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Located in the Large Magellanic Cloud, a star named VFTS 102 is spinning its heart out… Literally. Rotating at a mind-numbing speed of a million miles per hour (1.6 million kph), this hot blue giant has reached the edge where centrifugal forces could tear it apart. It’s the fastest ever recorded – 300 times faster than our Sun – and may have been split off from a double star system during a violent explosion.
Thanks to ESO’s Very Large Telescope at the Paranal Observatory in Chile, an international team of astronomers studying the heaviest and brightest stars in the Tarantula Nebula made quite a discovery – a huge blue star 25 times the mass of the Sun and about one hundred thousand times brighter was cruising through space at a speed which drew their attention.
“The remarkable rotation speed and the unusual motion compared to the surrounding stars led us to wonder if this star had an unusual early life. We were suspicious.” explains Philip Dufton (Queen’s University Belfast, Northern Ireland, UK), lead author of the paper presenting the results.
What they’ve discovered could possibly be a “runaway star” – one that began life as a binary, but may have been ejected during a supernova event. Further evidence which supports their theory also exists: the presence of a pulsar and a supernova remnant nearby. But what made this crazy star spin so fast? It’s possible that if the two stars were very close that streaming gases could have started the incredible rotation. Then the more massive of the pair blew its stack – expelling the star into space. So what would be left? It’s elementary, Watson… A supernova remnant, a pulsar and a runaway!
Even though this is a rather tidy conclusion, there’s always room for doubt. As Dufton concludes, “This is a compelling story because it explains each of the unusual features that we’ve seen. This star is certainly showing us unexpected sides of the short but dramatic lives of the heaviest stars.”
Original Story Source: HubbleSite News Release and ESO News Release. For Further Reading: he VLT-FLAMES Tarantula Survey I. Introduction and observational overview.
The blurb under the photo says it spins 100 times faster than the sun, the article says 300. The article says 1 000 000 miles per hour is 2000 000 km/h but its 1 600 000km/h. I kinda stopped at that point
Gotta agree with you, Mark. Incredibly sloppy and unacceptable. Wouldn’t accept this from – for example – an eight year old’s science report, why should we accept it from an ‘authority’ like Universe Today?
Great to see the errors pointed out above have been corrected. Any idea how they happened in the first place?
Well, whenever I make any errors, it is always due to too much beer! 😉
It is so amazing how this very same author just keeps repeating really fundamental mistakes in basic physics.
Ever thought or imagined what the maximum rotation is before a star rips itself to pieces? What about the really very basic physics of angular momentum, energy and its conservation?
You’d probably have thought it would have been better to express the rotation rate against this maximum velocity, which is close to a value of 1000 rotations per second. The fastest neutron star observed is around 760 rotations per second, making this blue giant more like lumbering giant than something spinning out of control.
Also isn’t angular rotation measured in terms of rpm or degrees per unit time? No measure like that appears here? I do ask. Why not?
interak_99 and Bertie Seyffert points here are really quite valid.
Yo, dude, give Tammy a break – she was just reporting on what the original article at ESO had stated.
Then it would be enough to copy and paste such an article from the original site to Universe Today — for which we do not need a “professional astronomy author” for. But things like this can happen. They sometimes happen to me. More focusing on the task and more critical thinking do help, as far as I can tell.
Actually, according to the original ESO article, the star, VFTS 102, “is rotating at more than two million kilometres per hour” – which is (in old money) over 1,250,000 m.p.h.! So the above article (and also the title) needs to be corrected again.
At first I wanted to get all rowdy about the fact that the rotational speed is given without any reference to the radius or angular velocity of the star, but then I did a little reading up on the star and refreshed my knowledge of how these things are measured. Just to try and cast some light on what they (the astronomers) actual mean by “rotational velocity”: it is the speed at which the gasses on the equator (i.e., on the surface) of the star is actually moving. Now, we all know this shouldn’t be the whole story, and that another piece of information is needed to completely convey the angular velocity. This is, however, where things get a bit… uhm… foggy. In the original article (look for it in the HubbleSite link) the authors basically use the observed rotational velocity to derive a lower limit of about 20 solar masses for the mass of the star. For a main sequence star, this translates to radius about 10 times that of the sun. (now, I know that this star went through some drastic accretion, and isn’t exactly main sequence. Just bear with me) Using my calculator this amounts to an angular velocity for the star of about 3.5 degrees per second (or about 0.6rpm). This can more impressively be understood as it rotating once around its axis every 100 seconds (again, just ball-parking here), on about twice every 3 minutes. Now, look at the seconds hand on your analog watch. This basically means that this star 20 times as massive as our solar system is going around at and angular velocity about as high as that seconds hand’s… Now consider that the sun accomplishes this same feat in about 25 days xD. Ok, I’m done 😀
I too, tried to recalculate these figures. Assuming that angular and linear rotational speed are directly proportional, and knowing that it takes the Sun 25.38 days to make one full revolution (360 degrees turn), an object rotating 300 times faster will complete one revolution in 25.38/300 days, or 2.03 hours – not 100 seconds you stated.
Here is another interesting calculation. The linear velocity at the equator of Sun is roughly 7,200 km/h. Three hundred times faster will make it 2,160,000 km/h for the star called VFTS 102. Then the star’s circumference is 2.03*2,160,000 = 4,384,800 km. Compare this to our Sun’s circumference of 4,379,000 km – pretty close! Yet the article claims the star is 25 times more massive than the Sun. Having almost the same circumference (and therefore diameter) as the Sun, it must be considerably denser. I am not sure how that would make it shine 100,000 times brighter than the Sun though, unless the star is made of totally different materials and converts it to energy in a totally different way.
If the mass of the VFTS 102 is 25 times greater than that of our sun it being packed in the same amount of space would cause it to emit more radiation in the form of heat and light. But it would again only be about 25 times its brightness if that. 100,000 times is just ridiculous. a star like this is essentially a giant neutron star. it is very unlikely that something that big could rotate that fast
The star is both much hotter and much larger than our sun, radiation is proportinal to T^4*R^2, wich pretty fast gets to big numbers.
It is the metric velocity at the circumspherence that is 300 times higher than our Sun’s, not the angular, but since the star as you surmice is much larger this also have to be taken into account.
Assuming the star is has 10 times the radius of the Sun (and i suspect it is even larger than that), then with 10 times higher metric velocity it would turn once in about 25/30 = 0.83 days. Still pretty fast, but not as fast as your watches second hand.
ok, ok. It was 1a.m when I posted, so forgive my hasty napkin calculation for being a bit… uhm… incorrect :D. BUT The angular velocity of the star is not 300 times that of the Sun. Doing the math: 7200km/h = 7200/(60*60)km/s = 2km/s which is 1/300 the rotation velocity of VFTS 102. Further the math tells us that the angular velocity of VFTS 102 is (assuming R = 10Rsun) (600km/s)/(60000km) = 0.01 rads/s (this is where I made the mistake in the original post :P). Seeing as one rotation is 2*Pi rads, the star takes 2*Pi/0.01 = 630 seconds = about 10 minutes to go through one revolution. So, yes, it’s not 10s, but 10 minutes :P. (the ten’s are coincidence)
Let me explain this very thoroughly.
An object rotating at a distance R from center, with an angular speed w, will travel with a metric speed v = w * R
Assuming Sun’s units as base, we have
300 = w * 10
Clearly w (the difference in angular speed) is 30.
The sun rotates in a little more than T = 25.38 days. T is proportional to 1 / w. VFTS 102 rotates one turn in 25.38 / 30 = 0.85 days (okej, rounded up as i included the decimals).
I agree that the radius R is uncertain, but at a lower end – increasing R would make the star turn even slower.
Your calculations are still off.
pfft, I’m just gonna stop trying to do math all together… xDxD You are, of course, correct. My mistake was that I had myself convinced the Sun’s radius was that of the Earth’s… I keep the Earth’s radius in memory for napkin math like this, but usually I’m trying to reckon stuff about the Earth :D. Using the Sun’s radius my approach yields a period of about 18 hours, which agrees with your 0.85 days (of course…)
I’m going to mourn my mathematical skills’ passing in a dark corner somewhere…
pfft, I’m just gonna stop trying to do math all together… xDxD You are, of course, correct. My mistake was that I had myself convinced the Sun’s radius was that of the Earth’s… I keep the Earth’s radius in memory for napkin math like this, but usually I’m trying to reckon stuff about the Earth :D. Using the Sun’s radius my approach yields a period of about 18 hours, which agrees with your 0.85 days (of course…)
I’m going to mourn my mathematical skills’ passing in a dark corner somewhere…
No worries.
You might be able to help me with one thing though, if you want.
If a star is rotating close to breakupspeed, and having a substantial luminous disk (as in this case) that does rotate at just under breakupspeed (the disk remains because the rotation+radiation pressure+gaspressure still allows it to stay in proximity) then the star and the disk will share many properties, like temperature, rotational speed. And the disk is likely a significant factor in size of the star.
Soo, how would the spectral measurements be separated. How could one know that the measurement is not actually of the disk itself, instead of off the star?
Just wondering if you have a clue.
Well, tbh I can’t find any reference to a luminous disk around the star in the actual papers… Granted, I didn’t read the older studies of this star (i.e. from 2007 and earlier), but those studies didn’t even speculate such high velocities (the highest being only about 450km/s). Thinking about it physically, it doesn’t really make a lot of sense to me that the star would have such an extended disk. The higher centrifugal force at the equator would only serve to oblate (flatten) the star more heavily, not create a circumstellar disk. It’s also not all that likely that the disk would still be around from the spin-up process, as the estimated time of about ~5000-20000 years since the companion star went SN is ample time for such a hot disk to cool and fall back onto the star. At the very least it would be too cool to contribute much to the emission spectra… If I missed something about the disk, please point it out, but I rather place my trust in the more peer-reviewed part of this story :D.
You can be correct, I also tried to read from the papers, but ill have to agree that the disk was merely an assumption.
My assumption was like this: Be stars are B- or O-type stars who rotate very close to breakup speed – this does lead to fragmentation in the outer layers that settle into the disk. By this star rotating at close to breakup speed, I assumed that it also have a disk and being a Be star, but that may not always be necessary.
Come to think of it, the characteristic of Be stars is that the disk has a spectral emission profile, while the star itself has a spectral absorbtion profile. Perhaps VFTS 102 simply doesnt have a luminous disk – absorbed by the star, or expelled from the system.
Thanks.
pfft, I’m just gonna stop trying to do math all together… xDxD You are, of course, correct. My mistake was that I had myself convinced the Sun’s radius was that of the Earth’s… I keep the Earth’s radius in memory for napkin math like this, but usually I’m trying to reckon stuff about the Earth :D. Using the Sun’s radius my approach yields a period of about 18 hours, which agrees with your 0.85 days (of course…)
I’m going to mourn my mathematical skills’ passing in a dark corner somewhere…
Why didn’t they express it in revs per sec.?
I agree they should have mentioned the time for a revolution, but the units are on the order of around 1 day, not on the order of seconds.
They can’t give a figure for the rpm’s with any sort of certainty due to the fact that they are unable to directly measure the radius of the star. The lower limits they give for the mass of the star are merely based on the fact that it hasn’t been torn apart by it’s fast rotation.
And 25Msun stars are normally 10 times larger (or significantly more) than the Sun. Besides, the temperature of the star can be measured and by that indirectly measuring the radius (with some uncertainty, as there is also a disk to include) and I am sure the scientists would have reacted if the star somehow was also the hottest star in the universe (required for a star with 100000Lsun if it is also very small).
Say you travel faster than light away from Earth, about 400 light years away and it only took you 100 years to get there, when you look back at Earth you would be able to see what the Earth looked like 200 years before you left.
But if you kept looking at Earth as you traveled faster than light on the way back, what you would see is time fast forwarding until you got back? to Earth 200 years after you left.
I know its not relevant to the article, but i was talking to friend about time travel and he reckoned by traveling faster than light you could travel back in time. In away it is possible if your definition of ‘Time travel’ would be to see into the past, but not physically being there.
Someone has stolen our starturn. x)
No matter how anyone tries to justify it, “miles/hr” is velocity, not rotational velocity. Sensationalism at the expense of science.
As per Kelvin Helmholtz, the sun can radiate 4×10^ 26 Jule / sec. or about 4×10^12 gm. From the report Incredible Spinning Star Rotates At A Million Miles Per Hour!, Universe Today, 5, December, 2011, I received The massive, bright young star, called VFTS 102, rotates at a million miles per hour, or 300 times faster than our Sun does. Centrifugal forces from this dizzying spin rate have flattened the star into an oblate shape and spun off a disk of hot plasma, seen edge on in this view from a hypothetical planet. The star may have “spun up” by accreting material from a binary companion star. The rapidly evolving companion later exploded as a supernova. The whirling star lies 160,000 light-years away in the Large Magellanic Cloud, a satellite galaxy of our Milky Way. Credit: NASA, ESA, and G. Bacon (STScI).
If this star radiate about 100 times mass than the sun per second, then the mass of this star will :
Let radiation = sigma dot = 100 (4×10^12 gm) = 4×10^!4 gm / sec. then the mass will = (Avo. No)^2 x root of sigma dot / one Curie unit x velocity of light = 6.5×10^33 gm, which is about 3.6 times larger than the sun mass. This equation taken from the book COMPLETE UNIFIED THEORY written by me. Again the star is far away, 160000 light year distance means 1.51368×10^23 cm. this distance almost equal to Avo. No. / 4 = 1.5055×10^23 cm, if we treated Avo. No as centimeter. According my theory, I calculated that 2xPi^2 x Avo. No. x ratio of mass of atom and electron = 2.166×10^28 cm which may treated as the radius of the universe. I calculated the distance of all planets from the sun and other stars, galaxies etc accurately with the help of this equation. Actually Complete Unified Theory is single and wonder theory in the scientific world. With the help of this theory, it is possible to calculate the many properties like the mass, radiation, structure, birth, from the particle to the universe.
Nirmalendu Das.
Email : [email protected]
Dated : 06-12-2011
Dear editor, flagged for personal, not science, theory.
I would have loved to, but the last time I checked I’ve not been a moderator of Universe Today. So, someone else made the job. 😉
wow, chipped chopped moderator in the metric stew.
i think the topic stream of this page/site/blog is tip top.
how about a addendum for salient metrics at the bottom of each article?
or would that be too much to ask for free?
i bet that star looks big and bright in the radio sky.
The tangential velocity of a body with spherical radius r rotating with angular velocity ? is v = ?r. It is probably best to state the star’s radius and its angular velocity first. From there a comparison between the tangential velocity of the sun and the star may then be made.
LC