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sithelephant t1_jdnv7hw wrote

A fun number to remember is that a circle with the diameter of the lunar orbit is very close to 1/64th the radius of earth. This means the earth covers about 1/4000th of the area which an asteroid has to pass through if it gets within the lunar distance.

So, if it goes past at 1 lunar distance, you have a 1/4000 chance of a hit (if it was random). 1/4 lunar distance, 16/4000 (1/250).

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Nowbob t1_jdo9nhq wrote

>a circle with the diameter of the lunar orbit is very close to 1/64th the radius of earth

???

What am I missing here? The lunar orbit is MUUUUUCH larger than the radius of the earth isn't it?

Wouldn't the lunar orbit being 1/64 the radius of the earth mean that the moon is whizzing past just barely overhead?

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Peat02 t1_jdo3zaz wrote

Wouldn't this be based on the volume of the lunar diameter sphere and the percentage that the earth makes up of that volume rather than the diameter?

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za419 t1_jdo52fp wrote

We'd have to do some orbital mechanics on this one. Most things in the solar system are roughly coplanar on the ecliptic, so the real shape is probably a section of the sphere a few degrees wide.

Or we could probably just guess and multiply the space available by a substantial number, because even that section is going to be pretty tall compared to the Earth. Space is big.

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Monnok t1_jdodiq3 wrote

Your Earth-volume : lunar-orbit-volume ratio works for the likelihood of finding the asteroid inside or outside of the Earth at any given moment.

But we can assume the object has an entire path of moments passing straight through on [basically] a line. An asteroid “looking ahead” directly at the round perimeter of the Earth might briefly occupy some point “in front” of the Earth before collision… but that’s still a collision path. And it’s never gonna get to any points on the other side. What it “sees” just is a flat disc in a flat disc. It’s either heading through the empty part or not.

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DoctorProfessorTaco t1_jdornit wrote

I think you mean 64x the radius of the Earth, not 1/64 the radius.

The radius of the Earth is 6,378km, I don’t think you mean that a circle with the diameter of the lunar orbit is 1/64 of that, or about 100km.

Additionally, even if you mean 64x larger, it’s still not quite correct. The average diameter of the lunar orbit is 768,000km. 1/64 of that is 12,000km.

What you probably meant to say was the average radius of the lunar orbit is approximately equal to 64x the radius of Earth, or that the radius of Earth is about 1/64 the average radius of the lunar orbit.

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reedef t1_jdo52pd wrote

You also have to account for the fact that the earth itself attracts asteroids even if they're not heading 100% towards the earth, though I'm not sure how large of an effect that is at interplanetary speeds

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OramaBuffin t1_jdoegks wrote

Not very high, asteroids are moving very quickly in general. Swinging by the earth will just slightly bend its trajectory rather than anything resembling sucking it in.

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reedef t1_jdp1shr wrote

So I started researching and Wikipedia actually has a calculation on this and for an asteroid with hyperbolic excess velocity of 12.5 km/s the effective increase of the cross sectional impact area is 80%. Not x10 or anything but not insignificant either.

The impact is going to be larger than that for objects that approach the earth from nearby orbits.

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