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)?
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
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.