^(B),
that is equal to A times N (times 10 to the n+m).
And the other nice thing about scientific notation is it turns
multiplication into addition because the exponents add:
A times N to–times 10^(m) plus B.
Similarly, if you take N times 10^(m) and raise it to the K,
that’s N^(k) times 10 to the m times K,
so it turns exponentiation into multiplication.
So, the nice thing about dealing with exponents is it
makes it one kind of arithmetic easier.
So, a specific example of this might be N times 10^(m),
the square root of that which is raising something to the ½
power. This is equal to N^(1/2) times
10^(m/2). That works if m is an
even number but if m is an odd number you get a crazy
exponent. This really has to be an
integer over here. So, what do you do then?
You say N times 10^(m) to the ½ is equal to 10 times N times
N^(m-1) to the ½ power. So, I’ve just moved a 10 from
here over to here, and then that is equal to 10 N
to the ½ times 10^((m-1)/2). And that’s how you recover when
you take a square root or something like that.
That’s how you recover an integer on this side.
We’ll have lots of practice in this kind of thing.
So, going back to the problem. 1 AU is equal to 1.5 times
10^(11) meters; 1 year is equal to 2.4 times
10^(1). That’s 24 times 6 times 10^(1)
times, 6 times 10^(1), times 3.6524 times 10^(2)
seconds. All right, so let’s do that
one, two, three, four, five–2.4 times 6 times,
6 times 3.6 times 10^(5) to the 5;
2.5 times 6 is around 15,15 times 6 is around 90,90 is a
little less than 100,3.6 a little more than 6.
Those two things multiplied together is around 300.
And 300 times 10^(5) is equal to 3 times 10^(7).
So, there are around 3 times 10^(7) seconds in 1 year.
That’s a good number to remember.
In mks units, G is this magical constant,
is something like 7 times 10^(-11).
That’s an approximation. It’s actually 6.6 something,
something, something times 10^(-11), but we’ll call it 7.
So, now we’re in a position to calculate what the solar mass is
in kilograms. So remember a^(3) equal
to GMP squared over 4π^(2).
So, let’s see, 1.5 times 10^(11);
that’s an astronomical unit cubed.
P squared, 3 times 10^(7);
that’s 1 year squared; 7 times 10^(-11) that’s
G over M over 4π^(2).
I’m going to do the arithmetic really fast.
I would like you to learn how to do this yourself rather than
relying on calculators so this is an example.
Let me blaze through this at warp speed;
then we’ll stop and consider what’s happened.
All right. On the right-hand side,
3^(2) is–that’s 3^(2) times 10^(7) times 2 is 14,
times 7 times 10^(-11). M over 4 times π,
times π. 3^(2) is 10.
10 times 10^(14) is 10^(15). And so on the top we’ve got
10^(15) times 7, times 10^(-11).
15 minus 11 is 4 so that’s 7 times 10^(4) on top.
π times π is 10 that’s 40. 4 times 10 to the 1 on the
bottom, 7 divided by 4 is 2, and so that’s 2 times 10^(3)
times m. Good, left-hand side,
all right 1.5 times, 1.5 times, 1.5,
that’s 1.5^(3) times 10^(33), why 33?
That’s 11 times 3. 1.5 times 1.5 that’s a little
bit more than 2. If you multiply something
that’s a little bit more than 2 by something that’s a little bit
less than 2 that’s 4, and so this is 4 times 10^(33).
Now we want m, right? So we’ve got to divide both
sides by 2 times 10^(3); m is equal to 4 halves times
10^(33) over 10^(3). That’s 10^(-33) minus 3.
That’s 2 times 10^(30) kilograms, got to put the units
in. Okay, how we doing?
Questions, comments, abuse. Yes sir?
Student: How accurate is that number?
Professor Charles Bailyn: How accurate is that
number? It’s accurate to about one
digit, which is why I only wrote down one digit of accuracy out
front. This is a pro–excellent
question thank you very much. This is appropriate because
remember I’m using 7 times 10 minus 11 for G,
where it’s actually 6.6. So, I’m already about 10% off
from there. I did my little calculation to
come up with one year equals 3 times 10^(7) seconds.
That’s about accurate to one digit or so.
And so the whole thing is done to one digit accuracy.
If you’re dealing with one digit accuracy,
it is true that 7 divided by 4 is 2.
It really is true, because if that wasn’t true,
then you would have to have more digits on your unit for
G or something like that. In particular,
let me make an official rule for this course.
Three equals π, equals the square root of 10,
all right. That will solve an enormous
amount of arithmetic problems and it will not get you into any
serious trouble. So, we don’t have to worry
about the .14159 and however many more digits you all
memorized it to. And when you multiply it
together you get ten. Yes?
Student: Are you expecting this kind of
calculation for problem sets? Professor Charles
Bailyn: Yes. The question was,
“Am I expecting this kind of calculation for problem sets?”
The answer is “yes.” Here’s the rule about
calculators. Let me put it this way:
You can only use calculators if I can’t tell that you’ve done
it. So, that means you can check
your work to make sure you’ve it right or something.
But if you start coming up with numbers like 7.1516397,
that’s eight digits of accuracy and I’m pretty sure you haven’t
worked it out yourself. So important,
no calculators on the tests, okay?
So, get some practice doing this kind of thing.
And this will–this I promise you will be useful to you in
everyday life because this is how you catch the politicians
doing screwy things with big numbers.
You do it in your head in scientific notation and you
figure out whether the answer is meaningful or not.
This whole business of significant digits,
I think, is badly distorted; by the way, it’s taught in high
school. In high school you,
and also I should say in laboratory courses sometimes at
the college level, you often get situations where
people say–give you a whole sheet of rules on how to figure
out how many significant digits you have.
This is nonsense. All you have to do is behave
like a human being. We say to each other,
I’ll meet you in the dining hall in ten minutes.
That doesn’t mean–that means something different from I’ll
meet you in the dining hall in eleven minutes and twenty-six
seconds. Even if the person happens to
show up in the dining hall in exactly eleven minutes and
twenty-six seconds. Ten minutes means I’ll meet you
there in ten minutes, we all know what that means.
I’ll meet you there in eleven minutes and twenty-six seconds
means you’re a character in a bad spy novel who’s just
synchronized his watch. So, this shows up in science
fiction too. I don’t know how many of you
are Star Trek fans, I certainly am [laughter].
And in all the different Star Trek movies [inaudible
comment]–thank you. In all the different–a friend
[referring to person who made comment].
In all the different Star Trek movies there’s always a second
in command who isn’t a human being, right?
A Vulcan or an android or some damn thing or another.
And to emphasize the non-humanness of these
characters, what they do is they make them use too many
significant digits. And so that makes them inhuman
and so the captain will say, “When are we landing on omicron
M?” The second in command will say,
“Well, we should assume standard orbit in 2.6395
minutes,” emphasizing somehow superior brain power or
something. But it’s nonsense because it
takes the guy ten seconds to say that sentence,
so what is this time calculated to a 100th of a second?
Does it start from when he begins the sentence?
From when he ends the sentence? What’s the other end of that
time interval? Can you say you assume standard
orbit to the 100th of a second? What does that even mean?
When you start beaming down? When you end beaming down?
Also, keep in mind it takes more than a 100th of second for
the sound to travel from his lips to the captain’s ears,
so the whole thing is just nonsense.
And so, you don’t need any special rules,
just behave like a human being; don’t behave like an android.
So, no androids. And that’s the only rule I’m
going to give you [laughter]. These two are the only rules
I’m going to give you about significant digits,
just do the right thing, okay.
All right, okay so this is all lovely but we haven’t actually
started to answer the question we started the class with.
Let me remind you where we began this disquisition on
numbers. The question was supposed to be
how far away from stars are planets?
So, we haven’t answered the question, when the question was,
“How close are planets to stars?”
This notation will confuse the people who didn’t make it to
class today; that’s probably just as well
too. All right, how close to
planets–are planets to stars? Now, we have a problem because
closeness–we usually think of this as being measured with
distances. And distances are a serious
problem in astronomy. Supposing you–Distance,
by the way, is more or less the same thing as size,
right? Size is the distance from one
side of something to another side of something.
So, size and distance are the same.
So, supposing you ask the question, “How big is the Moon?”
What is the size of the Moon? What is the distance from one
side of the Moon to the other? You go out one night and you
look up at the Moon and you do this [holding fingers apart],
one on each side of the Moon. You say, well,
it’s about an inch and a half. Now, obviously that’s
meaningless, because if you had done it here,
it would have been about a half inch.
And we all know that isn’t–that somehow is not the
right answer. If you look at the screen over
here, for example, and you hold out your hands
about eight inches from your–close one eye,
hold out–you can try this, hold out your hands about here,
you’ll find that that screen–the lit part of the
screen is about four or five inches across.
But if you go closer to your eye, it’s only about an inch
across. If you’re at arm’s length it’s
maybe six, eight inches where I’m standing here.
And if you try it at the back of the class,
you’ll get a different answer from what I just got.
So, distance is a problematical thing.
And so–But on the other hand, obviously, the size of the
screen hasn’t changed depending on where my arms were or whether
I’m standing at the back of the room or the front of the room.
So, something has to be measurable there,
but it isn’t distance. What it is, is angle.
What you can measure in astronomy are angles.
So, here’s what you really want to know.
Here’s a star; here’s a planet going around
the star, and here is you [draws diagram on overhead].
This is the international symbol for an observer;
it’s supposed to be a little stylized eyeball.
So, this observer is looking at this star-planet system.
And looks at the star and looks at the planet.
And what you can measure is the angle, which I’m going to call
α here. And if you simulate the angle
by putting your hands up close to your eye, you’ll get a small
distance; if you do it further away
you’ll get a bigger distance. But what you are measuring is
the angle. What you would like is a way of
being able to convert from that angle into a distance.
And now, the ugly specter of trigonometry rears its head.
Because that’s what trigonometry is.
It’s a way of turning angles into distances and vice versa.
And you may remember constructions that look vaguely
like this; let’s put some labels on it.
Here’s distance 1, here’s distance 2,
here’s distance 3. And this is the definition of
sine, the sine of α is equal to the opposite over the
hypotenuse, so that’s D_2 over
D_3. And now–so that’s how you do
it. Now, I promised you on Tuesday
that the sine was going to cancel out, and so it is.
Here’s how. If you use small angles it
turns out that the sine of α is equal to α,
if α is expressed in radians, which is a particular
kind of measure of angles. And so for small angles,
first of all, the hypotenuse and the longer
side are the same length more or less.
So, you’ve got a situation like this where D_2 over
D_1 is equal to α.
That is–let’s circle that one in red;
it’s going to be important. That’s called the small angle
formula. And once again,
you have to be careful of the units.
And once again, there’s two sets of units in
which you might possibly use this.
So D_2 over D_1 is equal to α.
If D_2 and D_1 are in the same
units, then α has to be radians.
A radian–there are 2π radians in a circle. You’ll recall there are 360
degrees in a circle that’s how you convert from degrees to
radian. But there’s a better set of
units. It turns out also that if you
do this, use this equation D_2,/D_1,
α, and D_2 is measured in Astronomical Units,
remember those? That’s the Earth-Sun distance.
And D_1 is measured in parsecs, which is equal to 3
times 10^(16) meters; that’s about three light years,
for those of you counting at home.
And α is in arc seconds. Then it also works out.
Let me tell–an arc second, let’s see, 60 arc seconds is
equal to an arc minute. Just so you know how these
units work; 60 arc minutes is equal to 1
degree. So, that’s a really small angle.
So, it turns out this set of units is particularly useful for
these kinds of planet calculations,
because distances from planets to stars tend to be a few
astronomical units. Distances from us to the
nearest star. The nearest star,
aside from the Sun, turns out to be about 1 parsec
away. So, distances to stars are some
number of parsecs. And so, you’ll get answers that
are a few arc seconds, and this is yet another way of
turning big numbers into small numbers.
You’ve used the right set of units so that everything’s
approximately one. Another way of compensating for
the weakness of the human brain. Okay, so now we can solve a
problem. We can actually solve
numerically the question of how far away from a star is our
planet? So, here’s an example.
Here’s a problem of the kinds that might show up on a problem
set. I told you that that object
orbiting Sirius has about a forty-year period.
So, let’s take a planet with a forty-year period;
that’s kind of between Saturn and Uranus in our own Solar
System. Around a star 3 parsecs away;
pc means parsecs, whatever they may tell you in
the humanities classes. Around a star 3 parsecs away,
what is the angular separation? Angular separation,
because that’s what we can actually measure.
We can’t measure the distance but we can measure the angle.
Okay, so this is a hard problem, why is it hard?
It’s hard for two different reasons.
First of all, well, you guys know how to
solve problems like this. You all went to high school.
What you do is you say, okay what do I know?
I have P=40, I have D=3,
I want α, so I look for an equation with D,
P, D and α in it.
I shove these two things into–you substitute them in for
P and D, and then I calculate α.
And you don’t have to actually know anything about anything in
order to do that; you just have to have a good
set of notes. And you go through all the
different equations until you find the equation that contains
all the variables you have, and the one that you’re asked
for. And then you can just plug it
in, you don’t have to know anything about anything.
This won’t work, right? Because if you go through the
lecture notes thus far, you will not find an equation
with P and D and α in it because it’s two
different equations that you have to put together.
And that requires you to actually know what you’re doing.
And that of course is much harder.
So, plug and chug won’t work. Chug fails because there’s no
one equation. This is a hard problem for
another reason, which is that I haven’t told
you something important. If you try and go through and
solve this you will discover that you don’t have all the
information you need, and in particular,
you need to know the total mass of the system,
the mass of the star. Most of the mass is in the mass
of the star. So, there’s missing information.
That’s one way to make a problem hard,
the other way to make a–if I don’t tell you everything you
need to know, it’s a somewhat harder problem.
The other way to make a problem hard–this is a classic,
be warned–I will do this at some point,
is to give you one extra piece of information that you don’t
need. This screws students up totally
because you go looking for the equation that uses the extra
piece of information, and there is no such an
equation and you don’t need it, then you’re cast into doubt.
I must need this somehow or he wouldn’t have given it to me.
Nonsense. In real life it is not the case
that all problems come to you with all the information you
need and only the information that you need.
That’s not how things work in real life.
And it shouldn’t be how things work in preparation for real
life, which is I guess what classes sort of ought to be.
So, missing information, too much information,
sometimes although not in this case;
I could have told you the temperature of the star,
for example. I’ll do that.
The temperature of Sirius is 10,000 degrees,
surface temperature. So, now I’ve given you both;
I’ve left out information and told you too much.
All right, so what is the mental process that you need to
go through in solving a problem which has these kinds of
difficulties associated with it? Okay.
So, how to think about such problems?
The first thing to notice is what you would do in the
ordinary plug and chug. What have you got,
what do you need? You have a value for the
orbital period, you have a value for the
distance, and what you want is an angle. Okay.
Now, so this doesn’t lead you to where you want to go.
The next thing you’ve got to do is make an assumption.
This is a star; all stars are more or less the
same. So, let us imagine that this
star has a mass more or less similar to that of the Sun.
So, the M is approximately equal to a solar
mass. How far wrong could you go?
In fact, you go wrong by about a factor of two,
but at a certain point you’re going to take the cube root of
two, and so it actually isn’t so bad.
So suppose M is equal to the mass of the Sun,
most stars are approximately. Ah ha, now we’re getting
somewhere, because now we have M and P.
And if you have put that equation–Kepler’s Third
Law–and interpreted it deeply into your brain,
and it has become part of your heart and soul and very being,
you will immediately realize that if you have M and
P, of course you can calculate
a. In fact, you probably realized
this first before you make the assumption in step two,
because that’s why you’re interested in the mass,
because you realize that if you happened to have M,
you would be able to determine a.
Next thing you have to do is, is interpret what this a
is; a is the semi-major axis
of the orbit of the planet around the star.
That’s something very close to–very similar to–the
distance between the star and the planet is a.
So, a in this equation is the same as D_2 in
the other equation, all right.
So, a is D_2 and the distance you have up
there is the distance from us to that whole system,
that’s D_1. And now, you can compute α,
right? That’s the thought process.
And you may have to go–sort of worry a little bit around here
before you have the “ah ha” moment of how to get the
problem. This is the moment when you’ve
solved the problem. The key to solving this is to
have an instinctive reaction to knowing that you’ve got a
period; namely, if I had a mass I could
figure out the semi-major axis. Or if I had the semi-major
axis, I could figure out the mass.
And that has to click. And the way that clicks is to
have this equation really–I was joking before–but really become
instinctive in a certain way, and there won’t be many
equations in this class. But it will really help you to
assimilate them in that kind of depth.
All right, so now in the few minutes remaining,
let me actually solve this problem numerically. P is equal to 40 years,
D is equal to 3 parsecs. Here’s something not to do:
40 times 3 times 10^(7), because that’ll give you the
answer in seconds. No don’t do that,
because both of the equations would rather have things in
Astronomical Units and years and things like that.
So a^(3) equals P^(2)M,
M is 1 by assumption, P is 40 so this is
1,600, that’s 40^(2) on this side, a^(3) that’s 1.6
times 10 to the 3. Now, we’re going to take the
cube root of both sides, that’s the cube root of 1.6
times, the cube root of 1,000. 3 times a third is equal to 1,
conveniently enough. The cube root of 1.6,
I don’t know that’s 1.2 or something but call it 1.
So, that’s 10, and so it’s 10 Astronomical
Units. Remember we’re doing this in
solar masses, years, Astronomical Units.
10 Astronomical Units, that’s how far away it is.
Now, we can do this [D_2/D_1]
where this is in Astronomical Units, and this down here is in
parsecs, and 10 divided by 3 is 3.
So, the answer is a is equal to 3 arc seconds.
Now, this is an important result because the–let me go
back to this picture. The angular scale over which
light is kind of thrown around by the atmosphere and by the
optics of the kinds of telescopes we can build is some
number of arc seconds. Good ground-based observations,
maybe half the light is inside 1 arc second;
the rest of the light is scattered all the way out to
maybe 10 arc seconds. From space it’s a little bit
better. But, light from observations of
stars is scattered over angular sizes of arc seconds.
So, you are guaranteed that there’s going to be a bunch of
light from the star right on top of where you’re looking for your
planet. And in fact,
it’s going to get worse, because one of the things about
this equation is D_1; the distance to the star.
We did this for Sirius. Sirius is one of the closest
stars; it’s 3 parsecs away.
Most–the center of the galaxy is 8,000 parsecs away.
So, if you take a typical star in our galaxy this number isn’t
going to be 3; it’s going to be 8,000.
And this number isn’t going to be 3, it’s going to something
like 1/1000th of an arc second. And the space telescope
scatters light over tenths of arc seconds, so you have factors
of 100 buried inside the light coming from that star.
So, that’s why you don’t care about more than one digit.
In fact, you probably don’t even care about anything other
than the exponents of this answer because why are you doing
this problem? You’re doing this problem to
figure out whether it’s plausible to think that you
might be able to see a planet as a separate dot,
separate from the star. And the fact that your answer
comes out in arc seconds or fractions of an arc second tells
you everything you need to know. You don’t need anymore decimal
places to realize that the idea of looking up and seeing a
planet is going to fail. So, that’s not the way that
we’re going to find planets. So, we have to come up with
different scheme and that’s what we’re going to talk about on