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Stellar distance using parallax

Stellar Distance Using Parallax. Created by Sal Khan.

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  • spunky sam blue style avatar for user Ethan Dlugie
    Aren't star charts usually drawn with West on the right and East on the left? I always think of it as you are lying down on the ground. North is still up, but east is now to your left, and west is to your right.
    (18 votes)
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    • male robot hal style avatar for user Ben Willetts
      Yes, star charts are usually drawn that way, for exactly the reason you suggest. But it is not intuitively obvious for most people who aren't used to looking at the stars, so I think Sal was just assuming the 'normal' way round for East and West because that's what most people deal with most of the time.
      (16 votes)
  • old spice man green style avatar for user Nick Kinnon
    Since the universe is expanding, wouldn't it be true that the star we are trying to measure would have moved away from the earth after six months making it so that the triangle is not an isosceles triangle making our calculations incorrect or is this distance negligible within the time of six months.
    (14 votes)
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  • piceratops ultimate style avatar for user Alexander Schott
    We measure the angle by a sextant for example right? How we get a fix center point to measure theta then if we got theta why we need to wait a half year to measure the other theta? Is that just to get a better estimation (precision?). You don't need the second theta to apply the trigonometry on that issue. How to be sure that your centered point is absolutly the same between those half years?

    Thanks for helping me understanding that issue. Sorry for english misstakes, I am still learning it (from Germany). One simple other question, can i do that double progressive form one sentence before? :)
    (12 votes)
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  • starky ultimate style avatar for user Andres Romero
    How do we measure those angles?
    (6 votes)
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  • blobby green style avatar for user Matthew Png
    What about the idea that during the 6 mths for earth to rotate halfway around the sun, the sun and the star have already moved relative to one another...be it due to their own orbits or due to space expansion. What about curvature of light? The light from the star to us would have probably curved a little due to any large bodies it passes nearby. Even if we had precision devises that could measure angles up to 0.00001 degrees, wouldn't variations due to orbit/expansion/curvature be significant?
    (4 votes)
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    • male robot hal style avatar for user Ben Willetts
      The sun and the star will indeed move relative to each other. But the amount they are moving is so tiny compared to the distance between them, even over six months, that it does not significantly affect the measurement. Remember that any star you are measuring with parallax will be in our own galaxy, so its motion through space will be quite similar to the Sun's anyway.

      As for curvature of light, any body large enough to significantly bend light around itself will be large enough for us to notice it (either because it shines under its own light, or we notice its effects on other light). So we can measure the effect and take it into account.
      (7 votes)
  • male robot hal style avatar for user Bis13
    Does "straight-up" mean the same thing as "zenith"?
    (6 votes)
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  • starky seedling style avatar for user NcWar
    Well is there any way to measure these angles at home lets say a star i might want to find the distance for?
    (3 votes)
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    • aqualine ultimate style avatar for user Mark Geary
      The parallax of the nearest stars is just under 1 arcsecond, which is about the limit of resolution for most amateur telescopes. It might be possible, if you had a relatively large amateur telescope with angle measuring equipment or imaging hardware, and persisted in taking measurements and/or images under near-perfect conditions for a year or more, you might be able to detect motion against the background.
      (5 votes)
  • piceratops ultimate style avatar for user Charlie McGill
    Did astronomers use this same method to first calculate the distance between the earth and the sun?
    (2 votes)
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  • duskpin ultimate style avatar for user Bodhi
    Would sending a telescope out to, say, Pluto, improve our accuracy?
    Have we done anything like this yet?
    (2 votes)
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    • male robot hal style avatar for user Charles LaCour
      The main reason to put a telescope into space it to get it away from the earth's atmosphere. Once it is outside the earth's atmosphere there is no real advantage for putting it out any further. One main drawback to putting it further away is that the power required for the signal to communicate with it increases with the square of the distance. Also the time to communicate with it also increases making harder to effectively communicate with it. The effort to maintain/repair the telescope also increases with distance.
      (3 votes)
  • blobby green style avatar for user Tafhimul Hasan
    I think Sal has made a wrong here. He has figured the direction of east and west in the night sky opposite. You know when you look earth from the above (like from space), the upper side (conventionally Russia) is north and your right hand shows the left. But if you watch the sky from earth, that's not true. Then, there's a mirror effect and right from the north defines east. Am I wrong?
    (1 vote)
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    • male robot hal style avatar for user Charles LaCour
      Think about time zones and when it is noon in them. When it is noon in England it will be noon in New York 5 hours later. The point where the earth is directly facing the sun, solar noon, moves west. So if you are looking down on the earth from above the north pole the earth will be rotating counter clockwise.
      (5 votes)

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.