I wonder if there ever could be a star (really small) which may orbit around a planet (really big)?


6 Answers 6


One thing to keep in mind is that objects that are bound gravitationally actually revolve around each other around a point called a barycenter. The fact that the earth looks like its revolving around the sun is because the sun is much more massive and its radius is large enough that it encompasses the barycenter. This is a similar situation with the Earth and Moon. If there were three bodies, where two bodies were of similar size (like a binary star system plus a massive planet) then an analysis of three body systems shows that there are stable configurations where the objects will be in very complicated orbits where it would be difficult to say one orbits the other.

Update: The short answer is yes, it is possible when you look at the complete dynamical system, for the reasons stated above. More evidence of this can be found in the study of regular star orbits where very complicated orbits are possible and can be stable. Currently the cut off for classification of a planet and a brown dwarf is 13 Jupiter masses, which is arbitrary to some degree. The lightest main sequence stars have a mass of 75 Jupiters. This will put the barycenter well outside the radius of either body for binary systems.

A quick check of the two body system using the equation:

$$R = \dfrac{1}{m_1 + m_2}(m_1r_1 + m_2r_2)$$

Setting $m_1 = 75$, $r_1 = 1$, $m_2 = 13$, $r_2 = 2$ gives:

$$\dfrac{75 + 26}{75+13} = 1.147$$

Indicating a barycenter at roughly $\dfrac{1}{7}$ the distance between the objects. More bodies will cause more complicated orbits, where again, it would be difficult to say which object orbits which. It should be noted that if the system was composed of 3 objects, 2 of which had similar mass, it would be possible to develop a system that appears to have two larger objects orbiting a third smaller object. A quick check reveals:

$$R = \dfrac{1}{m_1 + m_2 + m_3}(m_1r_1 + m_2r_2+ m_3r_3)$$

Setting $m_1 = 75$, $r_1 = 1$, $m_2 = 13$, $r_2 = 2$ $m_3 = 75$, $r_3 = 3$ gives:

$$\dfrac{75 + 26 + 225}{75+13+75} = 2$$

Whether such an orbit system is realizable when you consider the full dynamics of a natural system is debatable, but I am not aware of a specific proof that would rule it out.


It should be noted that there are new periodic solutions to 3-body problems when the objects have the same mass.


Anything the mass of a star is going to get hot like a star and fuse hydrogen like a star. In other words it will be a star not a planet!

While it's technically possible to have a rocky planet the mass of a star, in practice when stellar systems form there aren't enough metals available to build such a large object. Large objects are invariably built from hydrogen (and helium) and would therefore form a star.

There are plenty of binary systems with a star orbiting a white dwarf or neutron star, but even a dead star is still a star and not a planet.

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    $\begingroup$ I would say mass AND density of a star, if i have the mass of the sun stretched out over a billion light years it wont be dense enough to be a star $\endgroup$
    – user17607
    Jan 31, 2013 at 15:29
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    $\begingroup$ @RhysW: gather the mass together and gravity will take care on the density! After all, even the biggest stars started life as a nebulous cloud of gas. $\endgroup$ Jan 31, 2013 at 15:55
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    $\begingroup$ @JohnRennie tell that to Jupiter (and yes, I know there's people who consider Jupiter to be either a brown dwarf star or a protostar). $\endgroup$
    – jwenting
    Feb 1, 2013 at 7:09
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    $\begingroup$ @jwenting - and those people are wrong; brown dwarves start to be observed at masses at least 13 times Jupiter's. Jupiter could consume everything else in the neighborhood besides the Sun and wouldn't even double its mass. And Jupiter, in turn, is only 1/1000 the Sun's mass. $\endgroup$
    – KeithS
    Oct 3, 2013 at 14:57
  • $\begingroup$ @KeithS that's observed, not the theoretical limit. We're not sure what goes on inside that planet, it might be active enough to qualify. $\endgroup$
    – jwenting
    Oct 3, 2013 at 17:32

Typically, a star (or stellar remnant, such as a neutron star, white/black dwarf, or black hole) will be the most massive thing in the area, by far. Planets, even gas giants, are a small fraction of the mass of a typical main sequence star.

Now, as in Hal's answer, the relative mass of the planet and its star does make the center of mass, the barycenter, of the planet-star system a point that is different from the center of mass of either body alone. This will cause the star to appear to "wobble" as its planet moves around it. Tracking this wobble over time is how we have discovered most of the extrasolar planets we know of (which is why most of the exoplanets we know of are huge gas giants several times Jupiter's mass; the wobble's easier to see). However, as orbital motion's primary determinant is relative mass (another is relative distance, and the third is tangential velocity), the more massive star will be very close to the barycenter of the system, and the planet will be further away.

Our best-known example, Jupiter, the largest planet in our solar system, has a mass of about 1.9e27 kg. Our Sun has a mass of about 2e30 kg. In other words, Jupiter is about 1/1000 the mass of the Sun. Thus, while Jupiter does indeed have an effect on the position of the Sun as it orbits the Sun, the center of mass of the two objects is still much closer to the Sun than to Jupiter (pretty much on the Sun's surface, as explained by the comment). In fact, all 9 planets of our Solar system (throwing Pluto a bone here), all their moons and rings, and all other orbiting celestial objects such as asteroids and comets, all pooled together into one super-planet, would still be only about .15% of the Sun's mass. That would bring the barycenter of thus dual-body system out into open space a few million miles (depending on the distance between the two), but still far closer to the Sun than the super-planet.

While Jupiter is not even close to the most massive planet we have discovered (http://xkcd.com/1071/large/), we have not yet found a planet more massive than any star we've ever found, much less a planet more massive than its own star. The most massive non-stellar body we have discovered is about 55 times Jupiter's mass (still only about 5% of the mass of our Sun), and is a rogue body (not orbiting a star) that blurs the line between planet and star; it's dense enough to generate temperatures that cause deuterium combustion (not quite true fusion) and thus it produces its own thermal energy. Masses of this type are known as "brown dwarfs". As objects become more massive, they become progressively hotter, until they reach the threshold of true fusion at about 80 MJ and become "red dwarfs".

Thus, not only is there no known planet more massive than its star, it's thought to be impossible for any non-stellar body to gain enough mass for any true star to orbit it, without it becoming a star itself. As the mass of something like a gas giant increases, by attracting nearby wisps of gas, aging comets, etc, the density of the mass also increases as gravity does. This increases the core temperature of that body. Eventually, as with these brown dwarves, the temperature increases to a state of dense pre-fusion plasma, and then from there, things just continue to transition toward true fusion. It's conceivable that a planet could aggregate a mass primarily consisting of something other than hydrogen, that wouldn't fuse until much higher temperatures had been reached, but given what we know of our galaxy it is extremely unlikely for there to be enough of anything but hydrogen available to give a planet that kind of mass.

It is possible, though we haven't seen it yet, for a planet that's not quite a brown dwarf (maybe 40 MJ) to be found orbiting a red dwarf (about 80 MJ); this would meet our definition of a "planetary system" of a planet and star, and not a "binary system" of two stars. However, with the gas giant being only about half the mass of its star, the center of mass, and thus the barycenter of orbit, would be well out into open space between them, and you would, more or less, see them orbit each other. That's about as close as you could get to a geocentric system, and we have not yet observed it.

  • $\begingroup$ Just one tiny little nitpick for an otherwise great answer: the barycenter of the Sun-Jupiter system is pretty much on the Sun's surface - compare the mass ratio to the separation-radius ratio. $\endgroup$
    – user10851
    Jan 31, 2013 at 19:35
  • $\begingroup$ If, hypothetically, lumps of iron or heavier materials totaling over 50M<sub>J</sub> slowly came together sufficiently to form a solid mass without ever, as a collective unit, having contained any exothermically-fusible material or enough kinetic energy to trigger a supernova, would there be any reason they couldn't form a 50M<sub>J</sub> bona fide planet? $\endgroup$
    – supercat
    Apr 22, 2014 at 16:34
  • $\begingroup$ Hypothetically, yes, if you could find enough non-fusible material in the neighborhood and aggregate it, you'd end up with a superplanet. Keep in mind, however, that 50MJ is not much more than halfway to the mass of the typical red dwarf star. In order for this planet to be the closer one to the barycenter of an orbital system involving anything we'd call a true star, you're looking at a mass on the order of 150-200MJ. That would require all of the non-hydrogen material of several hundred Solar systems. $\endgroup$
    – KeithS
    Apr 24, 2014 at 15:44

John Rennie has already covered most of it in his answer, I just wanted to add a few explanatory notes.

Usually when stars are "born", they form from dense clouds of hydrogen and helium. Once the cloud gets dense enough, it starts to get really hot, and eventually fusion occurs.

The universe doesn't have an abundance of heavier elements, like calcium and iron. These are the by-products of stellar fusion, so for a planet to form, there needs to be (or have been) a star at some point to generate those elements. If the star that created those elements is still there when the planet is being formed, it is fully possible that the star has a smaller radius than the planet orbiting it (like a neutron star), but the star will also be much denser than the planet, ensuring that the centre of the orbit is nearer to the star than the planet. A star that has a smaller radius and a smaller density would have never turned into a star in the first place, it would have just remained as a nebula or a brown dwarf.

But as has already been pointed out, it's much more likely that a planet that massive will just get really hot and start it's own process of fusion, turning into a star.


Planets appear because of stars:

  • First you have some huge cloud of cosmic dust with a very big total mass
  • At some point some 3d party gravity field may disturb it and the cloud starts to have some center with bigger gravity which pulls more and more of other dust
  • At some point the gravity of the center becomes very strong and it pulls the particles around with more speed and the higher gravity is, the higher the temperature becomes inside
  • And then a thermonuclear reaction starts because of highly concentrated energy. This is a star. There is a chance that the new-born-star doesn't blow up and here were are.
  • During the process of cloud concentration, particles in other parts of the cloud also start gluing to each other (again because of gravity) - thus planets start to form.

So you have a 'big guy' in the center and 'small guys' in other parts of the cloud. If theoretically one of those small guys appear to be larger than 'big guy', then it would become a star instead, right?

So now - you can't have a star going around a planet because from the start it was larger. For instance, take a look at Solar system objects, the Sun is something like 98% of the mass.

And even if we could have a body that is heavier than a star and is located close, it would start to grab the higher layers of the smaller star. Which actually happens, but there are anyway two stars participating in this to form nova.


Theoretically YES: But, it has a constraint sticking to it. So, "Yes" slightly. This would be possible only when there already existed a system like such. If the mass of the small star (orbiting mass) is low (enough to be tidally locked with the "BIG" planet), the system would be possible.

But still, This is totally ridiculous. Because, such a thing cannot be formed on its own.

So Practically, it is NO for sure: To understand this, let's skim through their definitions. Though both the celestial objects have a common origin (Nebulae), a Planet is an object whose core hasn't fused enough hydrogen to maintain the reaction. So, it can be named as some kind of "inactive star". On the other hand, a Star satisfies all the necessary conditions, sustaining the fusion reactions.

When a planet has acquired enough mass, so that the core fusion reactions are sustained, it becomes a star. That's all. A star-planet system is somewhat different, because they have a common center of mass. In some unlucky way, a star is already in a negligible orbit around the the common center of mass, while the planet orbits a star (which can be considered that it orbits the planet slightly).

  • $\begingroup$ Maybe phrase it better? You've said theoretically yes, and then contradicted yourself at the end of the sentence. That's misleading. This isn't even "theoretically" possible. $\endgroup$
    – Kitchi
    Jan 31, 2013 at 8:29
  • $\begingroup$ Hey @Kitchi: I disagree with your quote "This isn't even theoretically possible". Why do you say so? $\endgroup$ Jan 31, 2013 at 10:46
  • $\begingroup$ Have a look at my answer to this question for a fuller explanation. :) $\endgroup$
    – Kitchi
    Jan 31, 2013 at 11:05
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    $\begingroup$ BTW, CrazyBuddy, mods don't have any special powers to make undeletions non-bumping. If indeed they are "bumping" in the fist place. $\endgroup$ Apr 19, 2013 at 22:43
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    $\begingroup$ "Nice pic" Befunky did that for me. I think I used the "Inkify2" tool for that. There are several other really neat effects in the free teasers. $\endgroup$ Apr 23, 2013 at 15:52

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