A SpaceLoop oompoc

October 25, 2020. In which I pretend to be Elon Musk, and launch rockets into space using empty cylinders. We’ll learn a little about atmospheric physics, rockets, and building giant cylinders, in order to sketch an order of magnitude proof-of-concept (oompoc).


  1. Introduction
    1. The basic design
  2. The physics
    1. The barometric equation
    2. Temperature and gravity*
    3. The size of the plunger
  3. Design considerations
    1. Vacuum vexation
    2. Stabilising the plunger
    3. Cylinder stress and construction
  4. Conclusion
1. Introduction

Elon Musk is the CEO and founder of SpaceX, a company which sends rockets into space. Along with engineers at Tesla and SpaceX, he also proposed a wacky, high-speed alternative to trains: the hyperloop, a pneumatic tube for people. In this post, I conduct an initial feasibility study for combining these two ideas, and launching rockets into space with vertical hyperloops. I call this the “SpaceLoop”, though I originally toyed with “hyperlauncher”.

Since this is merely a feasibility analysis, I will ignore most of the important engineering problems, and focus on the physics. And even though I am focusing on the physics, I will adopt a philosophy of order-of-magnitude (oom) estimates, which I call oomism. I state for the record that “oomism” and the “SpaceLoop” are very, very serious business.

Note: pictures to come.

1.1. The basic design

The basic idea of the SpaceLoop is simple. Take a giant cylinder, stretching from the ground to the top of the atmosphere, and put a “plunger” inside which can move up and down, just like a French press. If the cylinder and plunger are airtight, one can imagine evacuating the cylinder and placing the plunger at the bottom, just above the ground. One can then flood the the area beneath the plunger with air from outside the cylinder, at atmospheric pressure, which rushes in and pushes the plunger into the vacuum above it, from ground level to the top of the atmosphere, where the object, if it has acquired enough energy, is liberated from the earth’s gravitational pull. If you prefer household objects to technocrats, we can think of the SpaceLoop as a French press combined with a vacuum cleaner.

In the next section, we’ll discuss simple models of atmospheric pressure, and check that our scheme can, in principle, launch objects into space. This will constitute an oomist proof-of-concept, or oompoc. In the final section, we discuss some elementary issues of design and engineering.

2. The physics

Now that the high-concept pitch is out of the way, we have some physics to do. Can we launch anything? How light must it be? What physics are we neglecting? We’ll address all of these, though with a more or less oomist lens.

2.1. The barometric equation

In order to understand how much force is supplied to the plunger, we need to understand how pressure varies with height, and indeed, how high the atmosphere effectively is for our purposes. We will start with the simplest model: it is an ideal gas, in hydrostatic equilibrium, small enough that gravitational acceleration is constant. Let’s unpack these. An ideal gas satisfies the ideal gas law,

\[PV = Nk_BT \quad \Longrightarrow \quad P = n k_BT,\]

where $P$ is pressure, $V$ is volume, $k_B = 1.4 \times 10^{-23}$ in SI units, and $T$ is absolute temperature, e.g. in Kelvin. Finally, $N$ is particle number and $n = N/V$ is the number per unit volume

Hydrostatic equilibrium means that the mass of a small parcel of has is supported because pressure below is greater than pressure from above. If the parcel has height $h$ and area $A$, its mass $m$ is the volume $Ah$, multiplied by the density of particles $n$, multiplied by the mass per particle $M$. Using the ideal gas law,

\[mg = (n MAh)g = \frac{MPAhg}{k_BT}.\]

Consider an infinitesimally thin parcel, $h = dz$, where $z$ is the height above the ground. The difference in pressure above and below is $-dP$ (decreasing as $z$ increases), so that $mg = -A \, dP$ implies

\[\frac{dP}{P} = -\frac{Mg}{k_BT}\, dz.\]

This is a differential equation we can solve immediately by integrating both sides, and obtain the barometric equation:

\[P = P_0e^{-z/\lambda}, \quad \lambda = \frac{k_BT}{Mg},\]

where $P_0$ is the pressure at $z=0$, and $\lambda$ is called the scale height. So, in our oomist model, pressure decreases exponenentially, and is effectively zero after a few scale heights. Let’s calculate some actual numbers. The pressure at ground height is just atmospheric pressure,

\[P_0 = 1 \text{ atm} = 101.1 \text{ kPA}.\]

The scale depends on a number of things, including temperature, gravity, and the mass of air molecules. The last one is easiest. Air is mostly nitrogen gas $\text{N}_2$ ($28$ atomic mass units), made a bit heavier by oxygen gas $\text{O}_2$ ($32$ amu), leading to an average molecular mass

\[M = 29 \text{ amu} = 48 \times 10^{-26} \text{ kg}.\]

Neither temperature nor gravitational acceleration are really fixed as we go far away from the earth. But we’ll assume they vary slowly enough that the scale height can be computed keeping them constant, and see if this is reasonable. The average surface temperature of the earth is a cool room, $T = 15^\circ \text{ C} \approx 290 \text{ K}$, and gravitational acceleration at the surface is $g = 9.8 \text{ m/s}^2$. Putting it all together, we get a scale height

\[\lambda = \frac{k_BT}{Mg} = \frac{(1.4 \times 10^{-23})290}{(4.8 \times 10^{-26})(9.8)} \text{ m} = 8.4 \text{ km}.\]

This is pretty small! To get some sense of how quickly this drops, note that the pressure inside a vacuum cleaner is around $20 \text{ kPa}$. This occurs around height

\[z = \lambda \log \left(\frac{101.1}{20}\right) \approx 13 \text{ km}.\]

This seems like a reasonable place to cap off the atmosphere.

2.2. Temperature and gravity*

Perhaps, to be really responsible, we should drop our oomy prejudices for a moment and acknowledge that it gets colder and gravity weakens as you go up. How does that change things? Gravity is easiest. From Newton’s universal law of gravitation,

\[g(z) = \frac{GM_\oplus}{(R_\oplus + z)^2},\]

where $R_\oplus = 6.3 \times 10^6 \text{ m}$ is the radius of the earth, $M_\oplus = 6.0 \times 10^{24} \text{ kg}$ is its mass, and $G = 6.67 \times 10^{-11}$ is Newton’s constant, in SI units. This leads to an equation

\[\frac{dP}{P} = -\frac{M GM_\oplus}{k_BT}\, \frac{dz}{(R_\oplus + z)^2}.\]

Once again, we just integrate to get

\[P = P_0 e^{-\lambda} \exp\left[\frac{M GM_\oplus}{k_BT (R_\oplus + z)}\right].\]

To see if we really need to worry about this, we do a binomial expansion for $z \ll R_\oplus$ (which is all we will ever worry about)

\[\frac{1}{R_\oplus + z} \approx \frac{1}{R_\oplus}\left(1 - \frac{z}{R_\oplus}\right).\]

Plugging this in, we recover our original barometric formula. So, as long as we stick to heights much smaller than the earth’s radius, we don’t need to worry about changing gravity.

What about temperature? This is tricky, since we need some thermodynamics, and in particular, a further equilibrium condition to obtain a relation between $T$ and $P$. Let’s assume it isn’t raining, and the atmosphere is dry enough for equilibrium to be set by an adiabatic process, where $dQ = T\, dS = 0$. Then we can relate $T$ and $P$ using the $T\, dS$ equation for variables $T$ and $P$, using the ideal gas law:

\[0 = C_P \, dT - T \frac{\partial V}{\partial T}\bigg|_P \, dP = C_P \, dT - V \, dP,\]

where we used the ideal gas law, and $C_P$ is the heat capacity for a volume $V$ of air at constant pressure. Note that $C_P$ can be written in terms of the degrees of freedom $f$ of the gas as

\[C_P = \frac{\gamma N k_B}{\gamma - 1}, \quad \gamma = 1 + \frac{2}{f}.\]

Plugging this into the result for hydrostatic equilibrium, we find that

\[dT = \frac{V}{C_P} \, dP = -\frac{Mg(\gamma-1)}{k_B\gamma }\, dz \quad \Longrightarrow \quad T = T_0 -\left[\frac{Mg(\gamma-1)}{k_B\gamma }\right] z.\]

For nitrogen or oxygen gas at atmospheric temperatures, $f = 5$, consisting of three translational and two rotational degrees of freedom. There is a vibrational mode at higher temperatures we can ignore. Plugging these in, we get a rate of change of temperature, or lapse rate

\[L = \frac{Mg(\gamma-1)}{k_B\gamma} = \frac{(4.8 \times 10^{-26}) 9.8 \cdot 0.4}{(1.4 \times 10^{-23}) 1.4} \text{ K/m} \approx 9.6 \text{ K/km}.\]

It drops roughly $10^\circ \text{ C}$ for every kilometre we go up. Finally, if we substitute this result for temperature back into our differential equation for hydrostatic equilibrium, we find

\[\frac{dP}{P} = -\frac{Mg}{k_B(T_0 - L z)}\, dz.\]

Integrating both sides gives a modified barometric equation, obeying a power law rather than an exponential falloff:

\[P = P_0\left(1 - \frac{Lz}{T_0}\right)^{\Lambda}, \quad \Lambda = \frac{Mg}{k_BL}.\]

Does this change the “vacuum cleaner” height of the atmosphere? Not by much. It now hits $20 \text{ kPa}$ a bit earlier at $z = 11 \text{ km}$. The average height of the atmosphere is $12 \text{ km}$, so both estimates are reasonable.

2.3. The size of the plunger

Let’s see how much energy can be delivered to our giant plunger when we flood the base with air at atmospheric pressure. If $A$ is the area of the plunger, then as we travel up a height $H$, the total energy delivered is just the work done by pressure:

\[E = A\int_0^H dz \, P(z).\]

The result will depend on which version of the barometric formula we use. For the constant temperature case, the energy is

\[E_1(H) = A\int_0^H dz \, P_0 e^{-z/\lambda} = AP_0 \lambda \left(1 - e^{-H/\lambda}\right).\]

For a lapse rate $L$, we instead have

\[E_2(H) = A\int_0^H dz \, P_0\left(1 - \frac{Lz}{T_0}\right)^{\Lambda} = \frac{AP_0T_0}{L(\Lambda+1)}\left[1 - \left(1 - \frac{LH}{T_0}\right)^{\Lambda+1}\right].\]

Both have some maximum energy they can provide per unit area of plunger. In the first case, take $H \gg \lambda$, we obtain

\[\epsilon_1 = \frac{E_1}{A} = P_0 \lambda = 8.5 \times 10^{8} \text{ J/m}^2.\]

In the second case, our formula stops making sense when the temperature hits zero, at $H = T_0/L$, so we have

\[\epsilon_2 = \frac{E_2}{A} = \frac{P_0T_0}{L(\Lambda+1)} = 6.8 \times 10^{10} \text{ J/m}^2.\]

We will use the latter estimate because it gives us more energy! Ahem, I mean because it is more realistic. A rocket of mass $m$ at the earth’s surface has gravitational potential energy

\[U = -\frac{GM_\oplus m}{R_\oplus} = -mg R_\oplus.\]

If we want to launch this into space, we need to provide at least this much energy in order for it to escape the earth’s gravitational field. The minimum plunger size is then

\[A = \frac{|U|}{\epsilon_2} = \frac{mg R_\oplus L(\Lambda+1)}{P_0T_0}.\]

A typical rocket mass is on the order of $m = 10^6 \text{ kg}$, so the minimum area of plunger needed is

\[A = \frac{10^6 \cdot 9.8 \cdot (6.3 \times 10^6)}{6.8 \times 10^{10}} \text{ m}^2 = 900 \text{ m}^2.\]

This is only slightly larger than the area of the Falcon Heavy if laid flat! If we want to launch a $m = 1\text{ kg}$ cubesat instead, our platform can be a million times smaller, so the $100 \text{ cm}^2$ base of the cubesat itself is more than sufficient.

Of course, we are ignoring the mass of the platform. Assuming it has some areal density $\rho$, then we must instead have

\[A \geq \frac{m g R_\oplus}{P_0 \lambda - \rho g R_\oplus}.\]

This means that if $P_0\lambda > \rho g R_\oplus$, i.e. the plunger can launch itself, it can launch anything else, provided the object is light enough or the plunger big enough. This completes our oompoc that a French press cross vacuum cleaner can launch rockets.

3. Design considerations

In this last section, we’ll address a few basic design concerns, such as establishing and maintaining the vacuum, the firing mechanism, and construction.

3.1. Vacuum vexation

A simple question is: realistically, how big does the tube need to be? If it is only $12 \text{ km}$, there is still a pressure of $20 \text{ kPa}$ or so, not to mention mass. Even if we evacuate the cylinder, air will immediately flood back in at the top. But how quickly? Consider the slice of air of height $h$ just above the cylinder, at height $H$. We would like $h$ much smaller than the rate of change of $P$, so it is effectively constant. This slice mass and is subject to a pressure $P$ from above, so that it is subject to a combined pressure

\[P_\text{top} = P + nM gh.\]

A unit area part of this slice will fall (initially) according to

\[z(t) = H - \left(\frac{P + nM gh}{2nM h}\right) t^2.\]

The timescale for appreciable flooding of the top of the cylinder is thus

\[\tau_\text{flood} \sim \sqrt{\frac{2H}{P/nMh + g}}.\]

We can make this timescale large enough for practical purposes by making $P$ small. For analytic simplicity, let’s work with the constant temperature barometric formula. Consider $h \ll \lambda$, so that (after a little algebra)

\[\frac{P}{nMh} \gg g.\]

Then using the ideal gas law,

\[\tau_\text{flood} \sim \sqrt{\frac{2nhMH}{P}} = \sqrt{\frac{2hMH}{k_BT}}.\]

Let’s express our heights in terms of scale heights, $x = h/\lambda$ and $X = H/\lambda$. Then our earlier numbers give

\[\tau_\text{flood} \approx 0.04 \sqrt{Xx} \text{ s}.\]

The flooding timescale depends on the geometric mean of the cylinder height $H$ and parcel height $h$, or $\sqrt{Xx}$ in dimensionless terms. If we want a window of, say, $5$ minutes, we will need an absurd number of scale heights.

Although this flooding timescale increases when we take temperature lapse into account, it is still much too short at any reasonable height. It seems we must seal the top of the cylinder when we evacuate it, and open it again during launch. It will need to open quickly so that the payload arrives well before the flooding timescale; I imagine a sort of shutter mechanism, which can be progressively opened as the payload comes through. So now our French press-vacuum cleaner is combined with a camera!

3.2. Stabilising the plunger

I’ve been discussing “cylinders” this whole time, hedging on the shape of the plunger. It seems that, for reasons a symmetry, an old-fashioned circular cross-section should work best. If the platform had corners, for instance, these might form points of preferential vacuum leakage, while a circular platform means we have one uniform design problem for the edge. But the circle has additional advantages, as we will see in a moment.

If an instability develops, the plunger can tilt, or even flip, which would obviously be disastrous for the payload. There needs to be some way to counteract this. A simple mechanical constraint is a thick plunger which does not have the room to flip within the cylinder. This increases the mass, and obviously, we don’t want the plunger scraping the sides since friction will reduce launch energies. We want a smooth ride, with no tilt and minimal friction.

One entertaining possibility is to stabilise the plunger by spinning it, just like a frisbee. The payload need not spin, if placed on a gyroscopically isolated platform above the base of the plunger. The energy for spinning can be obtained during launch by deploying an array of angled flanges beneath the plunger, and any excess energy not used for rotation can be stored in flywheels to help depress the plunger on future runs. Regarding the rotation, the stabilised plunger could spin by means of bearings around its perimeter, with rolling friction instead of scraping. I leave the calculation of the energy dissipation from bearings to a longer paper expected April next year.

A final question is how the vacuum is maintained around the edges of the plunger. Another cute possibility is to use the vacuum against itself, adding a “collar” at the edge, a membrane which seals the “pocket” around the perimeter with the rolling bearings. The vacuum will press the collar against the walls of the cylinder and keep the pocket sealed! Depending on the interface between the collar and the walls, large amounts of sliding friction may be generated; this is another engineering problem I leave to future work. But in case air does enter the pocket, one could also equip the base of the plunger with a means of ejecting it. Again, we could use excess launch energy to power small vacuum pumps which, connected to the pocket, can flush out any stray air.

3.3. Cylinder stress and construction

Finally, we come to the elephant in the oom: the construction of the cylinder. We are envisioning a structure on the order of a scale height, more than $10$ times taller than the tallest building ever constructed. It must not only stand up, i.e. support its own weight, but resist buckling under the massive pressure of air outside it. Again, a circular design leads to a uniform distribution of this pressure. In line with oomism, we will ignore inhomogeneities, and think about how a perfect cylinder is compressed by atmospheric pressure, called cylinder stress. This is the source of many an engineering disaster, though usually the pressure is on the inside, e.g. the great Molasses flood of 1919.

Suppose the cylinder has radius $R$ and thickness $d \ll R$. For thin walls, the Young-Laplace equation applies, and the cylinder is compressed by a stress

\[\sigma = \frac{RP}{d}.\]

Here, the surface tension is $\sigma d$ (force per unit length), but the tensile stress is $\sigma$ (for per unit area). We now treat the cylinder as a bent sheet, subject to a stress $\sigma$. If the material has Young’s modulus $\mathcal{E}$, the strain $\epsilon$ or fractional change in length is

\[\epsilon = \frac{\Delta R}{R} = \frac{\sigma}{\mathcal{E}} = \frac{RP}{d \mathcal{E}}.\]

Thus, the change in radius is

\[\Delta R = \frac{R^2P}{d\mathcal{E}}.\]

This relationship only holds up to the yield point, a material-dependent value of the stress at which the response to further stress becomes nonlinear. Our goal now will be to see what materials we could use and how thick they will need to be, noting that the strain at ground level is

\[\epsilon = \frac{RP_0}{d \mathcal{E}}.\]

Let’s use our earlier calculation of the area required to launch the Falcon Heavy. If rockets don’t need to carry their own fuel, they will be lighter, but we neglected the mass of the plunger. Assuming these two terms cancel, the radius needed is

\[R = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{900 \text{ m}^2}{\pi}} = 17 \text{ m}.\]

For concreteness, let’s use steel, which can sustain a stress of around $\sigma_\text{YP} = 500 \text{ MPa}$, with a Young’s modulus of around $\mathcal{E} = 200 \text{ GPa}$. Then at the yield point, we have a strain

\[\epsilon = \frac{500 \times 10^6}{200 \times 10^9} = 2.5\times 10^{-3}\]

This corresponds to a thickness of at least

\[d = \frac{RP_0}{\epsilon\mathcal{E}} = \frac{17(101.1 \times 10^3)}{2.5\times 10^{-3}(200 \times 10^9)} \text{ m} \approx 3 \text{ mm}.\]

So, steel $3 \text{ mm}$ thick should resist buckling, and contract around $\epsilon 3 \text{ mm} = 8$ microns in radius. The pressures drop as we ascend, so this thickness can be reduced as we go up, and the radius slowly varied to ensure close to uniform thickness (though the Young’s modulus may be high enough to ignore this). For a cylinder of height $H = X\lambda$, we estimate the volume of steel require to resist buckling is at most

\[V = 2\pi R d H = 2700 X \text{ m}^3.\]

A cubic meter of steel costs around $\text{USD}4000$, so the raw materials for a SpaceLoop a couple of scale heights high will cost a mere $\text{USD}22$ million dollars, ignoring savings due to tapering thickness. This sounds very promising, but of course, the real problem is weight. Steel has a density of $\rho = 8 \text{ tonnes/m}^3$. The resulting strain at the base is

\[\sigma_\text{mass} = \frac{V\rho}{2\pi R t} = \rho g H.\]

Thus, at the thickness needed to resist buckling, the maximum height (in scale heights) is

\[X = \frac{\sigma_\text{YP}}{\lambda\rho g} = 0.75.\]

Before our steel cylinder reaches a scale height, it will collapse under its own weight!

Thus, we see that in order to fully exploit the power of atmospheric pressure for launching rockets, we have another engineering problem to solve: finding a suitably strong but light material to replace steel. A related problem is wind engineering, i.e. ensuring the structure will not be knocked over by the wind, a major design consideration for tall buildings. It may be that the structural reinforcements which solve this problem also help distribute the cylinder’s weight. But I’ll leave both issues for another time.

4. Conclusion

What started a couple of days ago as a light-hearted tribute to the pioneering spirit of Elon Musk has become, by this point, a circus of high-concept oomism. I hope you’ve enjoyed it; I certainly have. Please leave a comment or drop me an email if you spot any errors, or have suggestions for improving the design. Perhaps one day we will leave the world of kerosene and boosters behind us, and embark on a bold new age of rockets on frisbees sucked by vacuum cleaners up a French press and spat out through a camera shutter: a SpaceLoop to the stars. Then again, maybe not.

Written on October 25, 2020
Physics   Engineering