Tatooine-like planet discovered (w/ video)

Sep 15, 2011
Kepler-16b, shown in this artist's conception, enjoys double sunsets as it circles a pair of stars. The orange (type K) and red (type M) stars orbit each other every 41 days, while Kepler-16b orbits them both every 229 days at a distance of 65 million miles. Kepler-16b is similar to Saturn in size and mass, with a surface temperature of about -100 to -150 degrees Fahrenheit. Credit: David A. Aguilar (CfA)

(PhysOrg.com) -- Although cold and gaseous rather than a desert world, the newfound planet Kepler-16b is still the closest astronomers have come to discovering Luke Skywalker's home world of Tatooine. Like Tatooine, Kepler-16b enjoys a double sunset as it circles a pair of stars approximately 200 light-years from Earth. It's not thought to harbor life, but its discovery demonstrates the diversity of planets in our galaxy.

"Kepler-16b is the first confirmed, unambiguous example of a circumbinary planet - a planet orbiting not one, but two ," said Josh Carter of the Harvard-Smithsonian Center for Astrophysics (CfA). "Once again, we're finding that our is only one example of the variety of Nature can create."

Carter is second author on the study announcing the discovery, which appears in the Sept. 15th issue of the journal Science. He is presenting the finding today at the Extreme Solar Systems II conference in Jackson Hole, Wyoming.

Kepler-16b weighs about a third as much as Jupiter and has a radius three-fourths that of Jupiter, making it similar to Saturn in both size and mass. It orbits its two parent stars every 229 days at a distance of 65 million miles - similar to Venus' 225-day orbit.

Both stars are smaller and cooler than our Sun. As a result, Kepler-16b is quite cold, with a surface temperature of around -100 to - 150° Fahrenheit.

NASA's Kepler mission detected the planet through what is known as a planetary transit - an event where a star dims when a planet crosses in front of it. The planet's discovery was complicated by the fact that the two stars in the system eclipse each other, causing the total brightness to dim periodically.

Astronomers noticed that the system's brightness sometimes dipped even when the stars were not eclipsing one another, hinting at a third body. The additional dimming events reappeared at irregular time intervals, indicating that the stars were in different positions in their orbit each time the third body passed. This showed that this third body was circling, not just one, but both stars.

Although Kepler data provided the relative sizes and masses of the stars and planet, astronomers needed more information to get absolute numbers. The crucial missing information came from the Tillinghast Reflector Echelle Spectrograph (TRES) on the 60-inch telescope at the Smithsonian Astrophysical Observatory's Whipple Observatory in Arizona.

TRES monitored the changing velocity of the primary star as it moved around in its orbit. This yielded an orbital solution that set the scale of the Kepler-16 system. The team found that the two stars orbit each other every 41 days at an average distance of 21 million miles.

"Much of what we know about the sizes of stars comes from such eclipsing binary systems, and most of what we know about the size of comes from transits," said lead author and scientist Laurance Doyle of the SETI Institute. "Kepler-16 combines the best of both worlds, with stellar eclipses and planetary transits in one system."

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User comments : 42

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krwhite
4 / 5 (4) Sep 15, 2011
Is it furthest from the bright center of the universe?
CHollman82
3.7 / 5 (6) Sep 15, 2011
So... it's nothing like Tatooine... It's a gas giant
Nanobanano
2 / 5 (4) Sep 15, 2011
VERY interesting.

Would be interested to see like an animated computer model of what the orbital dynamics look like.

I think this orbit should be highly perturbed as the net gravitational acceleration felt by the planet should change by as much as 17% depending on the alignment of the stars.

It's shocking that it's stable...
Husky
5 / 5 (7) Sep 15, 2011
a gasgiant usually comes with a vast host of moons, perhaps there is our tatooine
GSwift7
4.4 / 5 (13) Sep 15, 2011
I think this orbit should be highly perturbed as the net gravitational acceleration felt by the planet should change by as much as 17% depending on the alignment of the stars.


not at all. It should be in a standard eliptical orbit around the center of gravity of the two stars. It could easily be far enough away from either of the two stars so that it would feel any "disturbance in the force" of gravity as a weak tidal effect.

Think, for example, about the outer gas giants in our own solar system. You could easily replace our sun with two smaller stars and still have Saturn and the others orbiting the center of gravity of the two stars just as if they were one.

btw, I am NOT your father. :)
GSwift7
4.5 / 5 (8) Sep 15, 2011
Here's a link to binary stars. Take a look at the section labeled "Planets" about two thirds down the page:

http://en.wikiped...ary_star

lol, They even mention Tatooine.

As you can see, some people think that planets are actually more common around binaries than around single stars.
Nanobanano
2.6 / 5 (5) Sep 15, 2011
Gswift7:

These two stars are far enough distance from one another than variation in their alignment to the planet nevertheless should alter net gravity by up to 17%.

Take a piece of graph paper and draw some circles, showing the orbits of the stars about a CoG centered on the origin. Now also plot points at every 90 degrees around the circle.

Now draw a hypothetical planetary orbit around the CoG at a to-scale distance.

The planet is not attracted to the CoG of the binary star system. The planet is accelerated along the vector sum of the individual component vectors of gravitation acceleration of each of the two stars independently, and these should each vary by up to 17% in magnitude, and also up to 9.5 degrees in direction, due to changes in relative positions.

Because of the inverse squared law of gravity, the sum of the gravitational force vectors for the third object in a 3-body is not the same as the overly simplified center of mass of the first two bodies.
Nanobanano
1 / 5 (2) Sep 15, 2011
I could draw a diagram on paper and show what I'm talking about, but how to show it here I got no clue.

But if you simply take the mass of the two stars and combine it as a center of mass and then try to figure gravity, you do not get at all the same thing as figuring gravity for each object seperately and then taking the vector sums.

Maybe we are mis-communicating, but center of gravity, center of mass, and net gravitational acceleration are not necessarily the same thing, nor in the same location in multi-body systems.
Nanobanano
2 / 5 (4) Sep 15, 2011
When the stars are aligned orthogonal to the line between the planet and the barycenter of the two stars, that is making a "T" shape, with the planet at the bottom and the stars on either side...

Then "some" of each star's net individual gravity actually interferes destructively with the other stars gravity, because you have two gravitational force vectors that actually diverge, by 18 degrees...

When the stars are in a straight line with the planet, then all of their gravity interferes constructively, i.e. a linear sum is equivalent to the vector sum, but because one star is much closer than the other, the inverse square law changes things too...

When you're somewhere in between, then the net gravitational vector is somewhere in between in magnitude and direction.

now this assumes that all 3 bodies are orbiting on approximately the same plane...If the planet does not orbit the binary on the same plane as the stars orbit one another, then it gets even more complicated...
Nanobanano
1 / 5 (2) Sep 15, 2011
Additionally, the relative mass of the stars changes things. If the two stars have different masses, then that changes how their gravity influences the planet, even in the straight line scenario or the orthogonal scenario.

If A and B have different mass, then during the alignments:

A-B-Planet

and

B-A-Planet

You have different net acceleration on the planet.

If A is 10% more massive than B, for example, then the configuration B-A-Planet exerts more net gravity on the planet than does A-B-Planet.
Nanobanano
2 / 5 (4) Sep 15, 2011
I find that for stars of identical mass, in a hypothetical system where the planet orbits at 10 units distance from the binary's center of mass, and the binaries each orbit at a distance of 1 unit from their barycenter, then the difference between simply (incorrectly) using the center of mass vs computing indivual force vectors and taking the vector sum is 3% when the stars are aligned with the planet.

And in that same hypothetical system, when the stars are orthogonal, the amount of net gravity lost to destructive interference is 2% of the gravity that would be computed by just using the center of mass, leaving 98% net gravitational acceleration.

2% and 3% is a pretty big deal for an orbit.

So in that simplified, hypothetical system, there's actually slightly more than a 5% difference in net gravitational acceleration between the linear alignment and the orthogonal alignment...
Nanobanano
3.4 / 5 (5) Sep 15, 2011
Oh well, fine...

Give me negative feedback, but then just do this:

Go do the math yourself and see, at least I did it myself.

I was giving a simplified example using round numbers so anyone who cares could more clearly see what I was describing.
TheGhostofOtto1923
4.1 / 5 (17) Sep 15, 2011
When the stars are aligned orthogonal to the line between the planet and the barycenter of the two stars, that is making a "T" shape, with the planet at the bottom and the stars on either side...
Does anybody here know enough about celestial mechanics to tell if this is typical quantumconundrum insane bullshit or not? If not I'll have to go with the assumption that its just more insane bullshit from someone who enjoys demonstrating how deranged he is in public.

Because that is QCs modus operandi in his 40 or 50 posts a day until he gets banned... again.
TheGhostofOtto1923
3.9 / 5 (14) Sep 15, 2011
Go do the math yourself and see, at least I did it myself.
I would if I could QC but I cant so I wouldnt presume to try.

Not knowing how doesnt seem to stop you though does it?

Does it QC? Why is that?
Nanobanano
3 / 5 (4) Sep 15, 2011
When the stars are aligned orthogonal to the line between the planet and the barycenter of the two stars, that is making a "T" shape, with the planet at the bottom and the stars on either side...
Does anybody here know enough about celestial mechanics to tell if this is typical quantumconundrum insane bullshit or not? If not I'll have to go with the assumption that its just more insane bullshit from someone who enjoys demonstrating how deranged he is in public.

Because that is QCs modus operandi in his 40 or 50 posts a day until he gets banned... again.


This is not a difficult math problem. What? You can't solve an inverse squared relationship and a few uses of square roots and pythagorean theorem?

All I had to do was show one configuration where vector sums different from the center of mass claim.

I managed to immediately show that at least the two simplest configurations are different.
Nanobanano
2.3 / 5 (3) Sep 15, 2011
Go do the math yourself and see, at least I did it myself.
I would if I could QC but I cant so I wouldnt presume to try.

Not knowing how doesnt seem to stop you though does it?

Does it QC? Why is that?


I do know how, you idiot, it's not even hard math for the special case of an exact linear alignment and the special case of an exact orthogonal alignment.

This is quite literally a high school physics problem.

You don't even need to solve the equations by hand for every configuration to prove what he said initially was wrong. You just have to show at least one configuration which MUST occur at least once per orbit is different than the center of mass alone would give...

I did that for two configurations.

Get over yourself, and try it with a piece of graph paper.

You did pass geometry in high school I hope, and you can do a vector sum, I hope, and you can do the force and acceleration formulae for gravity, I hope, so there.
TheGhostofOtto1923
3.9 / 5 (14) Sep 15, 2011
This is quite literally a high school physics problem.
In your mind. But your mind seems to have limited contact with the real world. I assume that the calcs for a binary system are daunting. But QC remains undaunted. How poignant.
Get over yourself, and try it with a piece of graph paper.
Get over yourself. Get back on your meds before you hurt yourself.

Days until nano/qc gets the banhammer - 5 and counting...
Nanobanano
2.3 / 5 (3) Sep 15, 2011
You really don't even need the gravitational acceleration formula, since the masses, "M1", "M2", and "m" stay constant, and G stays constant, all you need is (1/r^2).

Thus, it becomes a purely geometric problem.

For the linear alignment you have:

2/100 not equals to "1/81 1/121 = 0.02061"

When the barycenter is at 10 units from the planet, and A is at 9 units and B is at 11 units, using round numbers.

And for the orthogonal configuration, the square of the length of the hypoteneuse is 101, and "r" is the same as the hypoteneuse. So your linear force of gravity from EACH star's individual component is 1% less than half of what would be if you had a single star of double mass in the center.

However, your vector is not going to add entirely constructively, because the force vectors are not parallel. So then you have to figure out which portion of the vector is cancelling which portion of the mirrored one from the other star, cont...
Nanobanano
2 / 5 (4) Sep 15, 2011
...It turns out that about 1% is cancelled by the 1% coming from the other star, SO, it amounts to:

(100/101)/1.01 = 98.0296%.

And the calculation is NOT daunting, and certainly not for the purpose of this discussion. It's nothing more than vector sums.

You don't know enough to know that I know...

Sad...

Let's see:

http://en.wikiped..._problem

TheGhostofOtto1923
4 / 5 (16) Sep 15, 2011
Here - I found something which might be of help.
http://askthephysicist.com/

You might enjoy some nice music.
http://www.metrol...ana.html

Have a nice day.
Nanobanano
3 / 5 (6) Sep 15, 2011
Oh well, Otto, get a life.

You act like this is hard or something, and it's really not.

course I didn't even use vector notation, just screwing around on paper, but who cares?

My god, you can't do this math to check what I was saying?

If you can't, then you're not qualified to judge me, so STFU.

This doesn't require higher maths to solve, because you're solving for the NET instantaneous gravitational acceleration at various configurations, which must occur some time during the orbit.

Adding an extra star doesn't make this a phd level math problem. It makes it a vector sum problem, which you must have failed since you said you don't know how to solve that.
TheGhostofOtto1923
3.4 / 5 (18) Sep 15, 2011
You don't know enough to know that I know...
I do know your track record, and Ive dismantled your nonsense before by showing you what you dont know, as have many others. So I do feel qualified in passing judgment.
Nanobanano
2.4 / 5 (5) Sep 15, 2011
Otto, you're an idiot.

Because gravity changes as the inverse square of distance, teh vector sums of gravity will not be the same as the linear sums of the center of mass.

Why can't you see that?

http://www.physic...3l3a.cfm

and

so that might help you.
Nanobanano
2.6 / 5 (5) Sep 15, 2011
You don't know enough to know that I know...
I do know your track record, and Ive dismantled your nonsense before by showing you what you dont know, as have many others. So I do feel qualified in passing judgment.


Where?

You havent dismantled anything.

You have, however, managed to show that you don't know high school level geometry, because the special cases I chose here simplify to what was a 9th or 10th grade math problem when I was in school.

I don't need to solve, nor did I claim to solve, the entire fricken orbit of the system by hand.

To be right, I only had to show that two configurations which are each guaranteed to occur have different net gravitational accelerations...

I would be interested to know what you claim to have dismantled, I'm perplexed...
Starbound
3.7 / 5 (3) Sep 15, 2011
Kudos to Nanobanano for giving us the physics lesson. F_gravity = Gm1m2/r^2. Freshman physics, although I didn't really get it until Dynamics (Sophomore physics). Thanks for your time! Sincerely, a MechE grad.
TheGhostofOtto1923
3.9 / 5 (11) Sep 15, 2011
You know, it was the one from a previous incarnation of yours where you thought you knew what you were talking about in about 15 posts in one thread but I went to the effort of proving you didn't know near enough to say anything about it by providing links and then an actual scientist came along and politely agreed that I was right and that you didn't know what you were talking about. Again.

This was before I realized it was you, QuantumConundrum of the diarrheaic intellect disorder. That one. Remember? Come on refresh my memory. Something to do with energy wasn't it? Plus all those others.
To be right, I only had to show that two configurations which are each guaranteed to occur have different net gravitational accelerations...blah
In your mind. No matter, there'll be others which is why you get banned so much. Insanity def: jaywalking because you disregard the last 20 times you got runned over jaywalking.
TheGhostofOtto1923
3.7 / 5 (12) Sep 15, 2011
Kudos to Nanobanano for giving us the physics lesson. F_gravity = Gm1m2/r^2. Freshman physics, although I didn't really get it until Dynamics (Sophomore physics). Thanks for your time! Sincerely, a MechE grad.
This from someone who says this:
No news here. Everybody knows dinosaurs were wiped out when dinosaur scientists horribly botched a revolutionary new asteroid mining mission.
-?
I think I'll hold out for someone with a better sense of humor.

Remember if he follows you home you gotta feed him.

35 plus plus posts today (and counting)...
Thex1138
5 / 5 (2) Sep 16, 2011
The first trinary star system discovery would be amazing...
Guys I think the point is that this binary star system is there and the planet[s] have been orbiting enough to have an equilibrium of gravity, velocity, center of balance and momentum to maintain it... the math should be a derivative of that...

My god! It's full of stars!
Robert_Wells
not rated yet Sep 16, 2011
So... it's nothing like Tatooine... It's a gas giant


yea, as with a few articles a week on this site, they're mislabeled. obv referring to the binary system making it tatooine 'like' and not the physical description of the planet itself. they could have found another way to link this article to 'star wars' in order to boost up those hits, first of all "planet" discovered shouldn't have been used.

The first trinary star system discovery would be amazing...

The Verse man, The Verse. you my friend are not a browncoat.
Robert_Wells
5 / 5 (4) Sep 16, 2011
and yes i know The Verse is more than a trinary AND its fictional, just saying. ya know.

moving on

You know, it was the one from a previous incarnation of yours where you thought you knew what you were talking about in about 15 posts in one thread but I went to the effort of proving you didn't know....


i specifically remember someone going at it with him perhaps close to a year ago about the moon Titan. it was an article about whether it is possible that Titan has liquid water under its icy surface. QC messed the math up from the beginning and defended it post after post until it finally 'clicked' for him that he was wrong from his first post. if i remember correctly he had a few conversions wrong, took about 60 posts but he finally saw where he went wrong. just takes patience with him sometimes is all.
QC knows just enough to be dangerous.
Ricochet
not rated yet Sep 16, 2011
I'll wager $1 that scientists will eventually name some planets after Star Wars universe planets.
TheGhostofOtto1923
4 / 5 (12) Sep 16, 2011
QC messed the math up from the beginning and defended it post after post until it finally 'clicked' for him that he was wrong from his first post.
Yeah really.

I remember now - this one was about dry ice in antarctica. QC goes on and on about how it could be this and that, even though others are explaining how it's not possible. So I find a forum thread in 2 minutes where it was thoroughly discounted, by experiment, but QC the genius kept arguing with people here. I recommended he go to that forum and annoy people there.

How many times has this happened QC? how many times have you been banned because of it?
Ricochet
5 / 5 (2) Sep 16, 2011
I've learned to gloss over wall posts in these forums. I'm sure some people have valid information that simply cannot be explained in one post, but I figure that if someone wants to write that much stuff in a non-academic forum they should just become a writer for the site and call it good.

Until that happens, here's a cool laser video:
http://www.youtub...embedded
gimpypoet
1.8 / 5 (4) Sep 16, 2011
all theories are just theories,hypotheses are more educated guesses and no proof has been proven,still you fight and bicker to no end. is gravity the same in a dwarf galaxy or is it scaled to show it operates the same as everywhere?how can you say its the same as in a normal galaxy could you prove it? the math is interesting and could be checked,but who would "prove" it?you the bickerers? stop arguing and let all post their theories,then pick them apart using the scientic methods.until then i will laugh in relative silence until my laughter makes gravity push me from my chair with my hand reaching out to steady myself strikeing the keyboard which in turn causes my comment to be submited as a post on this here scientific website. get real,if anyone of us had credentials strong enough to back up or reduce to rubble any theory presented here, we would post here at all.google and wikipedia are not valid enough to wipe with,and are edited by people like us. my apologies to all I offended.
GSwift7
4.2 / 5 (10) Sep 16, 2011
Okay, I'm back. Good morning guys.

From the top: QC, your 17% figure is made up. You would need to know how far apart the stars are. In this case it's much less than that though. The two stars are relatively close. One is about 65% the mass of Sol and the other is much smaller, one of the smallest of this type of star we've ever seen. The combined mass of the two stars is less than Sol. The center of mass for the two stars is probably inside the larger star, and it moves around as the smaller star circles the larger one. You are correct in saying that center of mass is not the same as center of gravity, except that in this case that's wrong. You would be right if we were talking about relativistic or quantum physics, but this is (as you correctly said) a Newtonian problem. In Newtonian physics, center of mass is the same as center of gravity. So you got lucky, and you were both right and wrong in that case. You still obviously didn't bother to do a google before posting though.
FrankHerbert
1.8 / 5 (12) Sep 16, 2011
More like a Dune-like planet. My arms are crossed.
GSwift7
4.1 / 5 (9) Sep 16, 2011
continued:

Your treatment of the problem as a geometry problem with changing force vectors is flawed. It is more correct to treat the problem as a moving center of gravity problem. However, that's a technical point. The reality is that if the resultant change in tidal force on the planet was anywhere near 17%, the planet wouldn't be there. It would have either been torn apart or ejected from its orbit before it even formed.

Let me explain why your geometric solution doesn't work. You are greatly underestimating the ratio of the distance between the three bodies. The short side of your triangle, between the the two stars, is so small compared to the two longer sides of the triangle, that you can treat the short side as being zero. Let me elaborate on that though. The short side isn't measured between the physical center of the two stars. As the stars co-orbit, their combined center of gravity moves in an elipse. That elipse is the size of the short side of your triangle. It's smal
GSwift7
4.1 / 5 (10) Sep 16, 2011
this planet orbits the two stars at a distance about the same as Venus orbits Sol. I get the impression that you think that is close. In reality, that's still quite a good chunk of miles. That's about .7 AU away from the center of gravity of the binaries. That's about five and a half light minutes out. Think about that. Five and a half minutes at the speed of light. Are you starting to understand now? This isn't the "three body problem" you are trying to make it. Just as you correctly stated that it isn't necessary to use relativistic equations here, it is also not necessary to treat the two stars as seperate bodies in relation to the planet. In this case, just simplify the problem to a two body problem where one of the bodies wobbles so slightly that it is hardly noticable from the point of view of the planet. The movement of the smaller star isn't much of a factor here.

How do I know that? Because the planet is there. Observation > Imagination.
Ricochet
5 / 5 (2) Sep 16, 2011
More like a Dune-like planet. My arms are crossed.

Yes, well, I can see why you'd feel that way... hehe
gimpypoet
3 / 5 (4) Sep 17, 2011
both the planet and the smaller sun are in orbit around the larger sun. in other words, the smaller sun is like a planet,orbiting like the other planet in this system. no celestial mechanics problem. the center of ballance is equalized just like our system of 8 planets and related bits of debris and none of our planets are currently being flug out or have any problems even though there are higher mass planets in the outer parts. if the planet orbited is side the orbit of the smaller sun, forces "could "have destroyed or ejected it, but circleing both bodies is no different than the orbit of mars encircleing mercury, venus and earth, if mercury was a sun or venus was a sun. it is a three ring circus instead of an eight ring plus circus in our system.
Hillbilly logic. no math needed. if i remember right, the last planetary alignment was forcast by fearful people to be the end of us, causing all types of havoc and didn't. no problem we can see, no planets broke apart or out of orbit.
hush1
not rated yet Sep 18, 2011
Hillbilly logic - gimpypoet


lol
Resonance fanboy. What works with music (inversion) or celestially speaking, orbital planet swap brings the tent down on the solar system circus motion.
barakn
5 / 5 (1) Sep 19, 2011
http://kepler.nas...pler16b/ Star A .6897, star B .20255 solar masses. Separation .22431 AU (semi-major axis). Center of mass is therefore .22431*.20255/(.6897 .20255) AU = .0509 AU from star A. GSwift7's assertion that the center of gravity is within Star A (radius = .6489 solar radii = .00302 AU) is thus false.