I wonder how to determine the directions, in which the collision debris is launched when 2 asteroids collide. I am aware of: m1*v1 + m2*v2 = m*v = m3*v3 + m4*v4 + m5*v5 + ... and this works just fine for the masses and valocity, however I find it difficult to determine the boundaries of the directions and under what circumstances shatter be produced or the asteroids will just "merge".

All info is appreciated :)


I think that this problem doesn't have an exact answer. Some time ago, I talked about this with the astrophysicist Paolicchi (this is the asteroid named after him) who works on the field. The conclusion is that debris are produced at random and you can only impose some ("few") constraints globally, say on big branches of the asteroids belt or of planetary ring. There debris "termalize" after a big number of collision and remain at rest with respect to each others. In the case of just a collision the physics is complicated... I list just some points:

  1. the asteroids are typically non-self-gravitating objects, that is they're mainly taken together by (local) electric forces. Hence, they are not round and, typically, they doesn't merge, since matter globally is electrically neutral. The same is true for artificial asteroids;
  2. since they spin and they have complex shapes their collisions are very difficult to modelize. You can apply the conservation of momenta but you have no boundaries on the velocities of each single fragment;
  3. for some purposes, it could be useful to approximate the production of debris as proportional to the energy in the center of mass of the two colliding asteroids: $E_{\text{cm}}\propto n$ of fragments. Then assume a $n$-body decay, each with the same mass. Even in the case of $n\gtrsim 3$, you can only have a phase-space for these debris and some probability density functions for their production angles.
  • $\begingroup$ Yes, I am interested at first at a simplified model - where spherical acteroids with same structure and different masses collide and produce shatter debris. The problem I am facing is how to determine the maximum velocity vectors at which the debris could potentially be launched - e.g. a sphere with radius of (?) when 2 asteroids with the same mass collide frontally with v1(5, 0, 0), v2(-5, 0, 0) and not a sphere (but what?) when the same asteroids collide with v1(100, 0, 0), v2(-5, 0, 0) $\endgroup$
    – MerlinBG
    Dec 5 '13 at 11:54
  • $\begingroup$ Infinity... :) Consider for example the limit where the shatter debris are two very light-weight particles, other than the two original asteroids. Even further, suppose that these particles are photons! Their speeds are $c$ (ore close to..), while their total energies are fractions of the initial total energy: ${1\over 2}M{v_1}^2+{1\over 2}M{v_2}^2$. $\endgroup$ Dec 5 '13 at 12:14
  • $\begingroup$ What I meant is that the smallest the fragments, the higher their velocities... Such a collision can produce very fast projectiles! The only bound you have is on the total energy of the system, equals or lower than the energy of the two colliding asteroids in a certain reference frame, say your rest frame or the cm frame. $\endgroup$ Dec 5 '13 at 12:19
  • $\begingroup$ Can such a (purely mechanical) collision produce faster debris, than the original asteroids? My naive asumption was "no" (thats why I tried to figure out some "pattern"), but if this is not the case, I could "just" generate any number of random vectors, which adds up to the resulting "merged" particle's vector? $\endgroup$
    – MerlinBG
    Dec 5 '13 at 12:55
  • $\begingroup$ Well, in order to reject your naive idea, try to throw a glass on the floor; this is a two body collision as well as a two asteroids collision. You will see many light-weight glass fragments bolting away with faster speeds than that of the glass at moment of the impact... Now look at that mess! It can provide a good insight of how messy and complex are collisions of satellites! You also could be interested in what is called Kessler syndrome (en.wikipedia.org/wiki/Kessler_syndrome), if you didn't know it yet! $\endgroup$ Dec 5 '13 at 13:20

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