Current time:0:00Total duration:11:40

# Stellar distance using parallax

## Video transcript

In the last video, we
talked about how parallax is the apparent
change in position of something based on
your line of sight. And if you experience parallax
in your everyday life, if you look outside of the
car window while it's moving, you see that nearby
objects seem to be moving faster than
far away objects. So in the last
video, we measured the apparent displacement of
this star at different points in the year relative
to straight up. But you could also measure it
relative to things in the night sky at that same time of year,
that same time of day that don't appear to be moving. And they won't appear to
be moving because they're going to be way, way, way
farther away than this star over here. There might be other galaxies
or maybe even other clusters of galaxies or who knows. Things that are not
changing position. So that's another option. And that's another
way to make sure that you're looking at the
right part of the universe. So you could measure it
relative to straight up, if you know based on the time
of year and the time of day that you're looking at the
same direction of the universe. Or you could just find
things in the universe that are way far back that their
apparent position isn't changing. So just to visualize
this again, I'll visualize it in a
slightly different way. Let's say, this is
our-- let's say, this is a night field of vision. Let me scroll to the
right a little bit. Let's say, our night field
of vision looks like this. And I'll do it in a dark
color because it's at night. So our night field of
vision looks like this. And let's say, that this right
over here is straight up. This right here is if
we're just looking straight up in the night sky. And just to make the
convention, in the last video, I changed our
orientation a little bit. But I'll reorient us in a
traditional orientation. So if we make this
north, this is south. This will be west. And then this will be east. So when we're looking at
the star in the summer, what will it look like? Well, first of all, the sun
is just to beginning to rise. So if you could think about
it, this north direction-- we're looking at the sun from--
we're looking at the Earth from above. So the north will be the top
of this sphere over here. And the south will be the bottom
of the sphere, the other side of this sphere that
we're not seeing. The east will be this
side of the sphere where the sun is just beginning
to rise right over there. So what will be the apparent
position of this star? Well, it's going to
be towards the east. It's going to be
towards the direction that the sun is rising. So this angle right over here
is going to be right over there. So this will be the angle theta. So this is in the summer. And what about the winter? Well, in the winter in
order-- straight up-- in order for straight up to
be that same point in time, or that same direction of
the universe I should say, then, the sun will just
to be setting word. We're rotating around that way. So we're just going to
be capturing the last two glimpse of sunlight. So in that situation, the
sun is going to be setting. So this is our
winter sun-- so I'll do a slightly different color--
will be setting on the west. And now, the apparent
direction of that star is going to be in the
direction of the sun again. But it's going to be
shifted away from center. So it's going to be to
the right of center. Sorry, to the left of center. So it's going to
be right over here. And it's a little
bit unintuitive, the way I drew it
in the last video. Well, I won't make any judgment
on whether the last one is easier to visualize
or this one is. Over here, I just wanted
to make the conventions so that north is
up, south is down. But I just want to be clear,
over here, the sun is-- well, sun always sets to the west. So in the winter, the sun
will be right over there. This will be shifted from center
in the direction of the sun. So it'll be at an angle theta
just like that in the winter. Now, that's all review
from the last video I just reoriented
how we visualize it. What I want to do in this
video, given that we can measure theta, how can we figure out
how far this star actually is. So let's just think
about it a little bit before even give
you a theta value. If we know theta,
then we know what this angle is right over here. Because this is a
right angle, we're going to know this angle
right here is 90 minus theta. We also know the distance
from the sun to the Earth. And let's say, we're just
going to approximate here, it's one astronomical unit. It changes a little bit
over the course of the year. But the mean distance is
one astronomical unit. So we know that angle. We know a side
adjacent to the angle. And what we're trying to do
is, figure out a side opposite to the angle--
this distance right here, the distance from
the sun to the star. And this is, of course,
a right triangle. And you can see it right here. Here is the hypotenuse. So now, we just
have to break out some relatively
basic trigonometry. So if we know this angle,
what trig ratio deals with an adjacent side
and opposite side? So let me write down
my famous SOHCAHTOA. I didn't come up with it. So the famous SOHCAHTOA. Sine is opposite
over hypotenuse. Those aren't the
two we care about. Cosine is adjacent
over hypotenuse. We don't know what
the hypotenuses is. And we don't care
about it just yet. But a tangent is the
opposite over the adjacent. So if we take the
tangent of the angle. If we take the tangent
of 90 minus theta, this is going to be equal
to the distance to the star. This distance right over here. The distance to the star,
or the distance from the sun to the star. We can later figure out
the distance from the Earth to the star. It's not going to
be too different because the star is so far away. But the distance from
the sun to the star divided by the
adjacent side, divided by one astronomical unit. I'm assuming everything
is in astronomical units. So you can multiply
both sides by one. And you'll get the distance
in astronomical units. The distance is equal to the
tangent of 90 minus theta. Not too bad. So let's actually figure
out what a distance would be based on some
actual measurements. Let's say, you were
to measure some star. Measure this apparent-- this
change in angle right here. Let's say, you've got
the total change in angle right over here from six months
apart, the biggest spread. And you're making
sure you're looking at a point in the universe
relative to straight up. You can do it other ways. But this is really just
simplifies our visualization and simplifies our math. And you get it to be
1.5374 arc seconds. And I want to be very clear. This is a very, very, very,
very, very, small angle. Just to visualize
it or another way to think about it
is, one point, there are 60 arc seconds
per arc minute. And there are 60 arc
minutes per degree. Or another way to think about
is a degree is like an arc hour. So if you want to
convert this to degrees, you have 1.5374
arc seconds times one degree is equal
to 3,600 arc seconds. The dimensions cancel out. And you get this as being equal
to-- get the calculator out. This is being equal to
1.5374 divided by 3,600. So it's 4.206. I'll round because we only
want five significant digits. This is of infinite
precision right here because that's an
absolute quantity. It's a definition. So let me write this down. So this is going to be 4.2706
times 10 to the negative 4 degrees. Could write it just like that. Now, let me be clear. This is the total angle. This angle that we care about
is going to be half of this. So we could divide this by 2. So let me just do our
significant digits 4.2706 divided by 2, or I
could even say times 10 to the negative
4-- divided by 2 is going to be 2.1353 times
10 to the negative 4. So that's this angle
right over here. This angle, or the shift from
center we could visualize it, is going to be 2.1353
times 10 to the negative 4. So now that we know
that, we already figured out how to
figure out the distance. We can just apply
this right over here. So let's just take
the tangent-- make sure your calculator
is in degree mode. I made sure of that before
I started this video. --of 90 minus this
angle right here. So instead of retyping it, I'll
just write the last answer. So 90 minus this angle. And we get this large
number-- 268,326. Now remember, what
were our units? This distance right
here is 268,326. I should just round
because I only have five significant
digits here. 300 and-- although with trig
numbered, with the trig, when you're using,
involving trigonometry, the significant digits
get a little bit shadier. But I'll just write
the whole number here 268,326 astronomic units. So it's this many distances
between the sun and the Earth. Now, if we wanted to calculate
that into light years, we just have to
know-- and you could calculate this multiple ways. You could just
figure out how far an AU is versus a light year. So this is AUs. One light year is equivalent
to 63,115 astronomical units, give or take a little bit. So this is going to
be equal to-- the AUs cancel out-- this
quantity divided by that quantity in light years. So let's do that. So let's take this number the
we just got divided by 63,115. And we have in light years. And so it's about
4.25 light years. I'm messing with the
significant digits here but just roundabout
answer, 4.25 light years. Now remember,
that's about how far the closest star
to the Earth is. And so the closest
star of the Earth has this very,
very, very apparent, a very small change in angle. You could imagine as you go
to further and further stars from this, that angle,
this angle right here is going to get
even smaller and smaller and smaller. And all the way until
you get really far stars. And it would be, even
with our best instruments, you won't be able to
measure that angle. Anyway, hopefully, you found
that cool because you just figured out a way
to use trigonometry in a really good way to
measure angles in the night sky to actually figure out how far
we are from the nearest stars. I think that's pretty neat.