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Guest brak

meteorites striking a planet

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Guest brak

Hi,

 

This isn't an astronomy question really, but a pure geometry one (as

you'll see with all the unrealistic assumptions).

 

Imagine you have a perfectly spherical body in space (like a planet or

moon), and it is sitting still (not moving or rotating or revolving).

Meteorites, travelling in random directions, might hit it. Assume the

meteorites are travelling in straight lines and aren't influenced by

the gravity of the our little sphere.

 

Given that the meteorites are as likely to be travelling at any angle,

or through any point, as any other....what percentage of the

meteorites that end up hitting the sphere should strike the surface at

greater than 45 degrees, and what percentage at less than 45 degrees?

 

We debated this elsewhere and got nowhere. One person said it should

be exactly 50-50 (based on the ratio of the areas of the "flattened"

projections of the greater and less than 45 degree regions), another

said it was closer to 30-70 (based on the surface areas of those

regions).

 

Please set us straight.

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Lets reason in a 2d world, where your planet is just a circle. Without any loss of generality we may suppose that the meteores strike in a specific point (Since you assume meteors appear randomly everywhere and there directions are also uniformly distributed), due to symmetry argumentation. According to your assumptions the "hit angle" will be uniformly distributed also and therefore counting symmetry arguments the 1/2-1/2 response is surely the better one.

 

Mandrake

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If you assume that all directions have equal probability, then you can't use the circle analogy without allowing for the differences between a sphere and a circle. On a circle, you can treat the arc length of a section over the circumference of the circle as the probability of the random direction falling on that arc, but the points on a sphere (in 3-D polar coordinates) at which [math]\phi<\pi/4[/math] are more "scrunched up" than the points at which [math]\pi/4<\phi<\pi/2[/math], so you have to find the actual surface area of the sphere, cut it into whatever pieces are important, and the ratio [math]\frac{SA}{4*\pi*r^2}[/math] will give the correct probability.

 

Fortunately, the volume can also be used to give an equivalent result. So the ratio of the probability of hitting at less than 45 degrees to the probability of hitting at greater than 45 degrees is given by:

 

[math]\frac{\int_{0}^{1}\int_{0}^{2\pi}\int_{\pi/4}^{\pi/2}\rho^{2}sin(\phi)d\phi d\theta d\rho}{\int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{\pi/4}\rho^{2}sin(\phi)d\phi d\theta d\rho}=1+\sqrt2[/math]

 

So the "less than 45 degrees" impact is more likely by a factor of about 2.414.

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