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I found this crater equation $$ D=0.07 \cdot C_f \cdot (g_e/g)^{1/6} \cdot (W p_a/p_t)^{1/3.4} $$ on a website, where

$$ \begin{align} D &= \text{Crater Diameter}\\ C_f &= \text{Crater Collapse Factor (this is equal to 1.3 for craters > 4km on Earth)}\\ g_e &= \text{Gravitational Acceleration at the surface of Earth}\\ g &= \text{Acceleration at the surface of the body on which the crater is formed}\\ W &= \text{Kinetic Energy of the impacting body (in kilotons TNT equivalent)}\\ p_a &= \text{Density of the impactor (ranging from 1.8g/cm3 for a comet to 7.3g/cm3 for an iron meteorite).}\\ p_t &= \text{Density of the target rock} \end{align} $$ Can someone explain to me what the crater collapse factor is?

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  • $\begingroup$ I have the same doubts as you and if you have resolved your essay about craters or have any other information please mention it . $\endgroup$
    – Faiz Iqbal
    Commented Feb 20, 2016 at 10:32
  • $\begingroup$ Based on the arbitrary-seeming numbers and "crater collapse factor", it looks to be an empirical equation. Dimensional analysis doesn't work out. I wonder what the units are for Cf and W. $\endgroup$
    – jvriesem
    Commented Dec 8, 2023 at 18:15

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I can't find any description of how the equation you cite is derived, so I can only speculate. With that caveat, I would guess the factor of 1.3 is the ratio of the rim diameter to the excavation diameter.

The bolide will excavate an initial bowl shaped crater, and the diameter of this is the excavation diameter. Immediately after the impact various processes can occur, including a subsidence of the ground immediately outside the initial crater:

Crater formation

(image is from this paper)

The result of this is that the final crater diameter will be greater than the excavation diameter by about a factor of 1.3 (see for example this review). I would guess that this is what the author means by the crater collapse factor i.e. it describes the increase in the crater size due to subsidence of the ground outside the initial excavation crater.

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  • $\begingroup$ I'm hoping to test this equation out by using the 'drop the ball on sand' experiment. Do you think this will be enough? I also found another relationship where D^3 is proportional to the kinetic energy. Should I test that one instead? $\endgroup$
    – Luqman Halim
    Commented Jul 22, 2015 at 6:52
  • $\begingroup$ @LuqmanHalim: you obviously won't reproduce the 1.3 collapse factor because your sand won't show the faulting that you get in rock with large craters. It would be interesting to see the data though. Given that doing the experiments will be the time consuming bit, once you have the data I would try any equation you find. Bear in mind that your experiments will be far from typical of real impacts. Experiments meant to simulate real impacts use high velocity pneumatic cannons to fire the balls at supersonic speeds. $\endgroup$
    – John Rennie
    Commented Jul 22, 2015 at 7:17
  • $\begingroup$ I have done the experiment and collected some data. Based on that, I used the equation to find the crater collapse factor. The results I got was not satisfying at all as it's got no correlation to the other data I have. It seems kind of random to me but very close in value. This result is what got me confused about the crater collapse factor in the first place. $\endgroup$
    – Luqman Halim
    Commented Jul 22, 2015 at 7:39
  • $\begingroup$ You could shoot a gun into sand from various distances/a couple different types of gun. That might be more fun and closer approximation of meteors than than dropping balls. (Might want to bring your safety goggles), and maybe check with local laws on that kind of action too. $\endgroup$
    – userLTK
    Commented Sep 24, 2015 at 4:04

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