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Ratio of projectile to ejecta mass, shoot the moon. |
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| Mar7-12, 07:06 PM | #1 |
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Ratio of projectile to ejecta mass, shoot the moon.
Say we shoot the surface of the moon with various sized steel balls of size a cm and larger, say up to a km in size. Say the balls hit the surface of the moon moving at 20,000 mph. Is there a simple formula for the mass of the ejecta as a function of the mass of the projectile? Will the formula depend on whether the projectile hits dust or rock or a combination of the two?
Thanks for any help! |
| Mar9-12, 06:46 PM | #2 |
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After a bit of Googling, wow, what a find,
http://keith.aa.washington.edu/crate...ling/index.htm From, Crater Database and Scaling Tools, http://keith.aa.washington.edu/craterdata/ The math behind the calculator, http://keith.aa.washington.edu/crate...ing/theory.pdf |
| Mar9-12, 08:33 PM | #3 |
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Wow, nice find!
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| Mar11-12, 12:06 PM | #4 |
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Ratio of projectile to ejecta mass, shoot the moon.
Used the above calculator, seemed buggy for multiple calculations, and got the following result. For a steel projectile striking the moon at 10,000 mph perpendicularly, the mass of the projectile, M, was roughly proportional to the volume, V, of ejecta for sizes from 1 cm to 10 m,
M ≈ V above that size M^α ≈ V Where α is less then 1. Is the above an example of scale invariance? If I draw a picture of the projectile next to the volume of ejecta I can change the scale but not have to redraw the picture, one picture works for a range of projectile masses and ejecta volume? |
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