Traveling between co-orbiting bodies

[mabus83]

Hello all.

I'm new to this forum. I've downloaded and used Orbiter in the past, but only now registered to this forum, my reason for registering having less to do with Orbiter and more to do with my interest in orbital mechanics.

Currently, I'm in the first stages writing some hard sf, and I want to make sure that my figures are correct. Sites like Atomic Rockets are easy enough for me to understand, but the more in-depth textbooks such as Fundamentals of Astrodynamics are beyond me.

Without boring you with my half-baked story ideas, I'll cut to the chase. I want a smallish Kuiper Belt object with a Vasimir strapped on it on a trajectory that has it passing by other co-orbital KBOs. My iceteroid/spacecraft which is cannibalizing itself for fuel is traveling 1 AU per year (in addition to its natural orbital velocity). The spacecraft doesn't decelerate as it passes its co-orbital neighbors. Rather it sends out smaller vessels to rendezvous back and forth, making the spacecraft a cycler of sorts for colonies in the Kuiper belt.

Now this is what I want to confirm:

If, assuming all of the bodies are more or less co-orbital with the same natural orbital velocity, would the initial velocity work out to zero?

Therefore, would the delta v requirement be simply the speed of my vessel in m/s (since delta v = final velocity minus initial velocity)?

If so, would my delta v requirement just be 1 AU per year, or 4,743 m/s?

If my iceteroid cycler has a vasimir with an exhaust velocity of 294,000 m/s, then it should only entail cannibalizing a small portion of the iceteroid's mass. E.g. If consuming 10% of its own mass, giving the iceteroid a mass ratio of 1.111, the icetroid would have a delta v budget of 30,946 m/s, more than enough for its journey several times over.

Is this correct, or am I missing something important?

And if this is correct, how would it work for retrograde? Would the delta v requirement be the same since the 1 AU in question is simply relative distances between the co-orbiting bodies?

 

[Tommy]

The key factor is Relative Velocity, or the difference in velocity between two objects. The tricky part here is that Velocity is not just a speed, but also a direction. Any difference in direction (planar, or inward/outward) will increase the dV required by the smaller "go fetch" vessels.

Another thing to consider is that if your icesteroid is travelling faster than the onther KBOs, then it's orbit won't match - it will be larger. You might have a periapsis in th belt, but chances are your Apoapsis will be significantly outside the belt (depending on how thick the belt is).

 

[mabus83]

The key factor is Relative Velocity, or the difference in velocity between two objects. The tricky part here is that Velocity is not just a speed, but also a direction. Any difference in direction (planar, or inward/outward) will increase the dV required by the smaller "go fetch" vessels.

Another thing to consider is that if your icesteroid is travelling faster than the onther KBOs, then it's orbit won't match - it will be larger. You might have a periapsis in th belt, but chances are your Apoapsis will be significantly outside the belt (depending on how thick the belt is).

Thanks Tommy for your explanation. For now at least, I am not going to do the math on the smaller shuttle-like vehicles. I think I will leave it to the reader to assume that their mass ratios are sufficient to get the job done. However, the iceteroid craft is much more central to my story, some of the action is going to take place inside the interior, which will be like a scaled-down O'Neill colony, so I want to make sure I get things right for its delta v requirements and delta v budget.

As to the higher velocity of my iceteroid affecting its orbital trajectory, couldn't the iceteroid apply thrust at regular intervals at a direction perpendicular to its thrust axis to keep it close to its original orbit? If so, any idea how I could go about calculating this? Will what I thought would be a smooth co-orbital trajectory effectively turn into a bunch of single-tangent high delta v maneuvers? If that is the case, it looks like I will have to do some of the calculus-related calculations after all.

As to the thickness of the Belt, as I understand it, most KBOs (at least those discovered so far) are clustered around orbits resonant with Neptune's, so I am guessing the belt would be relatively thin, which would be problematic in light of what you pointed out.

Any suggestions or alternatives for getting a cycler to work would be welcome. What about putting the iceteroid in retrograde? Could it still stay co-orbital with the other KBOs then?

 

[Tommy]

As to the higher velocity of my iceteroid affecting its orbital trajectory, couldn't the iceteroid apply thrust at regular intervals at a direction perpendicular to its thrust axis to keep it close to its original orbit? If so, any idea how I could go about calculating this? Will what I thought would be a smooth co-orbital trajectory effectively turn into a bunch of single-tangent high delta v maneuvers? If that is the case, it looks like I will have to do some of the calculus-related calculations after all.

As to the thickness of the Belt, as I understand it, most KBOs (at least those discovered so far) are clustered around orbits resonant with Neptune's, so I am guessing the belt would be relatively thin, which would be problematic in light of what you pointed out.

Any suggestions or alternatives for getting a cycler to work would be welcome. What about putting the iceteroid in retrograde? Could it still stay co-orbital with the other KBOs then?

Don't know the exact number, but I think it would take about 10.000 m/s per year if you try the inward burns to stay in the belt. Retrograde would require HUGE amounts of dV for the "go fetch" vessels - four times your iceteroid's velocity.

I'd just slow it down a bit, there's really no need to be going almost 5km/s faster than the belt. Even at 1km/s faster, you should have plenty of fresh targets.

 

[mabus83]

Don't know the exact number, but I think it would take about 10.000 m/s per year if you try the inward burns to stay in the belt. Retrograde would require HUGE amounts of dV for the "go fetch" vessels - four times your iceteroid's velocity.

I'd just slow it down a bit, there's really no need to be going almost 5km/s faster than the belt. Even at 1km/s faster, you should have plenty of fresh targets.

Thanks again for the reply.

If it is only 10,000 m/s per year, that would be acceptable. As implied in my original post, I am willing to burn off about 10% of the iceteroid's mass (for a mass ratio of 1.111) which should allow for a delta v budget of >30,000 m/s. In that case, 10% mass burn off should cover orbital corrections for 2 1/2 years. My cycler is not intended to make a full circuit of the belt, only a 10 AU span over ten years, and since you'd get better mileage on the future burn-offs, I can probably do the one-way trip over 10 years with not much more than 20% burn-off of mass.

I know you said your 10,000 m/s was rough, but even if it's only good for back of the envelope calculating, could you share with me your figures?

As to retrograde, I was thinking maybe just cancelling out the natural orbital velocity at that distance, with Pluto's mean orbital velocity as a rough reference. That would be a retrograde of 4.7 km/s. My only thought is wouldn't any slow down also change my trajectory as it would for a speed up, bringing me closer to sun instead of farther out?

 

[Tommy]

Yes, slowing down would lower your orbit, so it's about the same dV as going faster. That 10/000 m/s is a semi-educated guess, though, so don't take it too literally without verifying!

If you are only going 10 AU, it should be able to set up a trajectory that starts at the outer edge of the belt, drops down til the halfway point, then rises again - and have it "straight" enough that you wouldn't need the inward burns, or at least minimizes it.

 

[martins]

Don't know the exact number, but I think it would take about 10.000 m/s per year if you try the inward burns to stay in the belt.

I seem to come to a different figure, but mine was just a quick back-of-napkin calculation, so I may have made a mistake (please correct then):

For Kuiper belt object, assume semi-major axis a = 40AU

=> orbital period T=2*pi*sqrt(a^3/mu) ~= 8e9s

You want to have a relative prograde speed of vrel = 1AU/year = 4.75e3m/s
This will reduce the period to

T2 = T - 2pi*a/vrel = 8.4e7s

In order to maintain the orbit despite the increased prograde speed, you need to accelerate towards the sun. Effectively, you have to add to the gravitational force vector as if you were orbiting a heavier sun. The mass of this "virtual sun" is given by

M2 = 4*pi^2*a^3/T2^2/G =1.8e34kg

This is several orders of magnitude larger than the true mass of the sun, so

dM = M2-M ~= M2

The force required for an asteroid of mass m is therefore

F = G*dM*m/a^2

and the acceleration

acc = G*dM/a^2 ~= 0.03m/s^2

The dV per orbit of the asteroid therefore works out as

dV = acc*T2 ~= 3e6m/s

 

[Tommy]

I'll take Martins estimate over my extremely rough guess, he's much smarter than I!

 

[martins]

I'll take Martins estimate over my extremely rough guess, he's much smarter than I!

Bad mistake :lol: - I just realised that my calculation was utter nonsense. In fact, T2 should be

T2 ~= 4e9

which would lead to

dV = 4.5e4 m/s per orbit,

or

dV ~= 350 m/s per (Earth) year.

Can this be right?

 

[mabus83]

Thanks Tommy and Martins.

I am reviewing over the figures. Math is not my strong suit, so I apologize in advance for what may be really naive questions.

For the equation that Martins first showed, I take it the M for GM = "mu" is the Sun's mass, right? That would be 1.988435×10^30 kilograms. Multiplying by the gravitational constant G, 6.67×10^-11 newton square meters per kilogram squared, gives us 1.326286145 × 10^20.

I was thinking the same on the semi-major axis. A lot of the known KBOs are more or less co-orbital with Pluto, it being the 2:3 orbital resonance with Neptune. As I said, I'm not good with math. To keep my units consistent, 40 AU has to be in meters, right? (Thank Kepler I've got Wolfram Alpha) That comes out to 5.983914828×10^12 meters. Cube that, it's 2.142674544 × 10^38, then divide it by mu (1.326286145 × 10^20), which works out to roughly 1.61554469x10^11, the sqrt of which is 1,271,040,791.55. Multiply by 2 and pi, gives me 7,986,184,826.3.

For T2, I assume a is the same as for the first equation since I want to keep to the same orbit and semi-major axis, so again it's 5.983914828×10^12 meters, right? According to Wolfram Alpha, 1 AU/yr works out to 4744 m/s, so a/vrel should be 1,261,364,845.69, which multiplied by 2 times pi would be 7,925,389,065.43. T minus this quantity gives me 60,795,760.87 for T2, which is different from what Martins got. Am I doing something wrong here?

---------- Post added at 02:14 AM ---------- Previous post was at 01:19 AM ----------

Although I'm quite sure my figures are already wrong at this point, I am going to follow them just to see where they take me. So with my result for T2 being about 6.07x10^7, square this, and divide this by G, I get about 5.541x10^25 for T2^2/G. Divide the cube of the semi-major axis by T2^2/G, gives me 3.867x10^12. Multiply by pi^2 and 4 gives me about 1.527x10^14. Is this right? I was a bit confused by the order of operations for the equation Martins provided. Is it (T2^2)/G or is it T2^(2/G)?

Since this works out to being less than the mass of the Sun, I will try the latter order of operations. In that case, 2 divided by the G would be 2.997×10^10, using this as the exponent of T2 gives me the insanely big number 6.079576087x10^7^29970000000. At this point my calculator and wolfram alpha have died on me as I try to figure (2.142674544 × 10^38)/6.079576087x10^7^29970000000. In any case, this can't be right either since whatever the answer is, it's going to be far less than 1.

So I am at a dead end here, it seems. If anyone can show me where I went wrong, I'd greatly appreciate it.

---------- Post added at 02:52 AM ---------- Previous post was at 02:14 AM ----------

Ok, I went through my figures for M2 again.

I'm not sure what I did wrong the first time, but this time I got 3.431x10^34, which is closer to Martins' figure, but still different.

This is the order of operations I did. Is this correct?

M2 = (4*pi^2*a^3)/(T2^2)/G

If so, (4*pi^2*2.142674544 × 10^38)/(3.696x10^15)/(6.67×10^-11) = 3.431x10^34

 

[martins]

For T2, I assume a is the same as for the first equation since I want to keep to the same orbit and semi-major axis, so again it's 5.983914828×10^12 meters, right? According to Wolfram Alpha, 1 AU/yr works out to 4744 m/s, so a/vrel should be 1,261,364,845.69, which multiplied by 2 times pi would be 7,925,389,065.43. T minus this quantity gives me 60,795,760.87 for T2, which is different from what Martins got. Am I doing something wrong here?

T2 = T - 2pi*a/vrel = 8.4e7s

This equation was wrong. Instead, it should be

T2 = 2*pi*a/v2

where

v2 = v0 + vrel

and

v0 = 2*pi*a/T

is the orbital velocity of the original orbit. Subsitute the values, and you should arrive at the value of T2 I got in my second post.

 

[mabus83]

From what I got, M2 - M = 3.4309812736×10^34 kg

This times G comes to 2.288x10^24.

For force (F), I multiply this by the mass of my iceteroid (which I worked out separately as 3.924x10^10 kg), and this total quantity by the square of 40 AU in m, giving me about 2.508× 10^9.

For acc (same as F except without the iceteroid mass in the equation), I get 0.0639107. Again, close to Martins' figures, but still different.

Multiply this by T2, gives me 3.885x10^6 per orbit (or 3,885,499.634234309 m/s). This is closer to Martins' first calculation of delta v than his second, corrected post. How do I get per Earth year? Making a fraction out of Martins' figures for per orbit/one year and multiplying that by my figure for per year, gives me 302,205.53 m/s delta v. This is 1000 times bigger than Martins' figure. Again, Am I doing something wrong?

---------- Post added at 03:37 AM ---------- Previous post was at 03:35 AM ----------

This equation was wrong. Instead, it should be

T2 = 2*pi*a/v2

where

v2 = v0 + vrel

and

v0 = 2*pi*a/T

is the orbital velocity of the original orbit. Subsitute the values, and you should arrive at the value of T2 I got in my second post.

Sorry, I didn't see your last post before I posted mine several minutes later. Thanks for the clarification, I will give that a try and check our figures to make sure that they match.

---------- Post added at 03:39 AM ---------- Previous post was at 03:37 AM ----------

By the way, this is probably a stupid question, but how did you convert the delta v per orbit to delta v per earth year?

 

[martins]

By the way, this is probably a stupid question, but how did you convert the delta v per orbit to delta v per earth year?

Since the magnitude of the acceleration is constant, dV is a linear function of dt, therefore

dV(1year) = dV(1orbit) * 1year/T2

where 1year = 3.16e7

 

[mabus83]

Ok, I got a 3.977x10^9 for T2, so my figures must match yours up to that point. Things start to go wrong for me with the equation for M2.

What was the figure you got for M2 since you corrected your calcuation for T2? I get about 3.565x10^10 for M2. I'm not sure if my order of operations is correct. I multiplied 4*pi^2*a^3. Then I did T2^2/G. Then I divided, so that my order was (4*pi^2*a^3)/(T2^2/G). This can't be correct, can it? It would make M2 less than M, giving me a negative number for dM.

---------- Post added at 05:03 AM ---------- Previous post was at 04:43 AM ----------

Tried again with a different order of operations, and got 8.014x10^30. This is a lot closer. Is this the correct figure?

---------- Post added at 05:24 AM ---------- Previous post was at 05:03 AM ----------

Ok. I went with the latter figure for M2 and got a dV of 354.06 m/s per Earth year, more or less the same as yours. No idea if the reasoning behind the use of these equations is correct. You're light years ahead of me on that subject, so until I get a phd in orbital mechanics, I will take it on faith. Thank you very very much for taking the time to walk me through this.

---------- Post added 07-09-11 at 01:30 AM ---------- Previous post was 07-08-11 at 05:24 AM ----------

Sorry, one more question. So, if the 350 m/s figure is correct, how does this figure into the total delta v requirement?

The prograde motion is just a one time acceleration of 1 AU/year (4744 m/s), whereas the 350 m/s orbital corrections have to be done regularly, so is the total requirement 4744 + 350n where n = number of Earth years?

As stated before, this is assuming the vessel does not intend to decelerate at any point and leaves the rendezvous with passing bodies to smaller shuttlecraft with their own delta v requirements considered separately.

Again, I want to thank Tommy and Martins, but considering that hopping between asteroids (on in this case, iceteroids) is a pretty common premise in hard sci-fi, I would think this issue would have been addressed before on these forums.