4
$\begingroup$

An entrepreneur has a dream. He wants to create the most amazing and magical amusement park in the solar system. He goes to investors with the blue prints of the park, and they think it's a good idea, but they are only willing to fund enough money to buy an asteroid in the asteroid belt, as the overpopulation in the solar system has made all other available land extremely expensive.

The entrepreneur is so elated that the investors accepted his pitch, that he immediately buys all of the materials to build the park. But in his excitement, he forgot one major detail. Due to the extremely low gravity on even the largest asteroids, most of his rides are nearly useless.

He panics. He already spent almost all of his investors' money. If he doesn't fix this problem he will go into massive debt. So, as a work around, the entrepreneur decides to use the remaining money to install enough mass at the center of his asteroid to simulate the gravity of the Earth.

What would be the consequences of an asteroid in the asteroid belt (say about 20-30 mile radius) gaining enough mass to simulate the gravity of the Earth?

Specifically:

  1. What consequences would this have on the asteroid belt itself?
  2. What consequences would this have on the rest of the solar system?
  3. And finally, is there any material dense enough to do this without increasing the size of the asteroid or would a black hole be necessary?
$\endgroup$
3
  • 1
    $\begingroup$ Surely it’d make more sense to just have the carts/carriages on the rides create their own locomotion via wheels or having them pulled down somehow, simulating the effects of gravity on a rollercoaster? $\endgroup$ Commented May 26, 2019 at 23:17
  • 1
    $\begingroup$ @LiamMorris Oh silly Liam. This would require an entire redesign of the park. All of the rides would have to be redesigned to account for the change in vertical g forces. It would require new legal agreements and health plans to allow employees to work in low gravity environments without long term damage such as muscle atrophy and bone mass loss. It would require compliance with new safety regulations to account for floating debris. No, no, installing an entire planet's worth of mass at the center of the asteroid is the only option. $\endgroup$ Commented May 26, 2019 at 23:38
  • 1
    $\begingroup$ Where exactly do you plan to get this new mass from? The total mass of the asteroid belt is only 4% of the mass of the moon. I'm guessing no one is going to let you "borrow" mass from a planet. The only unclaimed (and massive enough) source would probably be the oort cloud... which will also probably be the most expensive source of mass possible... $\endgroup$
    – conman
    Commented May 27, 2019 at 10:00

4 Answers 4

7
$\begingroup$

Let's first consider what it means to make an asteroid have the gravitational pull of the Earth. Let's assume that the asteroid is Ceres, which is the most massive asteroid in the belt. Ceres has a radius of 473 kilometers. Now, we know the acceleration due to gravity is 9.8 $\frac{\text{m}}{\text{s}^2}$. The formula is $$g = \frac{Gm}{r^2}$$

Rearranging gives $$m = \frac{gr^2}{G}$$

$G$ is about $6.673 \cdot 10^{-11}\text{N}\cdot\text{m}^2\text{kg}^2$. Substituting gives us $m \approx 3 \cdot 10^{22} \text{kg}$. For reference, the Earth is about 200 times more massive. Earth's moon is more than twice this massive.

Making the asteroid smaller would decrease the necessary mass. Mass is proportional to the square of the radius, so half the radius would give a quarter of the mass. However, density is inversely proportional to the cube of the radius and proportional to mass. So halving the radius would double the density. $$p = \frac{m}{V}$$

Or since this would be a sphere, $$p = \frac{3m}{4\pi r^3}$$ or $$p \approx 74 \frac{\text{tons}}{\text{m}^3}$$

This is more than even the densest element, Osmium (22.6). Neutronium can be this dense, but it requires the kind of pressures present in a neutron star to compress to this extent.

To your questions:

  1. Minimal. This is essentially the same as putting a body 40% the size of the Earth's moon into the asteroid belt. It may increase the deflection of some of the nearby asteroids, but it is unlikely to be enough to clear the belt.

  2. Minimal. It would be closest to Mars but still farther away from Mars than Mars is from Earth. And it's much smaller than the Earth (by any measure, but most importantly by mass).

  3. Probably not. It's not enough mass to stabilize neutronium. So a black hole it is. The challenge of course is building the outside of a strong enough material keep it from falling into the black hole. I'll leave it to someone else to determine if that's possible or not.

$\endgroup$
3
  • $\begingroup$ You are forgetting the main reason why there is an asteroid belt there, and not a planet: Jupiter. It won't calmly accept a massive neighbor. $\endgroup$
    – L.Dutch
    Commented May 27, 2019 at 7:53
  • 2
    $\begingroup$ This is less massive and farther away than Europa, which Jupiter tolerates just fine. Yes, Jupiter tore apart the accretion disk forming at asteroid belt distance and probably part of the accretion disk at Mars distance. Possibly even some of the accretion disk at Earth distance (hardly bigger than Venus now). But this wouldn't be formed from an accretion disk. It would be formed artificially. The tidal forces from Jupiter may complicate the shell, but the shell is already questionable. $\endgroup$
    – Brythan
    Commented May 27, 2019 at 9:20
  • $\begingroup$ Worth pointing out that the total mass of the asteroid belt is only 4% of the mass of the moon. Ergo, I'm not really sure where you are going to get the necessary mass to make your black hole in the first place. Maybe whoever owns Jupiter will let you a few 10e22 kg of hydrogen which you magically compress into a black hole? Otherwise you're going to have to start grabbing stuff out of the Oort cloud. $\endgroup$
    – conman
    Commented May 27, 2019 at 9:52
4
$\begingroup$

Frame Challenge:

Why I don’t think your idea wouldn’t work

Firstly, if “He already spent almost all of his investors' money” then he won’t be able to afford to fill the asteroid with enough material to simulate Earth’s gravity. Also, it may have unintended side effects such as making the orbit unstable or pulling in debris (as the asteroid likely doesn’t have an atmosphere to burn up debris due to it having such a low gravity).

Second, there might not even be enough material to simulate Earth’s gravity. If you wanted to simulate a planet’s gravity, you’d need a planet’s worth of resources to do it. You will not be able to get your materials from Earth as every material on Earth is contributing to its own gravity. If you start scooping out thousands of cubic miles of material from Earth and carting it off planet, altering Earth’s gravity, theres going to be a lot more people angry at you than just your investors.

Not to mention the logistics nightmare of trying to haul all the necessary material through space to your asteroid, the fuel consumption, the pay for the workers, mining equipment and all manner of things. Even if you manage to get all that done, you’ve then got the task of hollowing out the asteroid (and hoping it doesn’t collapse around you as you do) and then fill it up with a more dense material - all very costly, not good for someone on a limited budget and likely on a limited time scale.

If the goal is to get the rides working, simply have the rides create their own locomotion

A much more effective, easier, cheaper and actually plausible solution is to instead have the carts powered by fuel or electricity which drives wheels under the cart, causing it to accelerate. In fact, doing it this way, as there is no air resistance in space and there is less gravity pulling you down on the asteroid, you’d likely be able to go faster than you could on Earth, your ride could constantly be accelerating until you applied the brakes.

$\endgroup$
1
  • 1
    $\begingroup$ I agree with the frame-challenge and proposed solution (you can probably come up with even more fun "rides" in microgravity). Per some of the other answers that do the math though, it doesn't actually take an earth-mass to do it (more like a moon-mass). Obviously you wouldn't be getting the extra mass from the earth, although where exactly you are going to get this extra stuff from is a very good question. Oort cloud is probably the only plausible location, and the cost of that would depend on your hypothetical intersteller propulsion technology - probably very expensive. $\endgroup$
    – conman
    Commented May 27, 2019 at 9:58
4
$\begingroup$

Adding conventional mass is not going to work out well for your entrepreneur.

To simplify the problem, let's assume for the moment that rather than build into an existing asteroid, we're building our own. Our building material will be mainly tungsten with a touch of heavier elements to bring its density up to an even 20 g/cm^3, or 20,000 kg/m^3. We'll build a perfect sphere, because it makes the math much easier.

The surface gravity of a body is dependent upon its mass, its radius, and the gravitational constant. Because we're assuming a perfect sphere of uniform density and we know the desired surface gravity, we can solve for radius: about 1750 km, or some four times the radius of Ceres. You're most of the way towards Mercury's size, and actually outmass it by a slim margin.

So let's turn that around by fixing the radius this time, at 50 km. For earthlike surface gravity, the mass required is "only" about 3.7 * 10^20 kg - about a third of a Ceres, or a little more than a Vesta. (Taking 8 Flora as a guide, the asteroid itself probably contributes about 1% of this mass.) Because the mass is more concentrated and the radius of the surface is much smaller, we don't need nearly as much of it. Of course, that's a problem in and of itself: the material we're placing in the asteroid would need a density on the order of 200,000 kg/m^3, or somewhat denser than the core of the Sun. (And before you ask, no, you can't harvest Sun core material for this purpose: it's only that dense because it has the full weight of the Sun pressing in on it.)

Hence, a singularity. According to this answer, a 1 micrometer black hole would mass in the right neighborhood and should be containable-ish. (Moving it into place inside the asteroid is going to be an interesting time.)

Consequences for the rest of the solar system should be negligible. As noted, there are already objects of comparable mass inside the asteroid belt; one more shouldn't make a big difference. The planets all mass comfortably more than your black hole and will be more or less unaffected.

The asteroid belt will almost certainly be disrupted, unless you actually replace Vesta with your black hole. Other asteroids that pass near-ish to your new superheavy asteroid (keeping in mind that this is near in astronomical terms, not anything humans would recognize as "near") will have their orbits deflected. In the long run this will probably result in some of them being ejected from the belt, possibly impacting some unfortunate planet. Your inhabitants will want to make sure their asteroid impact prevention plans are up to date.

n.b. - I've left off my work in the surface gravity analysis because I'm not familiar enough with formatting to not make it a hideous blob of text. I'd be happy to take it step by step if needed.

$\endgroup$
1
$\begingroup$

In the worst case scenario, you've just destroyed the inner solar system.

Before we get to that however, let's deal with the black hole idea. Unless you bring it in from the outside, your black hole isn't going to give you the mass you need because a black hole is a region of infinite density, not infinite mass. That means creating a black hole in the middle of your asteroid just gives you a asteroid massed black hole and your rides literally fall into it and you've lost your investment. To add to the insult, your black hole with a mass that size won't sustain itself for very long so it would disappear as well a few seconds later. But, I digress.

The problem I have with this is not the fact that you've created that much mass in the asteroid belt; it's that you've done it without changing the orbital speed. As such, the asteroid, now earth massed, is going far too slow to maintain its orbit around the sun at the distance it is from it. That means that your asteroid is going to start falling into the sun.

I'm not the specialist in orbital mechanics here (HDE is) but I think you've just created an earth-massed comet. Let's assume that the asteroid doesn't come near any of the major planets for a while, disrupting their orbits, I have a sneaking suspicion that it may well eventually splash into the sun anyway. The coronal mass ejection that is triggered by this could easily wipe out life on earth if earth was in the blast cone. But, I'd argue that it's more likely to eventually hit one of the inner planets, possibly earth, and lead a collapse in orbit of both the planet and the asteroid, AND trigger a CME that could cause all sorts of havoc on the planets remaining.

Bottom line is don't do this. Please.

$\endgroup$
3
  • $\begingroup$ Orbital velocity doesn't depend on the mass of the orbiter. $\endgroup$
    – L.Dutch
    Commented May 27, 2019 at 4:10
  • $\begingroup$ @L.Dutch no, you're right. But, angular momentum does doesn't it? Therefore if you increase the mass of something without also increasing its kinetic energy, don't you cause it to slow (M=mv) and therefore fall into a more eccentric orbit? Is that wrong? $\endgroup$
    – Tim B II
    Commented May 27, 2019 at 5:39
  • $\begingroup$ @TimBII If he is just magically increasing the mass of the asteroid somehow, then whether or not the orbit changes depends on his magic. Presumably if you are able to "magically" increase the mass of an asteroid, you can also make sure the new object has the same orbital velocity. If you are doing things the normal way and "bringing in" mass from elsewhere in the solar system then same thing: presumably you are making sure to bring in mass with the same orbital characteristics, thus making sure that you don't alter the orbit. $\endgroup$
    – conman
    Commented May 27, 2019 at 9:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .