The data I have is only $v_\infty$ and $v_0$, velocity at a large distance and escape velocity respectively. I need to find at what impact parameter a meteor strikes the surface of a planet of radius $R$.

  • $\begingroup$ How is "impact factor" defined? $\endgroup$
    – xaxa
    Feb 6, 2017 at 12:30
  • 2
    $\begingroup$ I'm guessing the impact factor is defined the same way it is defined in a scattering experiment i.e., the perpendicular distance between the direction $v_\infty$ and centre of the planet. $\endgroup$ Feb 6, 2017 at 12:35
  • $\begingroup$ Probably textbook mentions something else? Perhaps the trajectory of the meteor is assumed to be tangent to the planet's surface?.. Otherwise it seems to me that the impact factor can be arbitrary. $\endgroup$
    – xaxa
    Feb 6, 2017 at 12:41
  • $\begingroup$ I think it is assumed to be tangential. $\endgroup$ Feb 6, 2017 at 13:00
  • $\begingroup$ Since the escape velocity is solely connected to a planet mass, you have just one parameter for the definition of a meteor dynamics. Then, the impact is arbitrary. You need to define your problem more closely. $\endgroup$
    – jaromrax
    Feb 6, 2017 at 13:03

1 Answer 1


There is a continuum range of impact parameters resulting in a striking. If the impact parameter $b$ vanishes they collide. If we infinitesimally change $b$ they still collide. Sou you should find what is the smallest impact parameter such that the asteroid does not strikes the planet. This will happen when the maximum approach $r$ between the bodies is less than the radius $R$ of the planet. When they are equal, the asteroid "scratches" the surface of the planet.

You can solve this problem just by using energy and angular momentum conservation (since there is only a central force). At the maximum approach of a conic orbit, $r=R$, there is not radial kinetic energy (the particle is in a returning point), so at this point $$E=\frac{L^2}{2mR^2}-\frac{GMm}{R}.$$ This shall equal the energy at infinity, $E=\frac{1}{2}mv_\infty^2$.

To eliminate the angular momentum $L$, just compute use the angular momentum of the particle at any point of the trajectory (there is only one point easy to compute it). This angular momentum is written in terms of the impact parameter $b$. Plug this into the above equation and solve for $b$. The answer is $$b=R\sqrt{1+\frac{2GM}{Rv^2_\infty}}.$$ For any impact parameter less than that, there will be a striking.

The last step is just to use the escape velocity to eliminate $GM/R$.


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