# Constructing an orbital trajectory that quickly returns to its origin

I'm working on a science fiction story that involves two spaceships engaged in combat while in orbit around a planetoid. My original idea called for spaceship A to trick spaceship B into firing a passive projectile such that it would orbit the planetoid and strike spaceship B. This was meant to happen within a matter of minutes, much faster than the spaceships' orbital period.

Of course, this was flawed because if the projectile travels with greater velocity it will simply have a larger orbit than the spaceships. Angling the trajectory closer to the planetoid would cause a slingshot effect, but that would be followed by a long, elliptical orbit, equally problematic.

It seems this scenario isn't possible without at least a third body. The story already calls for the battle to take place within a ring of asteroids (or rocky satellites, technically) also orbiting the planetoid. It occurs to me that the projectile could be made to slingshot around the planetoid, then slingshot around a large asteroid such that it quickly came back around and struck spaceship B.

The sheer improbability of this scenario doesn't concern me. Primarily, I just want to make sure such a trajectory would be possible. A secondary issue is that the planetoid is meant to be large enough to have small satellites orbiting it, but the flight of the projectile should be as quick as possible, only a few minutes in the original sketch. I'm worried that the "third body" asteroid can't be large enough to act as the second slingshot, but small enough to conceivably inhabit the ring of debris around the planetoid.

Is such a scenario possible?

First time poster. If my question isn't specific or constructive enough, please let me know how it can be improved. Feel free to edit my question to improve terminology, etc.

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Edited to point out that the projectile is strictly meant to be passive, without its own thrust. – Paul Bellora Mar 10 '13 at 20:53

All the following is simply a simple application of math. If I'm wrong anywhere (and I'm bound to be), please feel free to point it out to me or just edit the post, if you like.

The largest planetoid is Ceres, at almost 1000 km in diameter. If we assume that your planetoid was a bit larger than this, and that Ceres was the asteroid you're trying to slingshot around, we can do some maths and estimate a limit to see if it is truly possible to slingshot the rocket back into the spacecraft, using the largest gravity fields that we can.

At a mass of 940*10^18 kg, your asteroid will exert a gravitational force on your (perhaps) 100kg missile that is F = (6.27262 *10^12)/$r$^2. This will be for any value of $r$.

If you slingshot a body by a planet, the speed of the missile will then be twice the speed of the planet plus the missile's initial speed. So you should probably be orbiting against the revolution of the asteroid belt. Anyway, the final speed of the missile will be $2U + v$, where $U$ is the speed of the planet and $v$ is the speed of the missile. If we let this quantity be $w$, then you only have 2 variables now determining if you can slingshot back to the spacecraft: $w$ and $r$.

The critical condition here is that given a w (determining your kinetic energy) and an r (determining your potential energy), can we reach an energy quantity that:

a) enters the orbit of the asteroid such that it is captured in the orbit
b) escapes the gravity of the planetoid so the missile does not actually end up orbiting the asteroid
c) gains enough speed that it can slingshot back to the spacecraft in a few minutes.

The escape speed of the asteroid is:

$$v_{escape} = \sqrt{{\frac{2GM}{r}}}$$

So entering the orbit, your $v < v_{escape}$. But since you are exiting the planet, your $w > v_{escape}$.

Since $2U + v = w$, then we want everything expressed in terms of $v$:

So $$v < v_{escape} < w$$

$$v < v_{escape} < 2U + v$$

We're going to assume that the missile will be whizzing just above the surface of the planetoid (for maximum energy), so $v_{escape} = 0.51$ km/s and $r = 490$ km. Plus, we'll assume the missile is going at the fastest speed of satellite we have ever produced, which is $v = 70$ km/s. Substituting:

$$70 < 0.51 < 2U + 70$$

So, this can't happen. Your missile should be going slower than this. What we need is a speed that is less than 0.51 km/s when we enter, so let's go for 0.50 km/s.

$$0.50 < 0.51 < 2U + 0.50$$

To determine $U$, we need to estimate the distance your missile is going to be travelling. If we let $x$ be the distance of the spacecraft from the planetoid, and the planetoid is not moving, then if the missile is whizzing just above the surface of the planetoid, your missile will be travelling a distance of $2x + \pi r$. The missile will be travelling at different speeds throughout this course, so let's sum up the times required. Let's let the acceleration of the missile experiences be $a$.

The acceleration is the square root of the sum of the squares if the tangential and radial acceleration.

$$a = \sqrt{a_t^2 + a_r^2}$$

We know the acceleration at the surface of Ceres, which is our $a_r$. We also know $a_t$ because that's just the acceleration to get from $v$ to $w$.

$$a = \sqrt{(\frac{w - v}{t})^2 + (0.028g)^2}$$

Playing around with this equation, we will get

$$t = \frac{w - v}{\sqrt{a^2 - a_r^2}}$$

And this is the time it will take the missile to whiz by half the surface of Ceres. So summing up all the times:

$$t_{tot} = \frac{x}{v} + \frac{w - v}{\sqrt{a^2 - a_r^2}} + \frac{x}{w}$$ $$t_{tot} = \frac{x}{v} + \frac{2U + v - v}{\sqrt{a^2 - a_r^2}} + \frac{x}{2U + v}$$ $$t_{tot} = \frac{x}{v} + \frac{2U}{\sqrt{a^2 - (0.028g)^2}} + \frac{x}{2U + v}$$

Since we just added a new variable, let's define a new constraint. Let $t = 300$ seconds, for example. So it takes 5 minutes for the missile to whizz just above the surface of Ceres. Then we have:

$$300 = \frac{2U}{\sqrt{a^2 - (0.028g)^2}}$$

Let's also say that we don't want the entire trip to take more than ten minutes in total. So we let $t_{tot} = 600$.

$$600 = \frac{x}{0.50} + 300 + \frac{x}{2U + 0.50}$$

We can turn even more "knobs" with these variables and say that the spacecraft is 500 km away, so $x = 500$

Now having these equations, we can "solve" for $U$ and $a$. Because there's 1 equation and 2 unknowns, we're not really going to get just one answer.

$U = -0.607$

$a$ = 4.05*10^-3

So, that looks like a very slow acceleration. And Ceres seems to be going the other way around, away from you. This probably means that there's an error in the formulas. Or, your scenario is impossible.

I graphed the equation above (though I dont know how to do that here), with x = 500 and t = 300, and it showed me that you can't have a positive U unless your $t_{tot}$ is above 1300. And the graph looks the same if I change any of the constants I set.

I believe that this should be enough to give you something to play with and determine for yourself if it's possible or not. Besides, you get to decide where the ship is and how fast the missile is going. And anyway, the final equation that I got was way too complex for me to solve completely (and it has a lot of possible values, besides). I don't know any software that can help.

But the good news is, it's probably possible to slingshot the missile, but you have to allow a larger time for it to whizz by the asteroid. Feel free to change the dimensions of the asteroid in the formula. Some good data can be found on wikipedia.

Hope this is something though.

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Thanks for the great answer! Out of all them so far, yours is the most interesting because you actually try to construct the scenario. It's not conclusive, but you've definitely provided what I would need to dive in further if I decided to. Unfortunately a quick survey of the answers tells me the scenario is probably doomed. John Rennie's answer has a good point about the size of the planetoid vs. asteroid, which I was worried about. I would've liked him to present some mathematics like you did though. – Paul Bellora Mar 10 '13 at 21:12
It was a bit of a stretch from the beginning. You'd have to modify the environment specifically for something like this to happen (like increasing the density of the asteroid) which will just make this scenario seem contrived. Also, wouldnt a metal slug hitting that spacecraft be around the same danger level as an asteroid hitting it? – markovchain Mar 11 '13 at 1:30
Well it was always meant to be a stretch - I just don't want it to snap apart :) Here's what I think I'll do: there will be several large, metallic asteroids in a higher orbit. The slug will get fired outward from the planetoid, slingshot around an unspecified number of these large asteroids, and come back. Of course it's absurd, but mathematically possible, right? – Paul Bellora Mar 12 '13 at 0:44
That makes sense. I'm not entirely sure which simulators you used, but I'm sure their formulas have their own merit. These things are hard to fully simulate, though. You should do some reading here: en.wikipedia.org/wiki/Gravity_assist One important thing to note is, if you use multiple asteroids, your slug will have to be slow until the very last one. This means your asteroids until the last one have to be going very slowly (they're far out the orbit) while your last asteroid has to be faster (near the large planetoid). – markovchain Mar 12 '13 at 14:47
Good advice about the asteroid speed. I think I have enough info to at least fudge an absurd scenario, and to do more research if necessary. Thanks for all the help! I'll list you in the acknowledgements if the story's ever published :) – Paul Bellora Mar 12 '13 at 15:13

Your problem is that if the projectile is moving faster than the spaceships, which are in orbit round the planet, then the projectile velocity will be too great to be affected by the gravity of an asteroid. A large moon might do it, but then it wouldn't happen quickly.

How about having a small (e.g. asteroid mass) black hole in orbit round the planet. The point about a black hole is that even though it's mass is small you can get very close to it where it's gravity is very strong. You could even have the tidal forces shred the projectile so the spaceship gets hit by the hail of fragments.

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Interesting idea. I was hoping to avoid something as anomalous as a black hole, but that would certainly do the trick. – Paul Bellora Mar 10 '13 at 20:58

I don't think you can come up with a firing solution having orbiting period orders of magnitude lower than normal satellites IF it is a passive projectile - being there three, four or any other number of bodies. On the other hand, if it has a working rocket engine, you can orient the projectile towards the planetoid center with fired engines. But just the same, it is not very plausible scenario because you'd need hundred-fold radial acceleration to drop orbital period for an order of magnitude. The other problem is that such projectile would be visible by any means you can think of - IR scope, radar - you name it! I hope that spaceship B is an passive bucket full of robotic evildoers! :-/

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Yeah, the projectile is meant to be a metal slug - passive in other words. – Paul Bellora Mar 10 '13 at 20:48

The general idea of your scenario, I gather, is one of an attacker being destroyed by it's own weapon coming back to it.

In a terrestrial story you'd have to go for, say, a heat-seeking missile that is outrun just long enough, while flying in a specific loop, to fool the missile into retargeting on the origninal attacker.

Unfortunately, a scenario like that is no good in space. Any combat spacecraft will have a thrust-to-weight ratio that is far less than the thrust-to-weight ratio of a dedicated missile weapon. (Also, a technology advanced enough for space combat will have built-in recognition technology to avoid re-targeting the original attacker.)

I suggest another scenario:
The attack is performed with some laser weapon, and the defending ship has some totally new technology that can not only reflect 99.99% of the incoming light, but it also reflects precisely back to the source.

I don't know whether it's possible at all to develop an effective laser weapon for space combat. At worst any spacecraft is by itself already so reflective that any light simply reflects of its surfaces. Then again, technology that reduces radar cross section (known as stealth technology) uses surface materials that absorb as much electromagnetic energy as possible.

Currently, retro-reflection of laser pulses is used to determine Earth-Moon distance. Retro-reflectors have been placed on the Moon by the Apollo Moon landing missions. Astronomers have set up high power lasers that flash the general area where the retro-reflectors are positioned. Those laser beams spread out, I don't know how much. That spreading could mean it's impossible to build an effective laser weapon for space combat.

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Interesting alternative, but the projectile is meant to be a metal slug, without thrust of its own. So the problem isn't related to targeting but the trajectory of a passive projectile. Not sure what you mean about thrust-to-weight ratio - can you expand on why that's important if it still is? – Paul Bellora Mar 10 '13 at 20:53
@PaulBellora In space, when you want to turn (possibly turn through an entire loop) you use engine thrust. By contrast, for aircraft flying in an atmosphere the pilot uses the lift of the wings to fly in a loop. A space attack weapon will be a rocket engine with a payload of explosives. The bulk of that attack weapon will be its rocket engine. That rocket will have a lot of thrust for its overall mass. Big thrust with low overall mass enables a tight turn. By contrast, on earth: while an aircraft cannot outfly a heat-seeking rocket, it may be able to outturn that rocket. – Cleonis Mar 12 '13 at 22:41
What you're saying makes sense but I don't see your point. Are you saying it doesn't make sense for there to be a mass driver weapon in space? – Paul Bellora Mar 12 '13 at 22:48
Erm, I realized I had failed to read the detail of the weapon being a passive projectile. My apologies for the confusion. Passive projectile or smart missile, any scenario where an attacker is struck by it's own weapon will be astronomically improbable (a mini black hole just happens to whizz by??). It seems to me such a scenario will be like an episode of Red Dwarf. (There is in fact a Red Dwarf episode where to escape some disaster the crew must set up the equivalent of a billiards trick shot involving asteroids and what not.) – Cleonis Mar 12 '13 at 23:28