# Space debris impact

Given two pieces of aluminum, a sphere with diameter $$d$$ and a plate, how can we estimate the penetration depth given the initial velocity $$v$$ of the sphere? If that proves too complicated due to effects like melting, neglect plastic deformation of the sphere. Thanks in advance.

• Look up the "Thompson F-formula" and see if that's what you need. Sep 9 at 10:07
• @josephh can you provide a source with a derivation? Sep 9 at 10:17
• Hi Christian. There is this paper that references the original derivation first published by the US Navy. I haven't read it completely, but there are references therein that might help you find what you need. Good luck. Sep 9 at 10:31
• @josephh this applies only to "US homogeneous Class 'B' armor" and does not take into account material properties like the plasticity. Sep 9 at 15:32
• The general rule of thumb I learned (which is a hand-wavy order of magnitude estimate) is if the two materials are close in density and the impactor is moving at hypersonic speeds, the penetration depth is on the order of the size of the impactor. Sep 13 at 12:33

First, I gave a detailed list of useful references on cratering at: https://astronomy.stackexchange.com/a/16800/13663

Let us define some parameters for later use:

• $$V_{c}$$ is the volume of the crater [$$m^{3}$$]
• $$D_{c}$$ is the diameter of the crater [$$m$$]
• $$H_{c}$$ is the height/depth of the crater [$$m$$]
• $$V_{p}$$ is the volume of the projectile [$$m^{3}$$]
• $$v_{o}$$ is the speed of the projectile [$$m \ s^{-1}$$]
• $$m_{p}$$ is the mass of the projectile [$$kg$$]
• $$k_{p}$$ is the length-to-diameter ratio of the projectile [N/A]
• $$D_{p}$$ is the diameter of the projectile [$$m$$]
• $$\rho_{p}$$ is the mass density of the projectile [$$kg \ m^{-3}$$]
• $$\rho_{t}$$ is the mass density of the target [$$kg \ m^{-3}$$]
• $$\sigma_{t}$$ is a measure of the target strength [$$N \ m^{-2}$$ or $$Pa$$]

Then Lu et al. [2020] shows that the empirical (i.e., found by directly measuring) relationship between the volume of the crater and projectile is given as: $$\frac{ V_{c} }{ V_{p} } \approx 9.6 \ k_{p}^{-0.2} \ \left( \frac{ \rho_{t} }{ \rho_{p} } \right)^{3/2} \left( \frac{ \rho_{p} \ v_{o}^{2} }{ \sigma_{t} } \right) \tag{0}$$ for a concrete-like target being impacted by a rod-shaped projectile.

Other work has shown (discussed by Lu et al. [2020]) that the empirical relationship between the crater and projective diameters and between the crater height and projectile diameter are given by: \begin{align} \frac{ D_{c} }{ D_{p} } & = 9.11 \ k_{p}^{+0.267} \ \left( \frac{ \rho_{t} }{ \rho_{p} } \right)^{1/2} \left( \frac{ \rho_{p} \ v_{o}^{2} }{ \sigma_{t} } \right)^{1/3} \tag{1a} \\ \frac{ H_{c} }{ D_{p} } & = 1.37 \ k_{p}^{+0.267} \ \left( \frac{ \rho_{t} }{ \rho_{p} } \right)^{1/2} \left( \frac{ \rho_{p} \ v_{o}^{2} }{ \sigma_{t} } \right)^{1/3} \tag{1a} \end{align}

Note that in general, the volume ratio can be expressed as: $$\frac{ V_{c} }{ V_{p} } \propto k_{p}^{C_{0}} \ \left( \frac{ \rho_{t} }{ \rho_{p} } \right)^{C_{1}} \left( \frac{ \rho_{p} \ v_{o}^{2} }{ \sigma_{t} } \right)^{C_{2}} \tag{2}$$ where the exponents $$C_{j}$$ can be determined from empirical tests.

Unfortunately, there is no completely general, analytical formula because as $$v_{o}$$ increases, the interaction fundamentally changes (e.g., liquifying or even ionizing some of the target and projectile on impact, if energetic enough).

References

• Hughes, D.W. "The approximate ratios between the diameters of terrestrial impact craters and the causative incident asteroids," Mon. Not. R. Astron. Soc. 338, pp. 999--1003, doi:10.1046/j.1365-8711.2003.06157.x, 2003.
• Lu, Y., et al., "Hypervelocity Impact Cratering on Semi-Infinite Concrete Targets of Projectiles with Different Length to Diameter Ratios," Appl. Sci. 10(11), pp. 3910, doi:10.3390/app10113910, 2020.
• I tried your formula using the rough dimensions I assumed for mine, and it was a close match! Gives me more confidence in mine which uses the deformation energy method. Sep 19 at 18:40

The deformation energy or toughness of a material is the area under its stress-strain curve, pressure or energy per volume units. Let's call it $$U_0$$.

Assume the impact crater forms roughly a hemisphere of radius r, which is the depth we are trying to find. So equate the incoming kinetic energy with the energy needed to carve out a given hemispherical volume.

\begin{align} E_k &= U_0 \cdot V \\ \frac{1}{2} mv^2 &= U_0 \cdot \frac{2}{3} \pi r^3 \end{align} \qquad \quad r = \sqrt[3]{\frac{3 mv^2}{4\pi U_0}}

Unfortunately $$U_0$$ is not a commonly tabulated material property, but I did a rough estimate by assuming the stress strain curve was a triangle plus a rectangle, using for aluminum:

• Yield strength: $$276$$ MPa
• E: $$69$$ GPa
• Elongation at yield: $$0.4 \%$$
• Elongation at break: $$17 \%$$ Therefore $$U_0 = 46$$ MPa or MJ/m$$^3$$

I plugged in some estimated dimensions and weights from the Navy Paper referenced above, and they seemed to roughly match the penetration results, though I'd like to see more empirical data compared to this formula.

For a $$10$$ g projectile traveling at $$7600$$ m/s (the space station orbit), the formula gives an impact depth of $$30$$ cm.