• Support PF! Buy your school textbooks, materials and every day products Here!

Collisions Of Two Asteroids in the Main Asteroid Belt.

  • Thread starter AgentMoose
  • Start date
  • #1

Homework Statement



If there are two asteroids A and B of equal mass (and density of 3000kg/m^3), with the same semi major axis of a=2.2 AU with asteroid A having a circular orbit and no inclination, and asteroid B has an elliptical orbit with an eccentricity of e=0.05 and an inclination of i=5 degrees.
1) What is the collision speed during impact?
2) Based on the gravitational binding energy (GBE=(0.6)G(M^2)/r)), what is the minimum size an asteroid would need to be to survive the collision (without breaking up)?

Homework Equations



Given:
GBE= (0.6)G(m^2)/r
Not Given:
Va=√(GM((2/r)-(1/a))) where r=a at 2.2AU
Vb= √(GM(a/(r^2))(1-(e^2))) where r=a, e=0.05

The Attempt at a Solution



First I converted 2.2 AU to 3.2912x10^11 meters.

Then I tried to find the individual velocities of each of the asteroids using
Va=√(GM((2/r)-(1/a))) where r=a at 2.2AU, for the asteroid in circular orbit
and
Vb= √(GM(a/(r^2))(1-(e^2))) where r=a, e=0.05, for the asteroid in elliptical orbit

my results were:
Va = 20,077.2 m/s
Vb= 20,052.1 m/s

I know that I need to somehow take into account that one asteroid is colliding at a 5 degree incline, but I'm struggling on finding how best to do that to determine their impact speed. I attempted breaking it up into (x,y) components with vy=vsin(θ), vx=vcos(θ) with θ=5°, but wasnt sure if that was the right thing to do and was really stumped on where to really go from there.

For the 2nd part of the question, I think that once I find the collision speed, I could put that into the kinetic energy equation KE=(.5)m(v^2) and set that equal to the GBE equation and solve for r to determine the minimum size of the asteroid to survive the collision, but I'm unsure how to set that up, especially with only knowing the density, and not the mass or volume of either asteroid. Once I knew how to do that, then the actual solving for r part I think can do just fine. Thank you for your time.
 

Answers and Replies

  • #2
34,040
9,883
Vb= √(GM(a/(r^2))(1-(e^2))) where r=a, e=0.05, for the asteroid in elliptical orbit
Are you sure this formula is right for general r?
Same semi major axis means same energy, so at the same distance the asteroids should have the same speed.

I would calculate the velocity of B in the direction of A (conservation of angular momentum?) - once you have that, you can use the speed to determine all relevant velocity components.
 
  • #3
Well, that formula was the only one I could find among my materials that deals with the eccentricity of the orbit and the velocity of the asteroid. As far as the r goes, because asteroid A is in a circular orbit (therefore a=r), for a collision to occur in the first place, the r of asteroid B would also have to be 2.2 AU, which also happens to be its semi major axis as well. But I'm pretty sure it works for any r as far as I know.

Could you elaborate on what you meant in the second part of your reply, please? I'm not sure I fully understand what you mean.
 
  • #4
34,040
9,883
You can split the motion of asteroid B (at the collision point) in two components: one parallel to the motion of asteroid A, and one perpendicular to that.

What is the angular momentum of asteroid B? Can you split this into a component orthogonal to the plane of A, and one parallel to that? With conservation of energy and angular momentum, this allows to calculate the velocity components of B (as described above).


Concerning your formulas: there should be just one general formula, not two. That makes me suspicious.
 

Related Threads on Collisions Of Two Asteroids in the Main Asteroid Belt.

Replies
3
Views
1K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
0
Views
1K
Replies
5
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
6
Views
3K
Replies
2
Views
971
Top