# Could the end cap of the Pascal B (1) survive its trip through the atmosphere?

In 1957, the US had a nuclear test where a shaft was dug 152 meters into the ground. A 100 mm thick, 900 kg steel plug was installed and welded at the top of the shaft. Under it was 2 feet of concrete.

The bomb was detonated created a blast of less than 1 kt. The concrete was vaporized and the steel plug shot into the air at approximately 66 km / second (125,000 - 150,000 mph).

Some believe it became the first man made item to reach orbit, after all, all you need is approximately 7 km/sec to reach orbit, and this greatly exceeded that.

Others believe it was vaporized by the intense pressure of traveling though the atmosphere at this speed.

No one ever reported it coming down anywhere...

Is there a physics / mathematical way to determine what happened to this steel plug?

• It certainly didn't go into a closed orbit around the Earth as escape speed is only about 11 km/s. – dmckee Jun 25 at 21:13
• would you mean then that if it did make it into space, it would be traveling more or less in a straight line? I guess I am looking to see if it could have made it through the atmosphere or would the pressure have destroyed it. – Rick Jun 25 at 21:19
• The Test Site is big, and there is lots of nothing all around. If it came back down it could easily not be found. – Jon Custer Jun 25 at 21:41
• @JonCuster - at the speeds it went up, it would not have come down in the test zone. But my question still stands, could it have survived a trip through the atmosphere at the speeds it went up? – Rick Jun 25 at 23:30
• Let’s take the inverse - which meteoroids make it to the surface of the earth? science.howstuffworks.com/question486.htm - looks like it may have made it! – Paul Young Jul 3 at 2:31

I used the formula for meteoroid mass loss rate from https://www.spaceacademy.net.au/watch/debris/metflite.htm : $$dm/dt = (\Lambda A \rho_a v^3 m^{2/3} ) / ( 2 \zeta \rho_m^{2/3} )$$.

See below the legend and the values I chose (SI units, compare with the values at the link):

$$dm/dt$$ - mass loss rate

$$\Lambda$$=2 - heat transfer coefficient

$$A$$=1 - meteoroid shape factor

$$\rho_a$$=1.25 - atmospheric density

$$v$$=66000 - meteoroid speed

$$m$$=900 - meteoroid mass

$$\zeta=3\cdot 10^6$$ - heat of ablation of the meteoroid

$$\rho_m$$=7800 - meteoroid density.

I obtained $$dm/dt\approx 2.8\cdot 10^7$$. Thus, the meteoroid will lose mass comparable to its initial mass in $$\frac{m}{dm/dt}\approx 3\cdot 10^{-5} s$$. The meteoroid will travel $$v\frac{m}{dm/dt}\approx 2 m$$ within this time. Thus, the end cap probably did not survive in the atmosphere.

The results of the calculation seem strange. It is possible that the formula in the link is incorrect, or maybe I made some mistake, so take this just as an initial attempt to make an estimate.

• The mass loss rate depends on m (and v), so you can't just divide m by it to get the lifetime. Also, $\rho_a$ varies significantly with height and so also t. – Ryan Thorngren Jul 3 at 6:42
• @RyanThorngren : I was just making an estimate, so I estimated the time for the meteoroid to lose mass comparable to its initial mass, I did not mean that the meteoroid would lose all its mass over this time. Same for the atmospheric density - it does not change significantly over 2 meters. At this point I don't think numerically solving the differential equation is warranted, as we don't really know the precise values of the parameters. – akhmeteli Jul 3 at 6:51
• It took me a second to figure out that $2m$ meant "two meters" and not "twice the mass". (This is why it's better to write units in upright font and variables in italics.) – Michael Seifert Jul 3 at 7:52
• @MichaelSeifert : Thank you for the advice. – akhmeteli Jul 3 at 8:27

Building off of akhmeteli's excellent answer, I implemented the differential equations from the asteroid webpage in Mathematica. I then tried to tweak the numbers, within realistic bounds, to get the thing into space. In no realistic case was I able to get the thing more than a few hundred meters up before it completely burned away.

To maximize the distance travelled, we want $$\Lambda$$ and $$A$$ to be as small as possible; respectively, these correspond to the rate at which heat is transferred to the "asteroid" and the effective cross-sectional area of the object (taking into account turbulence). In addition, we want the heat of ablation $$\zeta$$ (the amount of heat require to vaporize a certain mass of the substance) to be relatively high, since this will reduce the rate at which mass is lost.

The parameter $$\Gamma$$ also has an effect; it describes the amount of drag experienced with the atmosphere. Interestingly, one can actually get the projectile higher by increasing the drag: a higher drag means the projectile slows down faster, but that means that the projectile can slow down enough enough that it doesn't burn up immediately.

My optimistic estimates are $$\Lambda \approx 0.15$$ (note that this number is used in the code example on the page) and $$A = 1$$ (which would be more streamlined than a sphere). I also used $$\zeta = 10\times 10^6$$ J/kg, since it was the highest "typical" value in the table.1 Finally, I used $$\Gamma = 0.5$$, an estimate given on that webpage for the lower atmosphere.

Here's the result of the simulation, with the parameters given above. The vaporization of the plug is complete at a height of 312 meters.

And here is the simulation for akhmeteli's parameters, with $$\Gamma = 0.5$$. The plug does not significantly change its velocity before it burns up; the final height is a little over 6 meters. As would be expected, this is within an order of magnitude of akhmeteli's back-of-the-envelope estimate.

If you tweak the unknown parameters of my "optimistic" case above, you can attain a height of 1 km if:

• $$\Lambda \approx 0.064$$ (more than twice as small)
• $$A \approx 0.031$$ (more than three times smaller)
• $$\zeta \approx 23.5 \times 10^6$$ J/kg (over twice as large.)
• $$\Gamma \approx 1.6$$ (much more drag—this slows it down sufficiently before too much of it burns away)

All in all, it seems unlikely that the plug got anywhere near space.

### Mathematica Code:

Feel free to tweak this code as you see fit. The code stops integration when either the mass of the steel falls below 1 gram, or the speed falls below 1 m/s. The code does implement a height-dependent atmospheric density via a simple exponential model, though it turns out not to be all that relevant for realistic parameters. The acceleration due to gravity is assumed to be constant.

Needs["DifferentialEquationsInterpolatingFunctionAnatomy"];
Λ = 0.15;(*heat transfer *)
A = 1;(*shape factor *)
Γ = 0.5; (*drag coefficient*)
ρa0 = 1.25 ;(*atmo. density *)
v0 = 66000; (*initial velocity*)
m0 = 900 ;(*initial mass*)
ζ = 7*10^6;(*heat of ablation*)
ρm = 7800;(*steel density*)
h = 7000; (*atmospheric "height"*)
a = Γ A ρa0 / ρm^(2/3);
b = Λ A ρa0/(2 ζ ρm^(2/3));
soln = NDSolve[{x''[t] == - a Exp[-x[t]/h] x'[t]^2/m[t]^(1/3) - 9.8 m[t],
m'[t] == -b Exp[-x[t]/h] x'[t]^3 m[t]^(2/3), x[0] == 0,
x'[0] == v0, m[0] == m0,
WhenEvent[{m[t] < 0.001, x'[t] < 1}, "StopIntegration"]},
{x, m}, {t, 0, 1000}]
{ti, tf} = First[InterpolatingFunctionDomain[x /. First[soln]]]
Plot[x'[t] /. First[soln], {t, ti, tf}, PlotRange -> {0, 66000},
AxesLabel -> {"Time (s)", "Velocity (m/s)"}]
Plot[m[t] /. First[soln], {t, ti, tf}, PlotRange -> {0, 900},
AxesLabel -> {"Time (s)", "Mass (kg)"}]
x[tf] /. First[soln]


1 It is not clear to me whether these are the appropriate units for $$\zeta$$; the page is unclear. They're dimensionally correct, though.

• So a 900kg steel plug, traveling at 125,000 mph only made it 6 meters? Would the blast pressure wave from below not assisted the plug by distorting / overcoming the resistance of the air pressure impacting the plug from above? – Rick Jul 4 at 12:48

The drag equation stands that the an object travelling through the air receives a force of d*v^2*a*q, if the aerodynamic coefficient (q) was 1 and the air density 1.2 then the cap would have received a pressure of 4961 Megapascals, steel can't stand that pressure.

Also the energy dissipated in 0.0001 seconds given by the same equation would be 31 gigajoules and it is required 0.61 gigajoules to melt 900 kg of steel so the only way the cap will survive the first 6 meters is if the 98% of the energy was dissipated to the air, which is hard because of the black body radiation at about 150000 kelvin (1000 J/(kg*k) * 6^3 kg (air around the cap) /31e+9 J).

Black body heat transfer in 0.0001 s would be 2 GJ .

• Does the fact that the steel will be surrounded by a shock layer change the pressure it will experience? If so, how? – Michael Seifert Jul 4 at 10:20
• the drag equation is based on dynamic pressure, which works for incompressible fluids, for making it work in air, the dynamic pressure needs to be multiplied, it is observed that at high mach numbers the drag coefficient is more stable and approaches a number. researchgate.net/figure/… – Sartem Cacartem Jul 4 at 15:24