# Tatooine-like planet discovered (w/ video)

"Kepler-16b is the first confirmed, unambiguous example of a circumbinary planet - a planet orbiting not one, but two stars," said Josh Carter of the Harvard-Smithsonian Center for Astrophysics (CfA). "Once again, we're finding that our solar system is only one example of the variety of planetary systems 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 planets comes from transits," said lead author and Kepler scientist Laurance Doyle of the SETI Institute. "Kepler-16 combines the best of both worlds, with stellar eclipses and planetary transits in one system."

Explore further

**Citation**: Tatooine-like planet discovered (w/ video) (2011, September 15) retrieved 25 June 2019 from https://phys.org/news/2011-09-tatooine-like-planet.html

## User comments

krwhiteCHollman82NanobananoWould 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...

HuskyGSwift7not 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. :)

GSwift7http://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.

NanobananoThese 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.

NanobananoBut 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.

NanobananoThen "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...

NanobananoIf 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.

NanobananoAnd 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...

NanobananoGive 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.

TheGhostofOtto1923Because that is QCs modus operandi in his 40 or 50 posts a day until he gets banned... again.

TheGhostofOtto1923Not knowing how doesnt seem to stop you though does it?

Does it QC? Why is that?

NanobananoThis 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.

NanobananoI 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.

TheGhostofOtto1923Days until nano/qc gets the banhammer - 5 and counting...

NanobananoThus, 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(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

TheGhostofOtto1923http://askthephysicist.com/

You might enjoy some nice music.

http://www.metrol...ana.html

Have a nice day.

NanobananoYou 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.

TheGhostofOtto1923NanobananoBecause 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.

NanobananoWhere?

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...

StarboundTheGhostofOtto1923This 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. 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.

TheGhostofOtto1923I 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)...

Thex1138Guys 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_Wellsyea, 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 Verse man, The Verse. you my friend are not a browncoat.

Robert_Wellsmoving on

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.

RicochetTheGhostofOtto1923I 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?

RicochetUntil that happens, here's a cool laser video:

http://www.youtub...embedded

gimpypoetGSwift7From 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.

FrankHerbertGSwift7Your 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

GSwift7How do I know that? Because the planet is there. Observation > Imagination.

RicochetYes, well, I can see why you'd feel that way... hehe

gimpypoetHillbilly 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.

baraknPlease sign in to add a comment. Registration is free, and takes less than a minute. Read more