Tuesday, 4 October 2016

energy - Proportionality constant in crater equations


I have searched for equations regarding craters and I came across two of them. The first one is from this NOAO website in the level two section. There, it says that, $$ D^3\propto E $$ where $E$ is the kinetic energy. It is not really an equation as there is a constant of proportionality, $K$, there. That's when I came across a second equation from this website that says, $$ D=0.07C_f\left(\frac{g_e}{g}\right)^{1/6}\left(E \frac{\rho_a}{\rho_t}\right)^{1/3.4} $$ where $\rho_a$ is the density of the impactor, $\rho_t$ that of the object, $g_e$ Earth's gravitational acceleration, $g$ the gravitational acceleration of the object, and $C_f$ the collapse factor. So these equations somehow contradict each other.


So I've been searching for the values making up $K$ using these two equations, but I'm wondering, is it possible to find $K$? Or do these equations not relate to each other at all?


I am writing my extended essay on this topic, so I did a 'drop the ball on sand' experiment. Can the more complex equation be used in this case? Also, I'm not sure what crater collapse factor means.



Answer



The issue is that there are multiple crater scaling laws, each with different assumptions, as Horedt & Neukem (1984) show (title of paper is Comparison of six crater-scaling laws).



Your first equation is a naive approach that the volume of the hole is linearly related to the kinetic energy, hence the $d= k\cdot E^{1/3}$ relationship. This particular relationship appears as a small-hole limit of Dence et al (1977) which appears in equation (2) of Horedt & Neukem.


The second equation you provided is from Shoemaker & Wolfe (1982) (conference stub, not an actual paper) and is equation (3) in the Horedt & Neukem paper. Here, we account for two other effects: the density and gravitational accelerations of the impactor and object. How this is derived, I cannot say because I was unable to find Shoemaker & Wolfe's discussion on it.


So the two equations are "contradictory" (the 2nd probably being more accurate than the 1st) because of assumptions made in the derivation. Thus, you cannot use your equation (2) to determine $K$ in your equation (1), you'll have to choose which of the two (or 6) you want to use.


Note that the 6 scaling laws in Horedt & Neukem show power-law of the diameter-energy relationships of slightly less than the 1/3 mark, most hovering around 0.28. Figure 1 from the paper, reproduced below, shows a plot of crater diameter versus impactor mass for the 6 different equations at a constant velocity (10 km/s) for solid-solid (solid lines) and ice-ice impacts (dashed lines).


Horedt & Neukem (1984) fig 1



I am writing my extended essay on this topic, so I did a 'drop the ball on sand' experiment. Can the more complex equation be used in this case? Also, I'm not sure what crater collapse factor means.



Yes, the more complex equation can be used in this case. Clearly $g=g_e$, so that factor can be eliminated. Then you'll need the densities of the impactor & object; this website says dry sand has a density between 1280 and 1600 kg/m3 and the ball's density can be computed by measuring the diameter & mass and using the typical $$ \rho=\frac{m}{\frac43\pi r^3} $$ According to Geological Implications of Impacts of Large Asteroids and Comets on the Earth (Amazon link), the collapse factor is a factor that accounts for an enlargement of the crater,




Terrestrial impact craters 10 km in diameter have all been enlarged by collapse of a transient cavity produced by the impact. The average enlargement is estimated at 30% ($C_f=1.3$).



It seems most sources I've seen suggest that for $D<\text{a few km}$, you can use $C_f=1$. Since I highly doubt you've created a >1 km crater in your experiment, the equation you want to use will be, $$ D\simeq4.4\times10^{-3}\left(E\rho_a\right)^{1/3.4} $$ where I've used $\rho_t=1440\,\rm kg/m^3$ (the midpoint of the density range), with $E$ determined via the conservation of energy and $\rho_a$ determined as above.


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