I'm imagining if we can live on a very special white dwarf which has very special characteristics:

  1. It has cooled down to a very low temperature

  2. Its rotational speed on the equator can almost cancel out the gravity (ie: the gravity is almost same as on Earth)

  3. Accretion occurs so that rocks accumulate on the equator of the white dwarf, and a ring shaped crust is formed, also there is water and ice from comets on the ring rock

  4. It has a star around the White dwarf which is in the habitable zone, just like the Earth around the sun

Can that type of white dwarf exist? Can we live on this type of white dwarf (either in short term or long term)?

  • $\begingroup$ For your purposes, how cool is a "very low temperature"? $\endgroup$
    – HDE 226868
    May 14, 2020 at 2:56

6 Answers 6


David Goldstein provides an answer on how fast the earth has to rotate to nullify Earth's gravity on the equator and the answer is a period of 5066 seconds or roughly 1 hour and 14 minutes

Bruce McLure describes the binary star system that you have in mind: the star Sirus and its white dwarf companion, Sirius B ("The Pup"). In "Your Weight On A White Dwarf", McClure computes how much a human being would weigh on white dwarf:

“Since the Sun is about 333,000 times more massive than Earth, that means you'd weigh 333,000 times more on a white dwarf having the Sun's mass but the Earth's size.”

Let's use McClure's number (333,000) and plug it in Goldstein's computation by dividing 5066 with 333,000 and the necessary rotational time that you need would be 15.21 milliseconds.

Mark Barton on the same thread as Goldstein points out that the Earth is already out of shape 1 parts in 300 because of the centrifugal force of its rotation. The "little rock" that you imagine as habitation would probably be like a very flattened disc (I'm imagining like Saturn's rings).

Assuming that in your world-building, humans have achieved some technology to endure these kinds of massive gravitational influences, the possibility of habitation would be on the same scale as the possibility of having human habitation on one of Saturn's rings. As a4android points out in the comments, if the equator is earth-sized, the rotational velocity will be faster-than-light. You probably need an equator length that accommodates Chandrasekhar's limit of of 1.4 solar masses so that the rotational time is slower-than-light.

  • 1
    $\begingroup$ Does this assumption still work? The radius of the two bodies is different so the same rotational speed gives a very different linear speed and without stopping and thinking about it I'm not sure how that modifies the centripetal effect. $\endgroup$
    – Tim B
    Feb 2, 2017 at 10:25
  • $\begingroup$ @TimB I'm assuming that the white dwarf is about the same size as the Sun since "According to the astrophysicist Chandrasekhar, no white dwarf can exceed 1.4 solar masses. If the star were any more massive, it would shrink even further, either into a neutron star or a black hole. In a close binary star system, a white dwarf sometimes can pull enough material from its companion star to exceed the 1.4 solar mass limit and explode as a supernova. "The Pup," however, is not expected to explode, because it's too far distant from Sirius." idialstars.com/pup.htm $\endgroup$
    – pageman
    Feb 2, 2017 at 11:17
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – pageman
    Feb 2, 2017 at 11:45
  • 6
    $\begingroup$ I checked and it seems the answer is wrong. Here it says rotation speed to compensate gravity is $v = \sqrt{rg}$. That's 4459 km/s at the equator. If you divide equatorial circumference by that it gives you 8.79 seconds for a white dwarf with Earth's radius and the Sun's mass. $\endgroup$
    – pablodf76
    Feb 2, 2017 at 14:13
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    $\begingroup$ Also note that you would not want to cancel out the effects of gravity; you want to cancel out all but the standard effects we feel (9.81m/s/s). It's a minor point that probably shaves off only a rounding error worth of speed, but I feel compelled to make it. $\endgroup$
    – GrinningX
    Feb 2, 2017 at 15:43

As explained above, there are at least two problems with the scenario.

  1. White dwarfs, since they have a very small surface area, cool down slowly (and the cooling itself decelerates over time). The inner part of the star stays very hot for a long time and radiates slowly through the crust. It's not clear how much time it might take for the crust of a white dwarf to cool down to Earth-like temperatures, but it must be a lot (in the order of trillions of years).

  2. The rotational speed needed to counteract the equatorial gravity of a white dwarf is immense, as you're assuming an object with roughly the same equatorial circumference as Earth but about the mass of our Sun. The only big solid bodies that rotate that fast are neutron stars, and they are much more compact than white dwarfs. Also, it's unclear how an Earth-size object could naturally gain such angular momentum. Neutron stars do because they shrink a lot in size.

You could circumvent problem number 1 by proposing that the story takes place in the very distant future. The main star your white dwarf would be companion of, therefore, won't be one of today's stars and might be among the last. The Stelliferous Era is supposed to last until 100 trillion years after the Big Bang.

I don't see how you could go around problem number 2 without very serious astroengineering.


Condition 2 and 3 are conflicting: if the rotation almost cancels the gravity at the equator, it means it is really easy for anything to escape. So no significant accretion can take place, since as soon as a significant boulder will stand up its top will be launched on the tangent into outer space.

  • $\begingroup$ I think "almost cancels out" means "results in earth-like gravity" (instead of the much higher gravity to be expected for a non-rotating white dwarf). $\endgroup$ Feb 2, 2017 at 12:48
  • $\begingroup$ OP amended the question after this answer was posted to say "the gravity is almost same as on Earth". $\endgroup$
    – user
    Feb 2, 2017 at 14:25
  • $\begingroup$ No, it doesn't. The effective surface gravity may be low, but the escape velocity is still ridiculously high. Even if the equatorial spin velocity is approximately equal to orbital velocity (allowing Earthlike gravity to remain may as well be a rounding error), escape velocity is still more than 4% higher. $\endgroup$ Aug 7, 2017 at 17:27

I believe, such object is not possible for our universe. Right now the coldest possible WD has 3,9 kK (3,900 K) in the photosphere. It's really hot place to live. The rotation speed for an object with radius of WD to compensate the gravity to 0,5-1,5 g should be very high, I don't know, how stable WD are, but pretty sure it will be destroyed at that speed. And if it was binary stellar system, with one star evolved to WD, it should lost its stability, so, again, not a very good place to live.


Such a white dwarf cannot exist.

The reason is simple, and does not require any calculations or deep understanding of physics.

White dwarfs are incredibly dense, orders of magnitude denser than ordinary "molecular" matter, which means that the Coulomb force whose outward pressure keeps the latter within its characteristic range of densities must be completely overwhelmed by inward pressure (resulting from gravity) in the former.

In other words, close to the white dwarf's surface, the gravitational field must still be strong enough to keep it as dense as a white dwarf. Edit: The faster the star is rotating, the stronger the gravitational field must be in order to compensate. Such a gravitational field would also compress a human to the density of a white dwarf (edit: assuming that the human was trying to stand on a surface attached to the white dwarf. Staying safe in a fairly close orbit might be feasible).

The same reasoning shows that one cannot stay alive close to (edit: and stationary relative to) the surface of a neutron star or other exotic star made of incredibly dense matter, which includes stellar-mass black holes, but notably does not include the supermassive black holes which exist at the centres of most galaxies, and which are far less dense than these other objects (at least, if one counts their size as being given by the extent of their event horizon).

  • $\begingroup$ You would have non-degenerate matter under the ring. I do not believe this is a showstopper. $\endgroup$ May 18, 2021 at 0:06
  • $\begingroup$ @LorenPechtel The question specified a ring-shaped crust, so presumably at least some of the white dwarf is exposed. It then follows that the "effective gravity" at the equator, on top of this crust (and with rotation taken into account), must be very close to the gravity at the white dwarf's surface, because otherwise the extremely strong gravity near that surface would quickly smooth out the unevenness. This "hydrostatic equilibrium" is why anything with surface gravity more than ~1% of Earth's is roughly spherical (and slightly flattened from rotation). $\endgroup$ Jul 31, 2021 at 19:58
  • $\begingroup$ For example, Mars' surface gravity is a bit under half of Earth's, and its giant volcano Olympus Mons is a bit over twice the height of Everest. The details depend on material properties, but with surface gravity > 10^5 times Earth's, it's hard to imagine a white dwarf supporting any unevenness more than a few metres in height. $\endgroup$ Jul 31, 2021 at 20:07
  • $\begingroup$ What's more, because of the physics governing matter inside white dwarfs, adding extra mass to one actually decreases its radius, so the surface gravity just grows more quickly! This mass-radius relationship continues until you have a neutron star and then a black hole, at which point the radius will finally grow again with mass, and do so fast enough for "surface" gravity at the event horizon to decrease again. $\endgroup$ Jul 31, 2021 at 20:13
  • $\begingroup$ You're missing part of the question: The white dwarf is spinning at just below breakup velocity. $\endgroup$ Jul 31, 2021 at 20:22

If it were spinning fast enough to cancel most of the gravity, the material would puff back out again into normal atoms. That would apply all the way down, since the pressure would be releived from the core material. So it would no longer be a white (or black) dwarf at all. An object as you describe is simply not possible.

If you want to cancel the gravity, the ship could be in orbit just a few feet above the surface of the dead star.


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