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Astronomers Find Planet Hotter Than Most Stars (nasa.gov)
175 points by el_duderino on June 5, 2017 | hide | past | favorite | 55 comments


It's also losing 10 million kilograms of mass a second. The universe never ceases to be mind boggling large.


It has a mass of roughly 10^27 kg (per the article, it is very comparable to Jupiter in mass).

Using a loss of 10 million (or 10^7) kg per second , we can calculate that the planet is losing 10^7 / 10^27 percent of its mass per second:

10^7 / 10^27 = 10^(-20) % = 0.00000 00000 00000 00001 % of its mass per second

If we change the time interval to one year, then we simply multiply the number of seconds in year (32 million, roughly), which moves the leading digit 7 places to the left.

The resulting answer is that the planet is losing 3*10^-13 %, or 0.00000 00000 003 %, of its mass per year =)


Again. The craziness of the universe. Thanks for doing the calculations I appreciate the context.


Is it close enough to the star for relativistic effects to make a difference here?


Very slight. Mercury is close enough for very very slight.


I try to visualize something like this and I can kind of feel my brain just trying to click off.


Same. It makes me want to see it all the more.

This is all proving to us, day after day, that the universe is much more complicated and weird than we could ever conceive let alone comprehend.


> Same. It makes me want to see it all the more.

Assuming your eyes don't completely fry instantly I have often wondered what the human eye would see being that close to such a bright hot star.

I assume complete whiteness and zero shadows but I honestly have no clue.


Imagine how, on a super sunny Earth day it's hard to keep your eyes open outside. I imagine it would be like that, but way worse.


> it would be like that, but way worse

I'm not sure I'd describe "being sublimated instantly" as just being "way worse" than having a hard time keeping my eyes open on a bright day but I guess you are technically correct :)

I wish we had the tech to simulate what's happening there in a way we can understand and see.


Ah yes :)

I was talking more about being "near" the star, haha.


Given that 1m3 of water is 1000kg in normal conditions, it is 10000m3 of water, which can be represented as ~21m cube. Not tiny, but nothing to click off for.


> KELT-9b is 2.8 times more massive than Jupiter, but only half as dense.

> Because the planet is tidally locked to its star

What does it mean for a gas giant to be tidally locked?


It means that if you average together the motion of all of the mass of the planet, it revolves once per orbit around the star. Which is the same as what it means for a rocky planet, if you think about it, it's just a little less obvious.

Another way to think about it is that the gas giant moves such that tidal friction caused by the gravity of the star is minimized... because that's why it's moving like that.


Makes me wonder what shape such a planet would take.


Very slightly elongated along the axis between itself and the star. Still nearly spherical and the deviation would be essentially ellipsoidal (not egg-shaped).


It's probably a bit more exciting than that when you include evaporation and the stellar wind! But yes, an ellipsoid is the basic shape, the tides elongate the planet both towards and away from the star, same as the tides on the Earth caused by the moon.


Yep. The original post even speculates the thing might have a comet-tail because of the evaporation.


I kind of imagine it looking sort of like a Hersey's Kiss shape :-)


"Tidal locking (also called gravitational locking or captured rotation) occurs when, over the course of an orbit, there is no net transfer of angular momentum between an astronomical body and its gravitational partner."[0]

[0] - https://en.wikipedia.org/wiki/Tidal_locking


One side of the planet is always facing its star. And I presume the gas swirling is on a much shorter/slower scale than the planet's rotation, so the notion of a side is meaningful.

https://www.google.com/search?q=tidally+locked&ie=utf-8&oe=u...


You have had several responses which are all basically correct. If you are still having trouble with the concept, think of the moon and how it always shows the same face to the Earth: you never see the "back." This is because the moon is tidally locked to the Earth.


I understand tidal locking as it applies to solid bodies. It was what "always shows the same face" would mean for something made of constantly moving and mixing masses of gas that puzzled me.

But, as you say, i have had several good answers - being tidally locked is about transfer of angular momentum, and gas giants still have angular momentum.


I imagine the gas flows in circles as if the poles were always perpendicular to the line between the star and the planet's center.

In other words, the gas on the dark side stays on the dark side, and vice versa, roughly.


I would imagine that you'd have a huge upwelling on the hot side, flow of gas to the dark(er) side, then cooling and sinking and flow along the inside over to the hot side.


Wouldn't those attributes mean the volume is exceptionally high? I thought Jupiter was near the limit for volume as increases in mass beyond that point just lead to more and more density.


As you heat gas it gets bigger. Jupiter is cool enough that it's got liquid and perhaps solid hydrogen in the middle (Juno will tell us the details soon!) I think this thing is so hot that it won't have a liquid or non-rocky solid core.


> It won't have a non-rocky solid core

Meaning a solid core, if it exists, will be rocky? How, at such high temperatures?

I'm uncertain whether I'm not misinterpreting your double negatives :)


High pressure does odd things even when it's hot, and I don't know off the top of my head how rocks work in that situation! I'm more of a black hole and accretion disk devotee.


I'd assume the minor spaghettification caused by extreme tidal forces us also a factor in its super low density.


I'd assume it loses material quite faster than a spinning one.

Here's a quick model: https://media.giphy.com/media/MBpKFBsQdeVIk/giphy.gif


I wonder how broken our astrophysical models will prove to be once we are more and more effectively able to peer outside our home system. There is a universe of examples out there, and I would find it slightly depressing if all of that complexity could be described by the theories of a swiss patent clerk.


Einstein's field equations are a system of extremely nonlinear partial differential equations. Actually using them to compute something is very nontrivial. And of course gravity is but one of the many forces at play here. You still have to account for thermodynamics, chemistry, nuclear physics, and other complicated effects to understand what's going on here.


Is the sun going to end up being called a "Dwarf Star"?

Also, this means nothing to me: "With a dayside temperature of more than 7,800 degrees Fahrenheit (4,600 Kelvin)"

Where in the atmosphere is this temperature begin measured? The temperature can vary wildly by altitude, latitude, and time of "day". As far as we can tell, as the atmosphere of a planet gets thicker the temperature keeps increasing indefinitely (eg check out fig 1): http://faculty.washington.edu/dcatling/Robinson2014_0.1bar_T...


Looking closer:

>"Given the high stellar luminosity and close orbit, the planet receives a large stellar insolation flux (Table 1). As a result, it has an extremely high equilibrium temperature: calculations, assuming zero albedo and perfect heat redistribution, give a value of approximately 4,050 K."[1]

That 4,050 K value is reported as the equilibrium temperature, which I understand to be the output of the Stefan-Boltzmann law:

  T = (I*(1-alpha)/(epsilon*sigma))^.25
  where,
  I       = Incident Flux
  alpha   = albedo
  epsilon = emissivity
  sigma   = SB constant = 5.670373e-8
According to the paper the planet receives 61.1 x10^9 erg/s/cm^2.[1] The solar constant is ~1360 W/m^2.[2] In the same units the planet gets 6.11 x 10^7 W/m^2, which is about 45 thousand times greater than what the earth receives. Anyway, plugging into the SB law I get:

  (6.11e7*(1-0)/(1*5.670373e-8))^.25 = 5729.377 K
So I am certain it is I rather than them with the error. Where did I go wrong?

[1] http://www.nature.com/nature/journal/vaop/ncurrent/full/natu...

[2] https://en.wikipedia.org/wiki/Solar_constant


I see now, I was calculating the temperature at the subsolar point, they are averaging over a uniformly illuminated sphere.

Since the illuminated side is a disc with area pi x R^2 and the sphere has surface area 4 x pi x R^2 we need to include a factor of 1/4:

  (6.11e7*(1-0)/(4*1*5.670373e-8))^.25 = 4051.281 K


In this case 4600K is what a black body at the planet's distance would be heated to by the star.


Thanks, actually take a look at the self-response I was writing as you posted this.


I just want to know where it is, jeez. Map of the galaxy please! (Assuming it's in this galaxy.)


After a extensive search I give up, but I found some data that may be useful.

As another commenter found, these star and planets are 650 light years away, that's very close for a star, very near us in the Milky Way.

This is a map of our neighborhood in a range of 5000 light year: http://www.atlasoftheuniverse.com/5000lys.html

For comparison, Polaris is 323–433 ly away and Rigel is 860 ± 80 ly away in the opposite direction. So 650 ly is a distance in between, but I still can't understand in which direction is this.

In a map of the full galaxy (x10 zoom out) http://www.atlasoftheuniverse.com/galaxy.html these star and planets are only a few pixels away from the Sun.

More data, in case someone know how to use an astronomy program to make a nice graphic: http://simbad.u-strasbg.fr/simbad/sim-id?Ident=HD+195689


> So 650 ly is a distance in between, but I still can't understand in which direction is this.

The SIMBAD link you gave gives you the RA and Dec (Right Ascention and Declination) which gives you a precise co-ordinate for where in the sky the object is. It's a bit difficult to wrap your head around the co-ordinate system but if you think about it hard enough you could translate to a map of the galaxy.

To be frank, most researchers don't look at maps of the galaxy for analysis, in general you're looking at things from the perspective of Earth.


Does the article specify the planet's distance from Earth? I can't seem to find this rather fundamental information.


Seems like the article indeed forgets to mention the distance. I guess the nature article[0] has all the information, but since I can't access that one right now I also found a Washington Post article[1], according to which the distance from Earth is about 650 light-years.

[0]https://www.nature.com/nature/journal/vaop/ncurrent/full/nat...

[1]https://www.washingtonpost.com/news/speaking-of-science/wp/2...


It doesn't but the paper's abstract mentions the HD identifier, so you can look it up on SIMBAD[1] to get the parallax. While parallaxes aren't the best measure of distance (they really aren't very good at all) you can get an okay estimation by taking [d = 1/p] where d is in parsecs and p is in arcseconds. Saving you the math, it's about ~188 ±20 parsecs (~564±60 ly).

[1]: http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=HD+195689&...


The nature article[1] mentions in the abstract that the identifier for KELP-9 (the host star) in the Henry Draper catalog is HD 195689. Here's the information about it from SIMBAD[2]. It also appears to have been referenced in 6 other papers, so I presume that such information already exists about it.

[1]: https://www.nature.com/nature/journal/vaop/ncurrent/full/nat... [2]: http://simbad.u-strasbg.fr/simbad/sim-basic?Ident=HD+195689&...


Any talk of a confirmed exoplanet will be in our galaxy.


Planets: so hot right now.


>He worked on this study while on sabbatical at NASA's Jet Propulsion Laboratory, Pasadena, California.

While on sabbatical? Wow! That's dedication.


Ummm . . . I think you misunderstand the meaning of "sabbatical." It doesn't mean vacation. It means freedom from teaching and administrative duties to give you time to go to another institution, learn new things, and work on interesting problems. This is exactly the sort of thing that I expect an academic on sabbatical to be doing.


I wonder why we don't see this in the corporate world more? When we need a change of pace we have to quit one company and start at another one, you'd think it would be in the companies best interest to have people go elsewhere to get their fix and then come back with their old knowledge plus any relevant new knowledge.


Apple used to have sabbaticals until 1997 or so (6 paid weeks off every 5 years). Anecdotally, I heard that many employees used the leisure time to look for new jobs, but that may have been due to the turmoils Apple was undergoing in the 1990s.


6 paid weeks off is hardly a sabbatical. For many jobs in the EU you get that many vacation days anyway each year.


That's about as much as people in EU get per year :)


In other words, sabbatical == getting sh*t done.


All true, though academics seem to wind up on sabbatical in the south of France with suspicious regularity.




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