The Ultimate Guide to Determining Star Distances: Methods Unveiled

[@dextersjab]

Clearest explanation of parallax and parsecs that I've found to date!

[@ANDeoband]

I am a Physics teacher from Pakistan. I am genuinely inspired by you, owing to your extremely practical explanations. Unfortunately, I teach online. Had this not been the case, I would've taught my students the same way as you do.

[@ldu97177]

Can someone explain clearly how the parallax angle (p) is derived purely from the distance of displacement of the star against the faraway stars? If all I know if the distance of the earth to the sun, and the distance the star has seemingly moved against the background, how is that distance translated to the parallax angle? All that is said in the video is that after we measure the amount the star has moved against the background after 6 months, we know the angle. How do we determine the angle if all we see is a distance of displacement? We can’t see angles as such.

[@TheLaw-mh4pb]

yes, this is my exact same question and NO ONE ever answers it. it seems everyone is completely missing this point and totally unaware of it

[@paulutato]

I had this question too and it’s raining somewhere, so. We have a triangle with star whose distance we want to measure at the left vertex 9:49, where two lines converge-one from Earth’s January position, one from June. This angle (θ) measures 2 arcseconds, with a 2 AU baseline between January and June. Bisect the triangle by drawing a line (the orange line in vid) from the star to the midpoint of the baseline, creating a right triangle with a 1 arcsecond angle (p) and 1 AU baseline (a vertical line at far right, also orange, in vid). To find the height (distance to the star), we use trigonometry: tan(p) = opposite/adjacent Now p is a very small angle. And for very small angles, tan(p) ≈ p. So we can swap out the tan(p) with just p in radians. So we need to convert p (1 arcsecond) to radians: To do so, we need to figure out how many arcseconds are in a degree and then convert to radians. 1 degree = 3600 arcseconds (just like 1 hour has 3600 seconds). So, in degrees, angle p is 1/3600 degrees. To convert degrees to radians, multiply by pi/180: 1 arcsecond = 1/3600 degrees * pi/180. 1 arcsecond = (π/648,000) radians = 4.848 * 10^-6 radians Therefore: 4.848 * 10^-6 = 1 / distance. Which is same as: distance = 1 / (4.848 * 10^-6) ≈ 206,265 AU Just as in the video, this distance is one parsec, showing how we convert arcseconds to radians and use trigonometry to find the star's distance.

[@bugpie9437]

Amazing video. Helped me make sense of my college lecture. Thanks Paul!

[@DreaScarlett]

Well done lecture. You made it interesting.

[@Abbu724]

Awesome, outstanding, Marvelous, STUPENDOUS Explanation!!!

[@PhysicsHigh]

Thanks

[@paulutato]

very cool explanation.

[@wispawelwis38]

7:06 How to get that angle from those pictures?

[@globalification]

For the past 1 year I was trying to figure out the parallax method. Only after watching your video I am satisfied that I really understood. Thanks a lot. Very simple and detailed explanation. Could you please share the link of your video for other method? ( luminasity)

[@aliahammedshawon9152]

Such an awesome explanation!

[@gothicbump]

Best video on this topic on KZbin This detailed explanation is what I needed

[@PhysicsHigh]

Thanks

[@Ptoki1]

8:53 how do you measure that angle though

[@marisaelenenadiejamusiccom3974]

In studying the subject I think it’s so much fun...Thank you for this video!

[@joserobertopacheco298]

Amazing explanation!

[@kevron01]

how do u measure the angle?!!?!?

[@marquezedwards3108]

Great explanation of parallax I’m 13 and I’m studying astronomy a lot and this is one of the last things I need to know and you simplified to make it less complex

[@PhysicsHigh]

Great stuff.

[@theodoresweger4948]

Very minor, you ref to approximate and used the symbol for exact. very well done appreciate your work Thanks.

[@turquoisejpg5671]

Great video and explanations

[@PhysicsHigh]

Thanks

[@lawrencelee8995]

What about the Sun’s Motion around the Galactic centre? While the Earth revolves half a circle around the Sun, the Sun itself has moved with respect to the galactic centre. If our target of observation is not moving at the same speed as our Sun so that any motion relative to the background stars are due to the Earth’s motion around the Sun, then the parallax measured will not correspond to a baseline of 2 AU, but 2 AU + the distance moved by the Sun instead. This problem has puzzled me for quite some time and I have no idea how to get it solved... Does this means that the method of parallax is applicable only to those stars moving in the same spiral arm as our Sun?

[@matyaskassay4346]

I assume the distance the Sun and other stars move compared to each other during half a year is negligible. Note that it takes around 250 million years for the Sun to make a full circle around the galaxy.

[@mykytashvets8730]

How is it done to make the perdincular of the baseline to exactly go toward the star you want to measure the distance to? Do you measure the angle everyday for 365 days and then choose the pair of angles that are 6 months apart and have the same value?

[@joneslu1377]

Thank you so much! Does it then mean that Alpha Centauri’s apparent motion between Jan and Jul as seen from Earth is 2x0.77=1.54 arc seconds?

[@司马缸-z6j]

how do you know how much angle shifted by just viewing the star shifting...

[@matyaskassay4346]

We measure apparent distances on the sky with grades. A full circle on the sky is 360°.

[@arunkumartm137]

Same doubt

[@stephaniemcguire]

How did you get 41.4 for the one side?

[@daniels4338]

We use parallax to determine that stars are extremely far away. To measure the parallax and determine the distance, we have to use as large of a baseline as possible because the stars are extremely far away..

[@theodoresweger4948]

if I got this correct 50000 parsec would be 160 thousand light years that to me seems incredible. Thanks so much for the information.

[@RM-lu1kx]

So we get a triangle with a base of 0,000316 Lightyear and a length of about 4,4 Lightyear, and that is for the nearest star. The shifting should be really really small right? And we have to assume stars are not moving. I cannot get my head around it.

[@thaburninator0904]

If we use this baseline measurement to check the distance of stars, how do we know our baseline measurement is good? Isn't measuring the distance to the sun difficult to do accurately, and if that baseline is off, would it result in...rough measurements? My brain is having a hard time with this one.

[@PhysicsHigh]

The diameter measurement is more than accurate enough for measurement. It’s approximately 300 million km. The uncertainty you talk of is minuscule at theses scales.

[@brandaccount2178]

How to find the angle as 1 parsec

[@PhysicsHigh]

Distance is 1 parsec when the angle variation (parallax) is 1 second (which is 1/3600 of a degree )

[@Last.Multiverse]

Conclusion of the video: Great knowledge= Great Math

[@martinwillemse8923]

I have an announcement about the Parallax, that from 2 points 300 million kilometers apart we get a kind of 3D image and we can measure the distances up to 200 light years away and we can measure about 500,000 stars their distance. This is possible because the Earth's orbit around the Sun has a diameter of 300 million kilometers and we can, for example, take a picture of an object in winter and summer and get the distance. It has come to my mind that we can also use pictures from a hundred years ago and now we can take new pictures of the same area and if we are perpendicular to the line we describe in space, in our circle around the nucleus of our galaxy, we get a base that is 2400 times larger than the 300 million kilometers of the orbit around the Sun and we can measure the distance 480,000 light-years away and the base also grows 840,000 kilometers every hour, because our solar system expands with this speed moves around the galaxy core and I wonder what you think about this and if you know if something is already being done with it, or if I'm coming up with something new?

[@zemotika]

I love this bro, this is so true- distances are changing every second since the universe is rapidly expanding… we know how far we are from alpha centauri but it’s constantly changing… also what do you think of the “Great Attractor”??? Because this object is also affecting the gravity of our galaxy and perhaps many nearby galaxies?

[@martinwillemse8923]

Dear @zemotika I have the idea that we are in a universe with shrinking atoms, that is due to the accelerating redshift, which can explain that. When atoms shrink, the wavelength also shrinks and you get a higher frequency, but the interaction between atoms also becomes faster and time and therefore shrinkage also goes faster. If you look into space with that idea and you see a galaxy with 50% redshift, then you're looking at double-wavelength light, which is emitted by atoms with a double diameter, where the interaction is only half as fast, and time and shrinkage is only half as fast, if you then look at a galaxy with 75% redshift, then you look at light that has been emitted with a wavelength 4 times as large and atoms with a diameter 4 times as large, the interaction there goes with a quarter of our speed just like the tide and shrinking and the shrinking takes twice as long as in a galaxy with 50% redshift and is therefore 3 times as far as a galaxy with 50% redshift and you have an accelerating redshift. Beste@@zemotika ik heb het idee, dat we in een heelal zitten met krimpende atomen, dat komt door de accelererende roodverschuiving, die dat kan verklaren. Als atomen krimpen, dan krimpt ook de golflengte en krijg je een hogere frequentie, maar word ook de interactie tussen atomen sneller en gaat ook de tijd en daarmee het krimpen sneller. Als je met dat idee de ruimte in kijkt en je ziet een sterrenstelsel met 50% roodverschuiving, dan kijk je naar licht met een dubbele golflengte, die wordt uitgezonden door atomen met een dubbele diameter, waar de interactie maar half zo snel gaat en gaat de tijd en het krimpen maar half zo snel, als je dan naar een sterrenstelsel kijkt met 75% roodverschuiving, dan kijk je naar licht wat is uitgezonden met een 4 maal zo grote golflengte en een 4 maal zo grote diameter, de interactie gaat daar met een kwart van onze snelheid net als de tij en het krimpen en duurt het krimpen 2 maal zo lang als bij een sterrenstelsel met 50% roodverschuiving en staat dan ook 3 maal zo ver als een sterrenstelsel met 50% roodverschuiving en heb je een accelererende roodverschuiving.

[@intrinsic2reasoning]

0:06 If you you knew what I know about them, 😒 you wouldn't want to even look at them.

[@profeluisegarcia]

The important point to be explained is HOW the small angle is measured

[@matyaskassay4346]

you mean the parallax angle? I guess they just have to measure how much the star has moved compared to it's background stars. A full circle on the sky would be equal with 360°.

[@goldeer7129]

I'm still trying to figure that out. I've made quite some research and the way the parallax method works is quite clear, but I still don't really grasp HOW we actually get the ANGLE. (Also the star is should not be making a right triangle as there's no reason to think it's exactly at the middle...)

[@Ptoki1]

did any of you figure it out yet?

[@wailani3263]

OOPS! My bad ! Diameter of earth's orbit, not of the earth. Of course earth's orbit is elliptical and the earth's position is not at the center of the ellipse, but in common parlance the radius is 93 million miles give or take ... because for a 1/2 of a base line, considering the change of position of the solar system in 6 months, is 2,2 trillion miles .. that makes the earth orbit diameter only 4 % of what the base line should be should actually be.. i guess making the star 96% farther from our solar system.

[@realistic.optimist]

Try decaf

[@wailani3263]

BUT! You are going by earth's diameter. Have you forgotten that our solar system has travelled @ 500,000 mile per hour for 4380 hours(6 months)!!! ... So, earth's position in relation to that star has changed by 2 200 000 000 miles (forget the earth's diameter!) Whoa! Put that in your parallax and smoke it ... that star is not so very far away!!!

[@lioncross9471]

Parallax you say? ....in brightest day ....

[@jayunited01]

Parsec

[@hokage774]

Without knowing trigonometry, you are wasting time here

[@Elo-hv3fw]

No. Without trigonometry, YOU are wasting time here.

[@hokage774]

@@Elo-hv3fw how did you guessed that i have not studied trigonometry? 😂

[@czealtamahawk1228]

So how far is the sun from the moon? And....how big is the sun and the moon?

[@klancyklancy7961]

This all wrong. How can you use a simply way to measure. The formula doesn't take into account how light travel through space it only takes into account how light travels through air. Also, it doesn't take into account how gravity bends light. It doesn't take into account the space time that is bending around the sun and earth. It doesn't take into account the temperature of space and how light travels through it. It doesn't take into account that after a certain point parallel lines are no longer parallel. It doesn't take into account the movement of any heavenly body or the rotation of the earth. This formula uses currency math (math used inside earths closed system, used for counting money, used to find area for square, doesn't take into account any force being imposed, all parallel lines are always parallel), it doesn't use language math (math which includes all forces being imposed, parallel are only parallel for a certain distance as long as no force is being imposed). In other words, you can't measure something like that using this equation because it doesn't take into effect any forces being imposed on the light that is traveling.

[@klancyklancy7961]

I just cant believe how you can past this as right.

[@klancyklancy7961]

Where is the speed of light traveling from the sun to the earth and the earth rotation in this formula D = (d/l)*L? Where is any of the forces being apply to the light of the stars we are looking at in this formula D = (d/l)*L? WHERE? WHERE? WHERE? WHERE?

[@duckman12569]

I bet you have a very nice telescope

[@PhysicsHigh]

Ha. Actually I don’t but wish for one. The opening shot was just done with dSLR and wide angle lens as time lapse.

[@duckman12569]

Could always go with a quality set of binocs, just sayin

[@vincentmcardell8183]

All your calculations are based on a stated distance from the sun to the earth but you don't tell us how we can know that this stated distance is correct. What if I had no knowledge of any distances to any of the celestial bodies? Tell me how I can calculate distances to the sun, moon, and stars beginning without any claims of known distances to them.

[@PhysicsHigh]

The distance to sun has been do a for a few centuries curious.astro.cornell.edu/about-us/41-our-solar-system/the-earth/orbit/87-how-do-you-measure-the-distance-between-earth-and-the-sun-intermediate

[@vincentmcardell8183]

@@PhysicsHigh I am asking you tell us how we can measure the distance to the sun, not how we can take it on faith that others have it right.

[@vincentmcardell8183]

@@PhysicsHigh You have measured that distance yourself, correct? You are not just taking it on faith that what you have been told is true, right?

[@vincentmcardell8183]

All I am asking is that you show us how to begin from a point of having no measurements at all, and then taking our own measurements of what we see in the sky, before we get to the point of doing mathematical calculations. Can you do that?

[@rap1df1r3]

@@vincentmcardell8183 I wouldn't expect any meaningful answer to that question, seeing how the distance has never actually been measured, only assumed based on a mathematical model.