Cross Section for Asteroid Impact

AI Thread Summary
The discussion centers on estimating the cross section for an Earth-asteroid collision, focusing on the effective target area that the asteroid "sees." The key concept is that the collision cross section S can be calculated by multiplying the maximum impact parameter bmax squared by π. Participants clarify that the escape velocity ve can be expressed as sqrt(2GM/R), which is crucial for simplifying the equations. There is confusion regarding the dependence of the final expression on the masses of the Earth and the asteroid, with emphasis on ensuring that the final result does not include these variables. The conversation highlights the importance of correctly substituting and simplifying terms to derive the collision cross section accurately.
ebjessee
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Homework Statement


In this problem, you will estimate the cross section for an earth-asteroid collision. In all that follows, assume that the Earth is fixed in space and that the radius of the asteroid is much less than the radius R of the earth. The mass of the Earth is M_e, and the mass of the asteroid is m. Use G for the universal gravitational constant.
MUG_ev_5_0.jpg

I've already solved:
E_initial = (m/2)v2)
L_initial = bmv
E_at surface of Earth = ((v_f)2*m)/2 +(GmMe)/R
L_at surface of Earth = (mvfR
bmax2 = R2+(2RGMe)/v2

Where I get confused is the actual significance of bmax:

The collision cross section S represents the effective target area "seen" by the asteroid and is found by multiplying (bmax)2 by π. If the asteroid comes into this area, it is guaranteed to collide with the earth.
A simple representation of the cross section is obtained when we write v in terms of ve, the escape speed from the surface of the earth. First, find an expression for ve, and let v=Cve, where C is a constant of proportionality. Then combine this with your result for (bmax)2 to write a simple-looking expression for S in terms of R and C.

Express the collision cross section in terms of R and C.

I know that a version of ve is sqrt(2GM/R), but I have no idea how to go on from here.
 
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Did you take your expression for ve and form v = Cve? What did that do for your bmax2 equation? Show your work.
 
Ok. We say that v=Cve
So bmax2 = R2 + (2RGMe)/Cve2

ebjessee said:
The collision cross section S represents the effective target area "seen" by the asteroid and is found by multiplying (bmax)2 by π

Then π bmax2 = π R2 + 2π RGMe/Cve2

Distributing pi to both sides of the equations helps me see that I'm now looking at circular area equations, so that helps to but bmax into perspective. I plugged in ve = sqrt(2Gm/R) which helped me reduce the previous equation to:

S = π R2 +π R2Me/Cm

That didn't work. I got a message the answer did not depend on m or Me.
 
Nevermind- just forgot to square C and use the same Me. Thanks!
 
You want to replace ##v_e^2## in your ##b_{max}^2## equation with your expression for escape velocity right away and cancel out what can be cancelled. I don't know how m snuck into your equations...
 
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