Take the 2-minute tour ×
Space Exploration Stack Exchange is a question and answer site for spacecraft operators, scientists, engineers, and enthusiasts. It's 100% free, no registration required.

Have any spacecraft been lost transitioning between Mars and Jupiter?

How do spacecraft navigate that field unscathed as well?

share|improve this question
    
Unscathed? As in safely just in the case a small rock hits them in that vast emptiness of the asteroid field? (It is not quite a dense as in SciFi movies). –  Hennes Dec 26 '13 at 9:27
    
I sort of made it sound as if it was the astroid belt in Star Wars, but I more meant the debris field that is somewhat there and how to plot that trajectory. –  klobucar Dec 26 '13 at 22:09

1 Answer 1

up vote 12 down vote accepted

No spacecraft has been yet lost to the asteroid belt. In fact, we have the opposite example of missing an asteroid when it was even targeted, like was the case with MINERVA lander of the JAXA's Hayabusa deep space probe, missing the 25143 Itokawa asteroid. Why haven't we lost any spacecraft due to collision with asteroids in the asteroid belt is also pretty straightforward once you start appreciating its volume and its average particle density.

Asteroid belt is a huge area, sparsely populated with asteroids. It is incredibly difficult to estimate it's mean density considering we most certainly haven't detected all its objects, with most simply being too small to easily detect and track from up to a few astronomical units away, but we did detect most of its largest to medium size objects and that should help us guesstimate its density a bit. We already have some answers to similar questions in other threads, for example:

But for the sake of argument, let's crunch some numbers here too, and I'll take a bit different approach not to leech on work of others.

If we define the asteroid belt region as a torus with the inner boundary at $2.06 AU$ (4:1 orbital resonance with Jupiter), outer boundary at $3.215 AU$ (keeping the snow line from the formation of the asteroid belt at $2.7 AU$, nicely in the geometric center of the inner and outermost boundaries), and its thickness follows average inclination to the invariable plane of all the known asteroids in the asteroid belt at $\approx 4 °$, multiplied by two so we get its total thickness.

   enter image description here

            Plot of inclination $i$ vs semi-major axis $a$ for numbered asteroids inward of about $6 AU$. The main belt region is shown in red,
            and contains $93.4\%$ of all the objects. For reference, Mars orbits out to $1.666 AU$, and Jupiter between $4.95-5.46 AU$.
            The diagram was created by Piotr Deuar using orbit data for 120437 numbered minor planets from the Minor Planet Center orbit
            database, dated 8 Feb 2006. (Source: Wikipedia)

Plugging these parameters in equation for the volume of a toroid (which is actually a leftover of subtracting the volume of the outer sphere with the inner one and then dividing it to a $8 °$ thick slice), we get the total volume of $4.56 AU^3$.

That's a much lower volume than $16 AU^3$ quoted in the similar estimate of the asteroid belt volume in an answer to a related question on Physics.SE, but keep in mind that I only accepted the area where majority of its asteroids are known to be, not its extremes. I'm doing this so we can appreciate highest risk of colliding with an asteroid in the asteroid belt, so the purpose is somewhat different. I also estimated the distance to the Sun boundaries somewhat differently, following same idea of keeping the area limited to where most asteroids would be (highest density).

OK, so now that we have this volume of highest particle density, let's try and describe it with to us a bit easier to imagine figures; Earth has a volume of $108.321 * 10^{10}\ km^3$. Given calculated volume of the main density region of the main belt:

This is regarding its volume. Trying to estimate its particle density though, the total mass of the asteroid belt is estimated to be $2.8 * 10^{21}$ to $3.2 * 10^{21}\ kg$, which is just $4\%$ of the mass of the Moon. We'll use $93.4\%$ of the upper estimated mass, describing our main density, so $2.9888 * 10^{21}\ kg$. That's nearly exactly $3 * 10^{21}\ kg$, so let's round it up to mean of the estimate.

Now, if we could assume similar composition, then if an average size object would be only 1 liter (1 cubic decimeter, or 10 * 10 * 10 cm) in volume and it would have a density of Moon rock ($3.3464 g/cm^3$, or to put it differently, would have a mass of $3.3464\ kg$), that would then be $896.49 * 10^{18}$ of such objects and they would occupy $1.965 * 10^{-11}\%$, or 0.00000000001965% of the total volume of the main density area of the asteroid belt.

These are just some guesstimates and I do hope I got all the decimal points correctly, but as you see, chances of colliding with an object in the asteroid belt are indeed astronomical. There is no reason to consume spacecraft's power to try and track asteroids potentially on a collision course with a passing-by spacecraft, especially since these objects aren't stationary to the vector of the spacecraft's travel and it would be incredibly difficult to detect them and infer their orbits to try and avoid them with the use of onboard thrusters.

To my knowledge, no spacecraft to date tried to create a detailed radar picture of objects in the asteroid belt ahead of the spacecraft to also try and avoid them in an extremely remote chance of being on a collision course. Navigation and collision avoidance of the spacecraft in relation to a single object on its path yes, for example the mentioned Hayabusa, and Rosetta and its lander Philae will try something similar in 2014. And some other deep space probes (I believe also Pioneer and Voyager probes) did use wide-angle radar imaging to try and detect objects in the asteroid belt for the purpose of cataloging and studying them, but to avoid collision, you'd have to establish their orbit extremely precise, otherwise you wouldn't know if you're avoiding the object, or actually helping your spacecraft to collide with it. A bit like deciding that running in the rain will get you less wet, even if you don't know its angle or frequency.

share|improve this answer
1  
"No spacecraft has been yet lost to the asteroid belt" OK but.. Spacecraft have failed, have all the failure reasons been explicitly identified? It could be the result of damage from micro-meteorites that caused the failure.. +1 for an (as always) comprehensive & excellent answer. –  Andrew Thompson Dec 26 '13 at 14:38
    
@AndrewThompson I personally am not aware of any whose failure could potentially be identified to have been caused by an impacting meteorite in the main belt, or a collision with an asteroid. Do you have any particular space probe in mind? –  TildalWave Dec 26 '13 at 14:47
    
Not really, the only one I could think of was the Jupiter bound probe whose main means of information transmission failed. But then A) I am not sure if the failure mode was identified, and.. B) NASA did a fine job of rescuing the mission (so it could hardly be considered 'lost'). ..In retrospect, is my earlier comment merely 'noise'? I am considering deleting it and just sitting back to enjoy the rest of the thread. What do you think? –  Andrew Thompson Dec 26 '13 at 14:55
    
@AndrewThompson Not noise but we do have a nice, cosy chat room for ... well, chat you know? :P I'll presume you meant Galileo with your last comment? Its high-gain antenna failed to fully deploy because they didn't reapply the lube in the cups along antenna's ribs that were meant to release it, or something like that. A collision would've likely caused a fair bit more significant damage. Plus, micrometeorites aren't really limited to the main belt, that could've happened anywhere. ;) –  TildalWave Dec 26 '13 at 15:44
    
Cool. Yes you picked the probe correctly, and again I was wrong. I'll leave the comments, but am basically expended on the subject, so no need for chat. :) –  Andrew Thompson Dec 26 '13 at 15:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.