So, the idea in a story is that in far far future there’s a trend that rich people, they are able to buy themselves planets. Tiny little planets with sizes 4-6 km (3-4 miles) in diameter. The company that makes them, seeks for an asteroid which is as round as possible, and then drills and makes a hole in it. In that hole, they put the white dwarf star matter in it. As far as I know, that’s the densest matter after the one inside a black hole and neutron star which are number 1 and number 2 when it comes to matter density.

That small object, which would not have any gravity detectable by a human being, would then have Earth-like gravity, and by installing few artificial magnets, they would create the magnetic field that would stop atmosphere being blown away by solar winds and radiation. The planet would also need to have a really thick atmosphere so that the pressure on the surface is like on Earth as well, but I am wondering what are your thoughts on that. Being that the planet is 5-6km in diameter and 16-20km all around the equator, it would look silly with an atmosphere that goes on 30-40km away from it :)

So, if you think there’a scientific solution for the atmospheric pressure to be the same without the cover having to be as thick as on real Earth, let me know.

My main question is, would the white dwarf matter be stable inside an asteroid with 5km diameter or would it blow it away? Maybe it needs the gravity of said star to be stable :) I mean it is logical.

The rich guys in the story have their own planets with its own little lakes, beaches, rivers, houses, little mountain, etc. They put it artificially in orbit around the Sun in habitable zones, and of course, the gravity of it influences life on regular Earth and people on regular Earth and all sort of weird stuff starts to unfold as more people start getting their own planets.

  • 5
    $\begingroup$ Sorry that I can't be bothered digging it up myself right now but have a look at Wil McCarthy's Queendom of Sol books, they built several small artificial planets that have a direct bearing on this issue, McCarthy was an engineer before he was a writer so while some of his stuff is fantastical (not necessarily the bits you think either) most of it plays by the rules we know. $\endgroup$ – Ash Jul 4 '18 at 17:36
  • 1
    $\begingroup$ Welcome to Worldbuilding.SE! When you have a moment, please take our tour and visit our help center to learn more about us. Your question is excellent for a first-post! However, it might have been asked before as this question. Please take a moment to review that question and determine if there is a difference. If there is, please edit your question to explain the difference to avoid closure as a duplicate. Thanks! $\endgroup$ – JBH Jul 4 '18 at 17:43
  • 3
    $\begingroup$ @ Branko Maksimovic Apart from the probable expansiion of the degerate matter once removed from the white dwarf, your scenario has the problem that the size and composition of the artificial planet has to be satisfy two different criteria at once. The artificial planets need to have one Earth gravity at the surface for the heath and comfort of the inhabitants, but need to have a high enough escape velocity at the edge of their atmospheres where gases escape. Surface gravity and escape velocity are calculated by different equations and it might be impossible to satisfy both. continued. $\endgroup$ – M. A. Golding Jul 4 '18 at 19:29
  • 2
    $\begingroup$ @Branko Maksimovic - continued. So try to calculate a size and composition of an artificial planet with both satisfactory surface gravity and satisfactory escape velocity in the upper atmosphere. And if that turns out to be impossible, you may have to use artificial gravity generators. Or build hollow cylindrical space habitats instead, that may be granted to the rich as fiefdoms by the Solar Emperor. $\endgroup$ – M. A. Golding Jul 4 '18 at 19:35
  • 11
    $\begingroup$ Don’t forget tidal effects!! If surface gravity at 3km is 1g, what is the gravity at 3.002 km? How about underground at 2.950 km? $\endgroup$ – Jim Garrison Jul 4 '18 at 22:32

Sorry in advance. You won't like this answer.

The "stuff" in white dwarf matter is more specifically known as degenerate matter. Basically, the more tightly you crush this matter, the more the valid quantum mechanical states for the electrons "fill up." For normal matter, there are so many more valid states than there are electrons, that we don't have this issue. For degenerate matter, you start to run out of states, and that means the matter cannot collapse any closer together. This creates a pressure that holds the white dwarf out against gravity.

This matter behaves funny. It turns out that this pressure does not depend on temperature, but rather depends on mass. The more mass you have, the smaller a white dwarf gets, rather than bigger. This strange process continues until you get to around 1.4 solar masses, at which point the hydrostatic pressures from this degenerate electron soup can't compete with gravity, and it collapses into a neutron star (which is held up by neutron degenerative pressure rather than electron degenerative pressure).

You happen to want a white dwarf with a small mass. Paradoxically, this means it is a very big white dwarf. If you took this matter from a heavy (small) white dwarf, it would probably have to expand as you brought it out to maintain that hydrostatic equilibrium (handwaving whatever magic you used to pull it out of the dwarf in the first place).

How big? Well, fortunately, because you are looking for a low-mass dwarf, you are also looking for one where the electrons are not approaching relativistic speeds. That means we can use the easier relationship $R\propto M^{-\frac{1}{3}}$. Using non-relativistic equations, a white dwarf with the mass of the sun would be approximately 0.014 solar radii in diameter, or 9737km. The mass of the earth is about 0.000003 solar masses. $0.000003^{-\frac{1}{3}}=69$, so the radius of your 1-earth-mass white dwarf would actually be 69 times bigger than the 1-solar-mass white dwarf, or about 670,000km. This is actually substantially larger than that of the Earth itself, indicating that at some point, the hydrostatic equations governing white dwarfs stopped being the dominant factor.

In other words, take 1 Earth mass of white dwarf out of the white dwarf, and the pressure will expand it outwards until it ceases to operate like a white dwarf, and starts operating like plain normal matter. You are going to need some handwavium to keep it compact.

You will also need some handwavium to keep the atmosphere. Science does not say the atmosphere will stick around. In fact, it says the atmosphere will flee your asteroid even more than it does on Earth. The acceleration of gravity decreases by the radius, squared. On Earth, the difference of gravity between sea level and 100km (the Karman line, the edge of space) is minimal. It's only about 3%. This is because the radius from your object to the center of the earth only changes from 6,371 km to 6,471 km. It's a pretty modest change. However, if your asteroid is only 5km, 5km+100km is a big difference. Gravity is going to be roughly 0.23% as strong at 100km as it is at 5km. If the gravity at the surface of your planet is 9.8m/s2, like it is on Earth, gravity at 100km is a mere 0.022m/s2! That won't hold much atmosphere at all!

  • 1
    $\begingroup$ The question is not the "surface gravity" at the edge of the atmosphere, but the escape velocity at the edge of the atmosphere. the artificial planets need to have one Earth gravity at the surface for the heath and comfort of the inhabitants, but need to have a high escape velocity at the edge of the atmosphere, and that is calculated by a different formula. $\endgroup$ – M. A. Golding Jul 4 '18 at 19:38
  • 6
    $\begingroup$ @M.A.Golding Escape velocity has the same inverse square of the radius term. The same logic applies. $\endgroup$ – Cort Ammon Jul 4 '18 at 20:04
  • 2
    $\begingroup$ Just a note regarding your last sentences: you woudln't need a 100 km atmosphere. If the radiation shielding is done with other means, a few dozen (or few hundreds) meters of atmosphere would be enough for the protagonists. $\endgroup$ – vsz Jul 5 '18 at 6:42
  • 3
    $\begingroup$ Why are you looking at 1 Earth mass to generate earth like gravity at the surface? The whole idea was Earth like gravity at 2.5km radius from the centre of the planet, not at earth radius from the planet, therefore much less mass (but higher density) is needed. $\endgroup$ – Hans Janssen Jul 5 '18 at 9:13
  • 1
    $\begingroup$ @P.Lord, I'm not saying that it's viable, just that the question was not about putting 1 earth mass into a 5km orb, but putting high enough density mass, but much less than 1 earth mass, into a 5km orb so it would have earth gravity at it's surface. $\endgroup$ – Hans Janssen Jul 5 '18 at 15:27

A first approximation

Let's calculate surface gravity. Assuming an object of density $\rho$ and radius $R$, the surface gravity is $$g=\frac{4\pi}{3}G\rho R\tag{1}$$ For a white dwarf, $\rho\sim10^9$ kg/m$^3$. If we want $g=9.8$ m/s$^2$, we find an $R$ of 35 meters. If you scale that up to a radius of even 2 km (a 4 km diameter), we find a surface gravity of 559 m/s$^2$. In other words, only a very small amount of your asteroid can be filled with white dwarf matter.

Let's go back to a different way of writing $g$: $$g(r)=\frac{GM}{r^2}=\frac{G}{r^2}\frac{4\pi}{3}R^3\rho\tag{2}$$ where I've substituted in $M=\frac{4\pi}{3}R^3\rho$. Here, $R$ is the radius of the blob of white dwarf matter. If we want $g=9.8$ m/s$^2$ and $r=2$ km, we find we need a blob radius of $R=520$ meters - roughly one fourth of the body's radius. I've assumed that the mass between the blob and the surface doesn't contribute much to $g$.

The big issue? There's a kilometer and a half of material between the blob and the surface, and the gravitational force at $r=R$ is going to be a lot. $g(R)$ will be about 145 m/s$^2$. Therefore, you need the material inside the asteroid to be able to withstand such forces. That's not easy. The pressure will be pretty large. Think carefully about this before constructing it.

A true white dwarf

Go read Cort Ammon's answer (and upvote it!). They talk about how, thanks to electron degeneracy pressure, white dwarfs obey a curious mass-radius relation: $$R\propto M^{-1/3}$$ In other words, the more massive a white dwarf, the smaller it is. Now, let's rewrite this as a scaling law, using Sirius B as an example. It has a mass of half a solar mass, and a radius of 0.003 solar radii. We can then write $$\frac{R}{0.003R_{\odot}}=\left(\frac{M}{0.5M_{\odot}}\right)^{-1/3}\tag{3}$$ Now, let's rewrite density. The mean density of Sirius B is roughly $3\times10^{10}$ kg/m$^3$. We then have $$\frac{\rho}{3\times10^{10}\text{ kg/m}^3}=\frac{M}{0.5M_{\odot}}\left(\frac{R}{0.003R_{\odot}}\right)^{-3}=\left(\frac{R}{0.003R_{\odot}}\right)^{-6}$$ using our mass-radius relation. Plugging this into $(2)$ gives us $$ \begin{align} g(r) & =\frac{G}{r^2}\frac{4\pi}{3}R^3\left[3\times10^{10}\left(\frac{R}{0.003R_{\odot}}\right)^{-6}\text{ kg/m}^3\right]\\ & =\frac{G}{r^2}\frac{4\pi}{3}R^{-3}(3\times10^{10}\text{ kg/m}^3)(0.003R_{\odot})^6 \end{align} $$ This then gives us $R=37.4R_{\odot}$. Cort Ammon got something within about a factor of 2, because they chose to not use general relativity (which didn't matter, honestly, for our purposes).

Degeneracy pressure

This mass-radius relation arises because a white dwarf is supported by electron degeneracy pressure, arising from the Pauli exclusion principle. Essentially, no two alike fermions (matter particles like electrons, quarks, etc.) can exist in the same quantum state simultaneously. Thus, when you compress a whole bunch of fermions, there's a pressure that resists the compression. In a white dwarf, this comes from electrons; in a neutron star, this comes from neutrons.

The mass-radius relation problem occurs in certain other bodies of degenerate matter, including neutron stars. For neutron stars, the mass-radius relation is not well-known because the equation of state (EOS), the equation relating pressure and density, is not known exactly. It's a very active area of research, both observationally and theoretically. Nonetheless, if you were to substitute in a neutron star or a quark star or some other body, you'd still have a problem.

Out of curiosity, let's try to calculate the minimum radius of a white dwarf. The maximum mass is given by the Chandrasekhar limit of about $1.44M_{\odot}$. Substituting this into $(3)$ gives $R_{\text{min}}\approx0.0021R_{\odot}$, or about 1467 km. That's not helpful. What it we push things even further, and try to find the smallest a neutron star can be? Well, the Tolman-Oppenheimer-Volkoff limit is essentially the analogue of the Chandrasekhar limit; it's about $2.25M_{\odot}$. Optimistic equation of state models give us a radius of about 9-10 km. Again, that's too high.

What about quark stars? These are hypothetical objects made primarily of a quark-gluon plasma, lying in about the same mass range as neutron stars. They're thought to be denser than neutron stars, and smaller, and, more importantly, from the little we know about their equation of state, the smallest ones should also be less massive. The problem, of course, is that they still aren't small enough. 6-8 km is reasonable for a somewhat small quark star. Additionally, we don't know very well how they behave; our constraints on the EOS are poor.

enter image description here
Figure 4, Rodrigues et al. (2011). Mass-radius relations for quark stars.

From the little I know about quark stars, the mass-radius relation depends on the ratio $\bar{\Lambda}/\mu$, where $\bar{\Lambda}$ is something called the renormalization subtraction point and $\mu$ is the much more familiar chemical potential. Setting $\bar{\Lambda}/\mu=2$ and $\bar{\Lambda}/\mu=3$ yield very different results, possibly differing by a factor of 2 (see Fraga et al. (2001)). If $\bar{\Lambda}/\mu=2$, we could see smaller quark stars.

That said, if we use some of the smallest radii fit by this optimistic value by Fraga et al., we find that, for $R=4$ km, $M=0.2M_{\odot}$, and so $g=1.66\times10^{12}$ m/s$^2$.

That's too high.

A black hole

We do have one more option. The more massive a black hole, the larger it is, and the less massive a black hole, the smaller it is. Say we instead put a black hole at the center of the asteroid, which has $r=2$ km. For our desired surface gravity, we need $M=5.87\times10^{17}\text{ kg}$. Calculating the Schwarzschild radius gives us $R_s=8.72\times10^{-10}$ m, which fits more than comfortably inside the asteroid.

Now, the black hole might evaporate via Hawking radiation, but it will take a long time - roughly $5\times10^{29}$ years, or 500 billion billion billion years. So it's going to stick around for a while. However, the gravity is still enormously strong, and it's going to accrete the rest of the planet pretty quickly.

  • 1
    $\begingroup$ that only leaves the little engineering detail of how to hold a BH centered inside a cavity in the asteroid. i suppose a charged BH could be handled electromagnetically. $\endgroup$ – ths Jul 5 '18 at 10:04
  • 1
    $\begingroup$ A charged black hole would be an interesting trick. Amusingly enough you could't go out and get "black hole stuff" like you might be able to with "white dwarf stuff," because you couldn't get it out. You'd have to create a made-to-order black hole. What a market that would be! $\endgroup$ – Cort Ammon Jul 5 '18 at 16:02
  • 1
    $\begingroup$ @ths I think, at that point, you're no longer handling the charged BH electromagnetically. Instead, you're letting the BH do it's thing, and handling the shell of a planet around it (much less mass). Still a heck of an engineering puzzle, but I think that means that if you had a few "columns" driven closer to the black hole with very strong charges on them, you might be able to create a stable configuration. $\endgroup$ – Cort Ammon Jul 5 '18 at 16:04
  • 2
    $\begingroup$ Running with the black hole theory, a 1g surface gravity would require a point source of 3.7e18kg at the center (source. Plugging that into a black hole calculator, which I used but did not vet in any way, that suggests that you could accomplish this with a micro-black hole that's about 10nm across (slightly larger, because you'll want it charged, but we're still talking nanometers). For an uncharged black hole (what the calculator does), its lifespan is 1.3e32 years. $\endgroup$ – Cort Ammon Jul 5 '18 at 17:12
  • 3
    $\begingroup$ That black hole would be less than a millionth of the mass of Earth, so it'd be easier to manage. I'd still recommend wearing gloves while handling the black holes. Personal Protective Equipment is very important for safety, both in the lab and in the factory! $\endgroup$ – Cort Ammon Jul 5 '18 at 17:14

The other answers have pointed out why you cannot convert a 5 km planet (asteroid) into a space habitat in the way you have described.

What you can do, however, is to construct an O'Neill cylinder. Hollow out a cylindrical asteroid, make it rotate, and live on the inside. This concept has been popularised in science-fiction novels such as Rendezvous with Rama (Arthur C. Clarke) or 2312 (Kim Stanley Robinson), featured (briefly) in films such as Interstellar, and has been the topic of questions on this site such as here, here, or here.

O'Neill cylinder
Source: NASA, via Wikimedia Commons

  • $\begingroup$ ...sorry, since you added book and movie references I just wanted to add an example from other medium. I deleted my comment. $\endgroup$ – xDaizu Jul 5 '18 at 14:59
  • $\begingroup$ Big problem here: The weather. On Earth we have the air at the equator moving 1000 mph while the air at the poles is standing still. This gradient is a major force in generating Earth's weather. The gradient across an O'Neill cylinder is far steeper. $\endgroup$ – Loren Pechtel Jul 5 '18 at 18:36
  • $\begingroup$ @LorenPechtel The energy source for the O'Neill cylinder is artificial and/or uniform. The climate is uniform. There is no gradient and therefore no wind unless artificially created. $\endgroup$ – gerrit Jul 6 '18 at 9:37
  • $\begingroup$ @gerrit No. Lets examine a cylinder 1km across. The surface is moving 70 m/s. On the opposite side it's moving 70 m/s the other way. Your gradient is 140m/s /km. On Earth you have 460m/s over 10,000km = .046m/s /km. The gradient in the cylinder is more than 3000 times greater. $\endgroup$ – Loren Pechtel Jul 7 '18 at 0:37
  • $\begingroup$ @LorenPechtel Oh, you are referring to shear forces. Differential rotating speeds on Earth are responsible for the coriolis force, which is important, but are not responsible for wind itself, which is due to pressure differentials. Vertical wind shear between surface (5 m/s) and upper troposphere (45 m/s) on Earth is 4 m s^-1 km^-1. The 140 m s^-1 km in an O'Neill cylinder would be vertical, whereas horizontal wind shear would be 0. And then there's the gravity differential. Weather would be very different, but I'm not convinced it wouldn't reach a liveable steady-state situation. $\endgroup$ – gerrit Jul 7 '18 at 9:31

I already think that the @HDE 226868 answer should be the accepter due to its always high-quality information. But there is a little thing that I do not agree with and changes the answer dramatically: heat.

A White Dwarf has a surface heat of around 6.000 to 30.000 K, that is a bit warm!. Your asteroid would smelt due to the high temperature...

That is why I suggest the use of a Black Dwarf, which is literally a cooled down White Dwarf. Obviously, due to its heat, you will need some of the highly advanced technology in order to cool down one, because its takes A LOT (gigayears) of time to cold down naturally. I won't tell you how to cold down one white dwarf, but due to this answer @HDE 226868 has made a question about it!

I am not sure if the change from a white to black dwarf would bring some other changes in its strange physics like the "curious mass relation" talked in his post.

  • 2
    $\begingroup$ But the white dwarf would clearly solve the atmopshere problem! Pesky hydrogen escaping into space? No problem. Just evaporate granite to replace it! =D $\endgroup$ – Cort Ammon Jul 4 '18 at 20:10
  • $\begingroup$ I've asked a separate question about the feasibility of artificially cooling a white dwarf. $\endgroup$ – HDE 226868 Jul 5 '18 at 14:15
  • $\begingroup$ @HDE226868, great! $\endgroup$ – Ender Look Jul 5 '18 at 17:15

Cort and HDE have already told you why this is not feasible regarding the planet from its core to its surface, and Cort elaborated on atmospheric escape as well.

If you handwave those away, you still got a lot of problems not related to the amount of gravity, but to the size of the asteroid.

On Earth, we have a lot of atmosphere. Your asteroid? Not so much. To keep the same pressure as here, you'd have an atmosphere volume to planetary surface ratio of $\frac{1}{1,000}$ that of the Earth. That's because volume scales down faster than surface area (Square-Cube law strikes again). The asteroid will have much less thermal insulation than here. The temperature difference between day side and night side will be incompatible with life as we know it. That difference will also cause hurricane-strong winds to be not events, but the natural state of the planet's atmosphere.

Also notice that the Sun has a considerable tidal influence on Earth. The Sun's tidal effect is 44% that of the Moon's own. On your planet, that influence will cause tsunamis even on lakes.

Geography and meteorology are to to your planet what biology is to Australia: quite the natural hazards.

  • $\begingroup$ No, the tidal effects of the Sun on an asteroid are minimal. $\endgroup$ – Edgar Bonet Jul 5 '18 at 10:54

Other answers show that according to what we know of physics, this can't really be done. Thus, you need some physics we don't know about yet (if your story can stand such a thing).

Assuming you don't want to make a super-dense material that transcends known physics, could you 'import' some gravity from somewhere else? Perhaps instead of putting any matter inside the asteroid, the builders put in a wormhole which is connected to (say) a black hole, or some other high-mass phenomenon. They could vary the amount of imported gravity by altering the size of the hole.

I realise this is a bit of a departure from good, solid, based in fact physics, so maybe too much of a reach to accept. Otherwise, sign up my great, great grandchildren for a personal planet - it sounds pretty cool.

  • $\begingroup$ So this would be on par with some magical future "gravity generators", but using a wormhole to a high gravity location as the medium rather than "ambiguous technology", right? An interesting follow up question would then be how to fine-tune the gravity level. Do we have several small wormholes that can be turned off or on as needed? Do we have control over where a workhole points on the other end so we can do on-the-fly adjustments to more ideal gravity sources? $\endgroup$ – Anthony Jul 5 '18 at 16:39

Other than using exotic states of matter, having true gravity the same as Earth is impossible. However, rotational gravity is very possible. Using the equation for rotational gravity, period in seconds equals 2pi multiplied by the square root of radius in meters divided by the meters per second squared, or this 1, for an asteroid 6 km in diameter you would need it to rotate once every 110 seconds. For an asteroid 4 km in diameter, you would need it to rotate once every 127 seconds.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.