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Elon Musk just commented in the post Falcon Heavy Test Flight press conference that reentering at interplanetary velocities means dealing with 'some of the heating things that scale to the 8th power'. He says this at 22:00 of the video below (it should start playing a few seconds before - to show the context of the comment). What forces was he referring to?

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    $\begingroup$ I would assume he refers to the cubic relationship between heat and velocity. 2^3 = 8. Twice the speed results in 8-fold higher heating rate. Possible related on Aviation StackExchange. $\endgroup$ – Thomas W. Feb 7 '18 at 10:37
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    $\begingroup$ proper Answer to the question on Physics StackExchange $\endgroup$ – Thomas W. Feb 7 '18 at 10:48
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    $\begingroup$ Thomas W - I Don't think that's what he's referring to. He clearly says that it scales with the eight power, not the cubic power and he was surprised that anything scaled with the eighth power. Lot's of things scale with the cube, lots of things scale with even the 4th power, but very few physical quantities scale beyond that. Radiative flux does scale with the 8th power of shock velocity. $\endgroup$ – ckmrk Feb 7 '18 at 17:10
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    $\begingroup$ @OrganicMarble adjusted the title accordingly. $\endgroup$ – kim holder Feb 7 '18 at 17:13
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    $\begingroup$ @uhoh, i added a reference to the exact time, but the video should start playing just before that when you press play. $\endgroup$ – kim holder Feb 13 '18 at 20:57
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@ckmrk essentially already gave the answer what Musk is probably referring to, and as his comment came a week ago with no answer in sight, I'm gonna expand on that.

Musk clearly references the 'heating' scaling as the 8th power of the entrance velocity, as this is the context.

Now the main heating of the spacecraft at entry comes from the shock luminosity that any object above local soundspeed will create in its front.
'Luminosity' here is the infrared heating from adiabatically compressed gas. Adiabatically compressed means its compressed so fast that it follows a certain thermodynamic relation, but essentially the gas will try to expand, and if it can't, it will heat up and radiate its additional temperature away.

The simplest model for this process are the 1D-Euler equations integrated over a Rankine-Hugoniot box. The result of the model are predictive equations for the gas quantities before $\rm Q_b$ and after the shock $\rm Q_a$.

We're interested here in mainly one of those equations, which is the one for the internal energy:

$$ \;\;\; e_b + \frac{1}{2} \, u^2_b + \frac{p_b }{\rho_b} = e_a + \frac{1}{2} \, u^2_a + \frac{p_a }{\rho_a}$$

Where we have the quantities $\rm e$ being the internal energy, $\rm u$ the flow velocity, $\rm p$ kinetic pressure and $\rm \rho$ the gas volume mass density. Now for the adiabatic case described before it is $\rm e = c_v \rho T$ and the pressure is found with the ideal gas EOS, giving us $\frac{p}{\rho} = \frac{k_B T}{\mu}$.
Now in the rest-frame of the spacecraft the gas is coming at it with orbital velocity, but pressure and internal energy can be neglected, compared to after the shock, where the gas is essentially at rest.

So we get $$ \;\;\; \rm \frac{1}{2} \, u^2_{orbital} = e_a + \frac{p_a }{\rho_a}$$

Now as explained, the r.h.s scales at proportional to T.

$$ \;\;\; \rm\, u^2_{orbital} \sim T$$

Next, as often done in those back-of-the-envelope calculations, one pretends that one actually has a more complex model than the Euler equations, and couples the kinetic gas Temperature to the photon radiation field, thus jumping into radiation hydrodynamics. This implies a Luminosity (as we astronomers say) or heating rate (as the engineers prefer) of $\rm L \sim T^4$ and we see trivially that it is

$$ \;\;\; \rm\, u^8_{orbital} \sim T^4 \sim L$$

This little exercise is probably done in all introductory classes for aerodynamics engineering, and that's what Musk remembered.
Also as mentioned, this is only the simplest model possible. Strong, additional energy dissipation sources, such as spontaneous chemical reactions and ionisation, change $\rm c_v$ and $\rm \mu$ and can therefore change the above scaling. At high entry speeds heating can be pretty extreme, then nonthermal radiative and non-equilibrium processes can even produce new radiative channels at cooling the gas and thus heating the spacecraft.

There are whole books about atmospheric re-entry and I find the whole physical complexity of the problem quite fascinating. I can recommend the a bit older book (1989) by J.D.Anderson 'Hypersonic and high-temperature gas dynamics', as reference.

Information candy:
The model discussed above also makes clear that 'friction' has not much to do with the heat generated at atmospheric entry, which I never tire of to point out to students. Purely frictional heat generation would have a different, weaker scaling.

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  • $\begingroup$ Very nice! So I should move my sentence: When Musk says "certain aspects" depend on the 8th power, he may be remembering an equation he once read in one of the rocketry books... to the beginning. Point being that real heating for real spacecraft does not really scale as $v^8$, but there exists an equation in a book... $\endgroup$ – uhoh Feb 14 '18 at 0:47
  • $\begingroup$ This feels right. I'm going to ponder it some more and await more comments and answers before deciding if i'll award the bounty. What is r.h.s, mentioned just after the 2nd equation? $\endgroup$ – kim holder Feb 14 '18 at 1:05
  • $\begingroup$ @kimholder: It means right-hand side (of the equation). $\endgroup$ – AtmosphericPrisonEscape Feb 14 '18 at 1:12
  • $\begingroup$ @uhoh: Well, simple models are always necessary starting points to understand a problem. You can't understand the full complexity of a solution, if you just throw the full model into a computer and close your eyes. And the simple properties of simple solutions are good enough as talking points. However I would be disappointed if Musk wouldn't know the even more extreme power-laws of gravitational destabilization. $\endgroup$ – AtmosphericPrisonEscape Feb 14 '18 at 1:14
  • $\begingroup$ Nice explanation! $\endgroup$ – Organic Marble Feb 14 '18 at 2:15
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tl;dr: While real radiative heating for real spacecraft does not really scale as v${}^8$ (it's way more complicated), when Musk says "certain aspects" depend on the 8th power, he may be remembering an equation he once read in one of the rocketry books he has borrowed and read over the years (originally near the bottom).


I'm going to venture a somewhat handwaving answer, which I normally try to discourage, but in this particular case, since the question refers to an off-hand comment (or as @AnthonyX nicely puts it "impromptu and unfiltered") by Musk who is likely on a mix of adrenaline, oxytocin, "...caffeine and a desire to help humanity colonize Mars' in the video after the successful FH launch, I'll wing it just this once.

Aerodynamic heating during reentry is a complex problem and while one can star with equations, it rapidly becomes a computational fluid dynamics problem, calibrated by comparison with an extensive set of measurements of physical processes at high velocity and/or temperature to get it right, or at least get it close.

Engineers may run many simulations at various sets of conditions (velocity, density) (cf. Table 3-8) in order to get a feeling for how heating rates depend on them. Then one can use that rate for initial, simple reentry trajectory tests. Later on, one would do full-blown CFD simulations of everything at once. It's sort-of like doing patched conics to get an initial trajectory, then a full n-body calculation when you want to refine.

Looking at a heating versus velocity plot over a certain, limited range of velocities, it may be easiest to express the slope as a power law. There may also be power laws in older analytical (formulaic) approximations as well. Power laws and exponentials are the two ways engineers often parameterize local variations because it's quick. For example, if something increases as the 8th power, then one knows if the velocity is increased by 10%, the heating increases by 1.1${}^8$ which, using a taylor series is factor of 1.8 or 2.1 using one or two terms:

$$ 1 + \frac{0.1*8}{1!} \approx 1.8, \ \ \ \ \ \ 1 + \frac{0.1*8}{1!} + \frac{0.1\times 8 \times7}{2!} \approx 2.08$$

So for complex problems, power laws are still used as quick ways to estimate in your head or at a blackboard how much something depends on something else. This is often done when arguing if something is important or not, a good idea, a bad idea, etc.

In these randomly selected NASA slides I found a plot of heating rate as a function of reentry velocity. I'm pulling it out of the context of a much more complex problem, but it serves the purpose.

To get the value for the power of a power law behavior, you use the ratio of the logs of the ratios:

If $y=x^p$, then

$$p = \frac{\log\left(\frac{y2}{y1}\right)} {\log\left(\frac{x2}{x1}\right)}. $$

After counting pixels, the power of the very steep, thiner red line is about 30, and the power of the upper, ending part of the thick red line at the top drops to about 5.3.

Radiative heating is a complex process, and organic molecules emitted from an ablative heat shield (if one is used) can go a very long way to reduce heat transport to the spacecraf spacecraft due to their very strong absorption of infrared light. But the plasma can get really, really hot, and so in addition to CFD one must do detailed modeling of the plasma as well. You can't just use the familliar $T^4$ power law alone when the properties of the plasma are also temperature dependent and there is a lot of self-absorption.

enter image description here


When Musk says "certain aspects" depend on the 8th power, he may be remembering an equation he once read in one of the rocketry books he has borrowed and read over the years:

enter image description here

See also Quora Did Elon Musk return the books he borrowed from Jim Cantrell? Hold your cursor over the spoiler alert box to reveal the answer. Hint - it's what you think.

No he never did

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  • $\begingroup$ I am in the odd situation of not knowing how to evaluate this answer, as it isn't my area at all. Some of it is useful but i don't know if it is the answer. $\endgroup$ – kim holder Feb 13 '18 at 23:37
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    $\begingroup$ @kimholder ya I understand. However there really may not be anything that truly does behave precisely as the 8th power of velocity in a realistic reentry situation. A premise of your question is that Musk was speaking precisely, and if you watch the video, that seems not to be the case. And even if someone find some equation somewhere that has a term to the 8th power, there's no way to know if it's that particular instance that he was thinking of at that moment. What I'm saying is similar to comment by AnthonyX below your question that I've linked to in the first sentence. $\endgroup$ – uhoh Feb 13 '18 at 23:41
  • $\begingroup$ I have been surprised by that interpretation. He concentrates on it for a bit, it isn't just in passing. He mentions how surprised he was by it. He has a degree in physics. $\endgroup$ – kim holder Feb 13 '18 at 23:43
  • $\begingroup$ @kimholder consider also that Musk tweeted a completely bogus orbit for the Roadster around the same time (Say it ain't so, Joe!). I'm pretty sure that he's just saying that heating has a strong dependence on temperature, and he's using the 8th power as engineering short-hand for "really, really strong". $\endgroup$ – uhoh Feb 13 '18 at 23:49
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    $\begingroup$ Almost no equation neatly fits what one observes in the real world. If it seems to, probably you just haven't looked closely enough, to enough decimal places. However, they get across the relationships between things. Reentry is one of the especially messy things. Still, the question was about an equation. $\endgroup$ – kim holder Feb 14 '18 at 4:08

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