Let’s say I have this handwavey “make tiny black holes and prevent them from evaporating in a burst of hawking radiation” tech, which I borrowed from a passing alien.
I also have a bunch of space stations I’ve built, which I want to give artificial gravity to. The space stations are basically spheres that can contain a black hole in their center without collapsing. My goal is to have earth-like gravity experienced at the surface of each sphere.
I know how big across each sphere is – lets say I built a bunch of them, each with a different radius. Some are a few miles across, one is the moon (I hollowed out the moon to get raw materials for all my stations, of course).
So finally, my question:
How can I calculate how much mass a black hole would need so that 1g is experienced at a given radius?
Eta: I should add, I don’t need answers fast - we are still digging out the moon.
Unless there is some general relativity weirdness, I suspect you can just use the gravitational force equation and use 1Kg for Mass1 and 9.8 Newtons as the force experienced. Then you can vary distance and/or mass of the Mass2.
F = GMm/R²
Here is an online calculator I found. It can even tell you the mass of the black hole in units of ‘Earths’
I figured it was going to be gravitational force but translating my question into that equation was escaping me. I think I was overthinking because of the whole weirdness you tend to associate with black holes, but in this case they’re just an arbitrarily small source of mass.
In fact I can confirm this is the correct formula to use because if we plug in Earth’s radius we find we need a black hole with the mass of the Earth – which is, of course, correct. A black hole Earth would pull on us with exactly the same force as the current Earth does, it just wouldn’t have a floor to push back against!
I should point out that Paul Birch of the British Interplanetary Society proposed using black holes as gravity generators in this way many years ago. He called these structures suprastellar shells, although I suppose a structure around a black hole might be best described as a ‘suprahole’.
If you have a black hole as massive as Phobos (10.6 × 10E15 kg), it would be a billionth of a centimetre in radius, but it would last 3.17429E24 years. You could put a shell around such a black hole with a radius of 270 metres, and experience 1 gravity on the surface.
We can reduce F = GMm/R² down to knowing M/R² is a constant calculated from the Earth’s mass and radius. Since M/R² = k we can calculate that M = kR² so double the radius you need 4 Earth masses. 1/10 the radius you need only 1/100 Earth masses.
Knowing the Earth’s mass is 6 x 10^24 kg (rounded) you can now calculate your M without needing to know k.
That made me wonder: what mass black hole would you need to get 1 g surface gravity and 1500 W/m^2 at some radius? 1500 W/m^2 is very roughly the energy the sun provides on Earth, so it seemed a good number to target.
Unfortunately, it leaves us with a black hole of mass 1.408e13 kg and a radius of only 9.79 m. Kinda crummy. You have to sacrifice either surface gravity or power. Concentrating the power doesn’t even help you much because it goes down so rapidly as mass increases. So you can have a planetoid with decent surface gravity but the total power will be negligible.
I was going to say how it wouldn’t be exactly like being on Earth on a station-around-a-black-hole with a radius of 200m and a hole sized to give one G, but it seems to be closer that I thought.
On earth, for anything under a hundred miles, you can ignore the curvature of the surface, but with a 200m radius station, you can’t. But it’s not that bad. A one meter wide shelf would need to bend by (if my calculations are correct) half a millimeter to be perfectly level according to gravity. A 4 meter wide room would be one centimeter off of level if it was perfectly flat. So, you’d need to account for it when designing buildings, but for most rooms a flat floor would be fine, and regular flat furniture is OK.
Also, gravity would be weaker as you go above the floor: about 2% less at your head than the floor. I don’t think that would cause any problems, as long as you’re not going more than a couple stories high and don’t expect to use spring scales accurately. (at 10m above the surface, you’re down to 90% of G). That effect would prevent the planetoid from having an unconstrained atmosphere of any density, so the station would have to be built for vacuum outside.
One major difference on such a tiny suprahole world would be that the top of the atmosphere would be subject to much less gravitation than the bottom of the atmosphere. This would allow the air to escape rapidly into space, causing the inhabitants to die from vacuum exposure.
To avoid this you’d need to put a roof over the inhabited space to keep the air near the ground.
This is actually great news if you are interested in leaving your planetoid too. All you need is a big tower (and probably more than one so your planet doesn’t wobble) along the equator. At a certain height that could be calculated the tower will moving at the same speed a vessel orbiting at that altitude would move. This would make landing on the planetoid more of a docking exercise than a true landing – you could approach, dock, walk down the stairs to the surface, do your business, walk back up the stairs, and head off again, all without fighting a planetary gravity well.
Here’s a question. Let’s say you build such a world, with a glass shell over it held up by those docking towers we mentioned. Let us also say that a small hole opens up in this shell.
Would all the air rapidly leak out? Or would the fact that gravity is still, for the most part, pulling the air molecules down towards the surface prevent an active spew of particles into the vacuum due to pressure, so you’d have more of a slow leak instead?
Eta: follow up question. Could you accomplish the same thing with powerful magnetic fields to keep the atmosphere contained? Bonus points if this would give you a constant planet-wide aurora
The air leak on such a world would be somewhere between an air leak from a habitat on an ordinary asteroid, and the negligible air loss from an Earth-like planet. I’d guess it would be much closer to the first one than the second.
I can’t imagine how you’d keep an atmosphere on an object from escaping using a magnetic field. You’d have to ionise the outermost layer somehow, as neutral molecular oxygen and nitrogen would not be affected by the flux.
