# Designing a binary asteroid pair

I am interested in building a little world involving a very realistic asteroid pair, orbiting each other, and would really appreciate some help with numbers and science for this. I would like two largish asteroids orbiting each other as close as is feasible.. (ideally I'd like a distance of up to 15km between them) Can you please make suggestions for:

- Reasonable mass, diameters and distance for the asteroids

They should be similarly sized (+/- 50%). I would like them to exert as much of a gravitational attraction on each other as they can, without being exceptional... maybe a few kilometers in diameter.. and reasonably close together. Please teach me or provide some simple math so I can provide size and mass and calculate orbital (around barycenter) elements.

- Estimates for how they rotate on their own axes

Are there any rotational constraints between the pair (like tidal locking?) What are reasonable rotation rates around their own axis? I am having a hard time visualizing how they would rotate.. would each asteroid rotate around a single axis which is static relative to the stars? Where would you place docking ports on a rotating asteroid to simplify docking maneuvers?

• I like the question but have a reservation about it. See the discussion here: meta.worldbuilding.stackexchange.com/questions/4062/… – Tim B Nov 4 '16 at 16:19
• This is an excellent hard science question, and one that I would love to attack with the program I used here. However it is much too broad. Perhaps you could split your three bolded bullet points into separate questions? As of now, I am voting to close. – kingledion Nov 4 '16 at 18:38
• Is it really too broad? It is asking ONE question.. please provide numbers and suggestions. I then broke that down into the three things I want numbers and suggestions for: size/mass, rotation/movement, and Other. It just seems like useless duplication to split this into separate questions (which I'll gladly do if y'll decide it so) as I'd just be copy pasting the context three times with the bold bit added to each at the bottom... – Innovine Nov 4 '16 at 23:20
• I have edited the question down in size and scope, is it better now? – Innovine Nov 5 '16 at 7:54
• HS tag about info cite - Consider alternatively the science-based and reality-check tags. Avoid using this tag as the only tag on a question. Just to ensure you fully aware about the tag – MolbOrg Nov 5 '16 at 9:37

All 2-body systems end up tidally locking each other. In fact, the Earth is slowly going into tidal locking with the Moon, it just takes a lot longer for the larger body to lock with the smaller one (the moon is of course already locked). In your case, it is expected that a binary system of asteroids would be tidally locked.

I simulated the asteroids orbiting the sun at one AU, where the sun has position {0,0,0} and a velocity which offsets that of the asteroids.

Let's assume our asteroids have an average density of $~1000\frac{kg}{m^3}$. Then the mass of asteroid of $3\left(km\right)$ diameter with a volume of $\frac{4}{3}\pi 1.5^3\left(km^3\right)\approx14\left(km^3\right)=14\times10^9\left(m^3\right)$ has a mass of $14\times10^{12}\left(kg\right)$

We can estimate the relative velocity by means of an idealized circular motion, in which case $a=\frac{v^2}{r}$. We also know that the acceleration due to asteroid 2 is $$G\frac{m}{r^2}=6.674\times10^{-11}\left(N \frac{m^2}{kg^2}\right)\frac{20\times10^{12}(kg)}{\left(1.5\times10^4\right)^2(m^2)}=5.93\times10^3\left(\frac{m}{s^2}\right)$$ And thus: $$v=\sqrt{ar}=\sqrt{1.5\times10^4(m)\times5.93\times10^3\left(\frac{m}{s^2}\right)}=0.3\left(\frac{m}{s}\right)$$

This gives us a good aprroximation, and with some testing, it turns out that $0.4\left(\frac{m}{s}\right)$ gives us a good orbit:

So, let's use the following statistics:

## Asteroid 1

Mass: $14\times10^{12}$

Position:$\left(AU,0,0\right)$

Velocity:$\left(0.4\frac{m}{s},29780\frac{m}{s},0\right)$

## Asteroid 2

Mass: $20\times10^{12}$

Position:$\left(AU,15\times10^3,0\right)$

Velocity:$\left(0,29780\frac{m}{s},0\right)$

Meaning they have an initial distance and velocity relative to eachother of $\Delta s=13\times10^3$,$\Delta v=0.4\frac{m}{s}$.

## Behaviour

Our asteroids have a steady near circular orbit around the sun, the orbit takes about 390 Earth days:

The distance between the asteroids fluctuates slightly, both in the short as long term, this is more realistic than perfectly circular orbits. The long term fluctuation is due to the influence of the sun, this gives the distance between the asteroids over 2 earth years:

## Tidal locking

The asteroids rotate very slowly around eachother, and will almost certainly be tidally locked.

## Eccentricity and Stability

This configuration is stable over single digit years, the precise configuration needs to change for stability to be maintained over more years.

For decades long stability, the eccentricity actually seems to decrease, regardless of the initial situation, however when the orbit eccentricity becomes too low (the orbits become highly circular), the orbits become suddenly less stable, here is the plot over 200 years:

As you can see, after 70 years or so, the asteroids start to draw closer together. Increasing velocity accuracy, draws this out,

Let me know if you need anything else, I love this stuff.

My guess is this means it's really difficult to find a stable orbit between such two small objects, however by increasing the accuracy of the velocity, it should be possible for the orbit to be stable for a few thoasand years, though such a configuration would be highly unlikely to occur naturally.

• nice. influence of other bodies in solar system maybe too ? )) just them self they will be pretty stable - we see that on earth-moon and others. what about Jupiter influence it pretty significantly influence our sun, saw that in my simulations) – MolbOrg Nov 6 '16 at 12:00
• @MolbOrg I'll try adding Jupiter, and run a longer term simulation. – Feyre Nov 6 '16 at 12:03
• Something must be seriously wrong in the numbers above. No small objects can orbit each other every few seconds at a distance of 15km. .. lets just have one stationary, and the other going around it.. thats a path of 90km, in 20 seconds? Thats close to escape velocity for a planet, not a small rock. You're off by many zeros somewhere. – Innovine Nov 6 '16 at 18:43
• what size is each asteroid, in km, and what is the distance between them (in km), and the time taken to orbit each other (in seconds)? tip: they might be going around the sun at 15km/s, but certainly not each other. – Innovine Nov 6 '16 at 21:19
• Your asteroid size is 10^6m.. thats insane, thats bigger than the largest known asteroid. I said maybe a few kilometers in diameter, not thousands of kilometers. I also never said 15000km anywhere, in any edit. Again, you're out by x1000. – Innovine Nov 6 '16 at 21:23