Mass of the sun

Mass of the sun calculation with NASA data and multivariable calculus, pretty accurate!
In this video, unlike the meme, I actually calculate the mass of the sun ☀️ For this, I use spherical coordinates, as well as a density function of the sun (according to data provided by NASA); surprisingly the answer we get is very close to the actual mass of the sun! Enjoy this sizzling adventure!

[@blackpenredpen]

WTF the mass of the ☀️

[@ryanaldrinputong4318]

Meanwhile, flat earthers calculated the mass of the sun using cylindrical coordinates

[@blackpenredpen]

Oh btw, so you will put 2+2 on the midterm right?

[@yrcmurthy8323]

Ya

[@weerman44]

I love random videos like these! It's amazing that NASA found that formula for the density :D

[@koenth2359]

Very nice! From a physics point of view however two things feel a bit odd. - Convention is to use rho for density, not for normalized radius; - It is far, far, more difficult to find the mass distribution in the Sun, than its total mass, so this really feels 'backwards'. Historically, the mass of the sun was found using Kepler's and Newton's laws, and experiments determining the gravitational constant G Even though the absolute distances of planets to the sun were not known accurately yet, the relative distances were known quite well (essentially by triangulation), so it could be established that the expression below for MG gave indeed the same value for different planets (Kepler 1609-1619) 4π^2r/T^2 = MG/r^2 => MG = 4π^2r^3/T^2 Here r is the mean radius (average of perihelion and aphelion), and T is the revolution time (siderial). For Earth, once the distance to the sun was known reasonably precisely (Cassini found 1.4×10^11m in 1672), and using Newton's understanding of gravity (1680's) one could find MG ( here I use a more modern value): T = 365×24×60×60= 3.16×10^7 s r = 1.50×10^11m MG = 4π^2r^3/T^2 = 1.33 × 10 ^20 [m^3 s^-2] Historically, the big challenge was to find G with any accuracy. In 1798 Cavendish measured the force between known masses (certainly not easy!) and called that 'weighing the earth'. Translated into SI units he found G=6.74 ×10-11 m3 kg-1 s-2. Nowadays we know it is G=6.67×10-11 m3 kg-1 s-2 1.33 × 10 ^20 / 6.74 ×10-11 = 1.97E30 kg The error in the result is within 1%, probably due to using only 3 digit precision in the various data

[@lithostheory]

2:24 lack of (physical) intuition of a mathematician... of course the center of the sun is MORE dense than the outer layers.

[@NAMEhzj]

Yes, it doesnt seem that obvious from the formula but this is what the graph looks like: www.wolframalpha.com/input/?i=519r%5E4+-1630+r%5E3+%2B+1844r%5E2+-+889r+%2B155+on+%5B0,1%5D&assumption=%7B%22MC%22,+%22r%5E4%22%7D+-%3E+%7B%22Variable%22%7D

[@Mau365PP]

@@NAMEhzj cool

[@JobvanderZwan]

Read the formula more closely: r⁴ and r² are positive, r³ and r are negative, and r³ has a bigger constant than r⁴. So initially, r³ will dominate, but when you go out further r⁴ will dominate. The question is to which unit r is set. If it is normalised to [0,1] as NAMEhzj did, the core is densest. If it is using meters, it looks like this: www.wolframalpha.com/input/?i=519r%5E4+-1630+r%5E3+%2B+1844r%5E2+-+889r+%2B155+on+%5B0,6900000000%5D&assumption=%7B%22MC%22,+%22r%5E4%22%7D+-%3E+%7B%22Variable%22%7D

[@lithostheory]

@@JobvanderZwan r is a fraction of the total radius so it is a unitless quantity and runs from 0 to 1. (Otherwise the units would not work out in the formula (unless the constants have crazy untis...)).

[@wolframalpha8634]

Meme - a - matics

[@Theraot]

The mass of the sun is exactly one solar mass

[@schumensch]

Hmmmm just a few million billion trillion kilos off. No big deal 😂😉 Great job!

[@duckymomo7935]

Let M = mass of the sun Therefore, the mass of sun, ☀️, is M

[@codemagenta2681]

Btw. The Model density doesn't converge at r=infinity. It even isn't continous at r=1, if you say that the density is 0 for r>1. There are actually more elegant models than that, which might be numerical less good, but they descripe the actual physics better.

[@NAMEhzj]

Why would it have to converge at r -> infinity? why would it have to be continuous for r> 0? Just define the density as a function of r where r is in [0,1] because for bigger r that isnt the sun anymore... So that then is continuous everywhere (even at 1, continuity depends on the domain) and you can integrate it perfectly fine...

[@shiina_mahiru_9067]

There is something makes me feel 'the triple integral over the sun' so funny 😂 My first thought on it was knowing the distance between sun and Earth, mass of Earth, and assume circular orbit, and use the gravitational force formula and circular motion to approximate it

[@manojprabhakar9111]

But distance between earth and sun varies continuously

[@nischay4760]

@@manojprabhakar9111 it varies very little though.. i think approximating it as a circle should be good enough

[@shambosaha9727]

We can do it without assuming circular orbit... just use Kepler's third law.

[@cipriannegoescu5847]

This dude is high on math all the time. Truly inspiring!

[@IoT_]

Looks like someone wants to get another PhD in physics 😁

[@drpeyam]

Never 😂😂😂

[@bruzanHD]

Missed the opportunity to write RdrR... hardee har har

[@szymon5830]

With no joke i actualy (a year ago)had a homework to calculate the mass of a Sun, but subject was not math it was physic.

[@catholic_zoomer_br]

Mass of the sum == number of subscribers Dr. Peyam deserves

[@shambosaha9727]

Well... in femtograms though

[@LucasEccard]

OBF- Brazilian Phisics olimpics. The question this year was, briefly: knowing that the ligh takes 500 seconds to reach Earth, from Sun, estimate the mass of Sun. This was the funniest question of the test, and the coolest: it can be done! PS: You also had the values of constants on the first page of the test, like speed of light, etc.

[@tamasvaszary9812]

It is so obvious man! gravitational force equals the centripetal force and you can write the velocity of the Earth in terms of distance from the Sun and angular velocity. The distance is ct of course... the problem takes 1.5 minutes actually

[@jimdetry9420]

I thought, how would I answer the question? If you know the orbital period (p= year) and distance to the sun (r~93,000,000 miles) and the gravitational constant (easy to look up ~6.7E-11 m^3/(kg s^2)), convert 1 year to seconds, the distance to meters and use M=r(v^2)/G (where v=2 pi r/p). You get ~2.0E30 kg

[@Mau365PP]

Well, now math teachers have proof that it can be done (and under 15 min). *no excuses !!!*

[@gnikola2013]

What? Physics with Peyam? Oh nvm it's just some multivariable calculus

[@davidgould9431]

To everyone who thinks this is physics, not maths: the error was ("less than one: pretty precise I would say" (13:57)) about 50% (1.9ish + 0.9ish = 2.8ish). This puts it firmly in the field of astronomy (except the order of magnitude was actually right). Inspiration for this observation: Professor Mike Merrifield on Sixty Symbols, who's made a number of jokes about his field's estimating ability. Unless, that is, you consider physics's 120 orders of magnitude problem with the cosmological constant (www.csicop.org/sb/show/the_problem_with_the_cosmological_constant) ― but I'm sure we can somehow brush that under the carpet. Great video, by the way. :-)

[@gordonchan4801]

2:20 I think it should be denser as we approach to the centre.

[@tatjanagobold2810]

wooow peyam this was awesome! Exploring the physics world ;)

[@mohamedosama9188]

Very Very Fantastic

[@Niyudi]

A question from the Brazilian physics Olympiad 2018 was: "Light takes about 8min20sec to get from the sun to earth. Estimate the mass of the sun." It's meant for high schoolers and it's actually not that hard, given you actually study physics haha

[@tamasvaszary9812]

It is so obvious man! gravitational force equals the centripetal force and you can write the velocity of the Earth in terms of distance from the Sun and angular velocity. The distance is ct of course... the problem takes 1.5 minutes actually

[@jkn6644]

2:10 Max density is of course at the center of the Sun: 155 kg/liter. That is rather heavy plasma. Sun is elliptical, but I think error is in this density approximation. spacemath.gsfc.nasa.gov/weekly/6Page102.pdf

[@wavexphonk]

Opens physics paper: Question, Calculate the mass of the sun. Me: Do i look like i care...........

[@ffggddss]

Not bad? That's terrible! It's off by a factor of almost 3/2 ! Incidentally, the radius of the Sun, to more places, is R = 6.955·10⁸ m = 6.955·10¹⁰ cm which would increase the discrepancy. You can do *much* better just by applying Kepler's 3rd Law in the form: GM = ω²a³ using the universal gravitational constant (G ≈ 6.6743·10⁻⁻¹¹ m³/kg·s²), and the measured semimajor axis (a ≈ 1.496·10¹¹ m) and period (T = 2π/ω ≈ 365.256·86400 s) of Earth's orbit. I *REALLY* think the NASA density model has a typo somewhere. But thanks for showing this! BTW, the oblateness of the Sun is very small, and wouldn't introduce anywhere near this much error into that calculation. PS: There IS an error in that polynomial. When I do the sum of fractions at the end, I get a negative result: 519/7 - 1630/6 + 1844/5 - 899/4 + 155/3 = -1.8071... Hmmm, looks like it's exactly 2.5 less than yours - that difference would vanish if the next-last term were 889/4, rather than 899/4. Did you maybe make a copy error there? Another thing - that density at r=0 is 155 g/cm³ = 155 kg/L - that's really dense! But that is what we should expect at the center of the Sun! Then at the surface, r=1, where it should go to 0, or very close to 0, if we replace 899 with 889, we get: (519 + 1844 + 155) - (1630 + 889) = 2518 - 2519 = -1 so I'm suspecting maybe that cubic coefficient should be 888, not 889. Or some other coefficient may have to change by ±1. That would, however, make the mass result a bit higher, increasing that discrepancy. One of the disconcerting features of that density model, is that it's an alternating sum of large numbers; not very good for overall computational accuracy. Fred

[@drpeyam]

Well, I dunno, it’s at least what’s written on the following page, do whatever you want with that information... spacemath.gsfc.nasa.gov/weekly/6Page102.pdf

[@ffggddss]

@@drpeyam Ah, so my first suspicion was correct - you wrote "899" on the board, where it should have been "889." Other than that, you're right; your final result is what the model on that page gives; and the model density at the surface does come to -1. Finally, it's impressive only because of the wacky polynomial they gave for a model; a difference of large numbers is always trouble when accuracy is desired. Regardless, it is a fun, and educational exercise; I *do* appreciate your work here. Fred

[@drpeyam]

Oh, you're right, my bad! Still doesn't change by much, so maybe the model isn't perfect ;) Thanks for the kind words :)

[@leonardromano1491]

The density function does not seem very good to me. f(1) is not zero so it is not even smooth and it does not vanish for infinite radius. You can get better approximations using the mean distance from a given object in great distance from the sun (so that the newtonian approximation is good) and its dynamics.

[@coreyplate1001]

The sun's surface, is, in fact, not smooth, and the density formula is clearly meant to be used within the boundaries of one solar radius, given appropriate approximation of the radius at various points on the sun's surface. It's not meant to model the sun as if it expands infinitely into the universe.

[@rogerkearns8094]

What's the mass of the sun? Just a M☉...

[@korayacar1444]

What's the distance to the Sun? Just an AU.

[@cedricp.4941]

The density goes negative when r->1

[@michaeleiseman4099]

Yes, I noticed this too! If the expression for the density were d=519r^4-1630r^3+1844r^2-889r+156 {0 1 and the density at the center of the sun is 156 gm/cm^3. Perhaps this is what he meant?

[@AndDiracisHisProphet]

and this makes sense

[@pierreabbat6157]

This is backward. To calculate the mass of the sun, you take the astronomical unit and the year and plug them into the Kepler-Newton formula. To calculate the density versus radius, you take the mass and the radius, and you have to know astrophysics to figure out the equations.

[@Gold161803]

How is the density measured from Earth?

[@hOREP245]

Where did you find that original density equation?

[@drpeyam]

NASA, haha

[@reggyreptinall9598]

NASA knows the density of the sun, that sounds like a joke.

[@awabqureshi814]

Beautiful

[@rob876]

Such a poor model for the density of the sun. You'd think that the least requirement for the model would be that it gives the correct mass for the sun. What went wrong? Is it perhaps a discrepancy in how the radius is defined?

[@drpeyam]

I think I also made a small sign error somewhere, but the density given isn’t 100% perfect

[@JamalAhmadMalik]

And I thought m stood for math when it was for mass!

[@federicopagano6590]

good job my heart

[@cameronkhanpour3002]

Although I love the application of the math you employed into solving this problem, I don't believe the answer you got as accurate as it might seem. The mass you calculate is 2.86*10^30, and the actual mass is 1.989*10^30 so you explain these to as close because they are only about 1 away from each other, but it actually 1*10^30 difference, which truly shows the difference between the estimated mass and the actual mass.

[@coreyplate1001]

It's a relative error of 43%, which is quite large when you're weighing, say, the amount of an enzyme you want to add to your organic starting materials in the manufacture of a medication, but, not so much for most purposes related to our understanding of the sun.

[@AlexandPedro]

Do the mass of the Earth next!

[@drpeyam]

Haha, exact same approach actually, just different density!

[@Vincentsgm]

And radius

[@AndDiracisHisProphet]

@@drpeyam i think more interesting is the mass of earth's atmosphere

[@gian2kk]

Why is the mass the triple integral of density?

[@mikhailmikhailov8781]

Imagine an integral as the sum of products f(x)dx. It has to have the same dimensions as f(x)dx. So if you integrate a function where f(x) is in kg/m^3 and x and dx are going to be in meters. Thus you will get some quantity in kg/m^2. Integrate two more times and you will arrive at kg alone, which is what we need.

[@atrumluminarium]

Triple integrals over x,y,z give the volume and by definition ρ=m/V

[@koenth2359]

I just thought about integrating shells of area A(r) = 4 π r^2 with mass dM = f(r) dV = f(r) A(r) dr = (519t^4 -1630t^3 +1844t^2 - 889t+155) 4 π r^2dr where r=Rt and dr =Rdt, so 1 M = 4π R^3 ∫ 519t^6 -1630t^5 +1844t^4 - 889t^3+155t^2 dt 0

[@rob876]

The volume of an infinitesimal element is dxdydz so you have so "sum" over x, y and z (three "sums").

[@guitar_jero]

So... if we could fill a 1 liter bottle with Sun’s core it’d weigh approximately 155 kg?

[@coreyplate1001]

For a moment, but, being as it's made up of oxygen, nitrogen and carbon in ionic form, it would erupt from the bottle with violent force and expand to a more stable density

[@user-rv9vk8by5i]

Oh, silly mathematicians... He just estimated the mass of the sun but he doesn't realise it has gravity, which pulls it together, creating huge pressure in the centre and the densest part is the core To compare: the earth, tiny as it is, is big enough to create pressure which forces iron atoms to be so close together that they're a solid in the inner core, while the outer core is cooler and has less pressure, but it consists of liquid iron Tldr 155 g/cm^3 is the density of the surface of the sun, not the sun's core

[@coreyplate1001]

@@user-rv9vk8by5i the density is 155g/cm^3 in the center, and the sun isn't massive enough to fuse iron : P

[@shivimish9962]

how do you model the density variation of some thing that is not even visible?#mind=blown

[@AlwinMao]

The equations of stellar structure: how density, temperature, and pressure relate while considering gravity and the outward flow of energy. Because we can measure the luminosity and mass and radius of the Sun, we have all we need to figure out its internal structure. Each layer of the Sun must be a certain density so that it can hold up the layers of the Sun above it, while fitting the constraints of luminosity/mass/radius that we can measure.

[@HomelessBumm]

And who decided that?

[@mariomassy4751]

Really the best video thanks !

[@jfr9964]

Or you could just take the Earth’s mass and orbital velocity, but...

[@Culmen222]

286 is 44,44 ish % more than 198. Like a guy not weighing 80 but 115 kg. I guess the NASA formula was simplyfied. Interesting calculation, hard to imagine those huge numbers. Actually Otto Walkes once made fun of Rammstein's "Sonne", joking "Hier kommt die Tonne". (here comes the ton) Tonne also means heavy person in German. kzbin.info/www/bejne/hpqTkptjeNKhaNk

[@matthiashannesson7239]

This is perfect.

[@AlwinMao]

Anyone interested in fitting the polynomial yourself using the original source, this is data provided by Astrophysicist John Bahcall. www.sns.ias.edu/~jnb/SNdata/Export/BP2004/bp2004stdmodel.dat

[@AlwinMao]

If you'd be interested in doing another version of the video you could approximate the density as exp(-10*r + 5.46). When I looked at the NASA fit vs this data, it is rather inaccurate at higher radii. Higher polynomials in the exponent provide a better fit but are harder to integrate by hand ;) A physical note: atmospheres naturally have exponential forms because layers have to support themselves but also the layers on top of them, which requires exponentially higher density on the way down.

[@hail0]

uhhh sir is this gonna be on the test???

[@drpeyam]

Of course!!!

[@yogeshsingh3597]

Fun😎😎😎😎😎😎😎🥰🥰🥰🥰🥰🥰

[@williamdavis2505]

You should question the polynomial model of the density of the sun. Try plot (519r^4−1630r^3 +1844r^2 −889r + 155, {r = 0.4 to 1}) in Wolfram Alpha. A negative density for r > 0.95? That can not be correct. Density approaches zero at around r~= 0.52? Strange, and also suspect. Try the same plot for {r = 0 to 1}. The maximum density for r > 0.52 is about 2 g/cm^3. The maximum density for r < 0.52 is 155 g/cm^3, achieved at the center, r = 0.

[@coreyplate1001]

It makes sense if the density of the sun really does behave in such a way. There are many forces interacting in the sun, including its own gravity, the hydrogen fusion engine, the CNO cycle, the gravitational forces of the solar system, which, in aggregate are quite significant, and the interactions are even more complex, with quantum mechanics, fluid dynamics, thermodynamics, and many other types of interactions apparent. The sun has significant layering, thanks to collecting oxygen, nitrogen and carbon collecting at the center, the helium collecting in a shell around it, which will not burn in earnest until the hydrogen engine ceases to efficiently push back against the force of the sun's own gravity, the hydrogen shell, and the area of the sun where hydrogen ions escapes as a part of the solar wind. The only part which I find must be incorrect is the negative density, which implies negative mass, which while not impossible in a broad, cosmological sense, is probably not present in the sun in any fashion.

[@williamdavis2505]

Corey Plate thanks. Did you run that Wolfram Alpha code? I agree there should be no such thing as negative mass (although Matt at PBS Spacetime has gone a bit crazy about this recently). How do you explain a layer of almost zero density at r ~= 0.52?

[@abhisheksinghsoam6014]

Sir I have found a way. To. Calculate the mass of sun I want you to check it whether it is correct or not actually in my area there is no one whiich can calculate that so. Please help me if you are interested please comment back. And. I. Can garantie you. That you will love the solution 😊

[@LokolcThe]

Now prove the sum of fractional part of π is infinite.

[@davidgould9431]

Assuming every digit appears infinitely many times, then the sum is obviously ∞* (-1/12), so it must be -∞ :-D

[@LokolcThe]

@@davidgould9431 You are right!

[@andresfv6490]

Where's Oreo !? >:W

[@bandamkaromi]

what the hell big fat stuff is sun.. :/

[@derfret1365]

Uhm... who does the error calculation?

[@coreyplate1001]

It's about 43% error

[@theSASarethebest]

Lol

[@XdarkmutantX1]

Wonderful, but can't you substitute in Newton's law and get the answer?

[@wolframalpha8634]

LOL then the video wouldn't last a minute

[@yourlordandsaviouryeesusbe2998]

Haven't watched the full video yet, bit are you talking about Newton's law of radiation or Newton's law of gravitation? I know that the law of radiation can also be used to find the mass, but it will produce a large margin of error.

[@wolframalpha8634]

@@yourlordandsaviouryeesusbe2998 he means the gravitational one

[@hOREP245]

LMAO it would be a lot simpler and give almost the correct answer.

[@Silverwind87]

3

[@aidangm7419]

YOU FIGURED IT OOOOOOOOOOOOOUUUUUUUUUUUUUUUUUUuuuuu u u u t