Built on Facts

Bacon in the Asteroid Belt

A reader in the thread on Snow Crash came up with an interesting exercise. In the book, Hiro Protagonist has a pretty awesome car. He’d better – he uses it for time-critical deliveries for the Mafia.

The Deliverator’s car has enough potential energy packed into its batteries to fire a pound of bacon into the Asteroid Belt.

A metaphor, like the decibel thing we discussed earlier? Maybe, maybe not. Let’s work it out and see.

The first thing to notice is that there’s several levels of complication involved depending on how much detail we want. Roughly speaking there’s two sources of gravity to overcome. There’s the earth and the sun. The bacon doesn’t have to go far to get away from the earth, but the force the earth exerts is pretty steep during that distance. The force of the sun is much smaller, but it’s not going to diminish all that much between the earth’s orbit and the asteroid belt. We could stop there by calculating the changes in potential energy associated with the translation up the gravity wells and we’d have a figure.

And it might even be mostly right, depending on how we interpret things. In reality you’re more than likely not just launching the bacon from a standstill radially away from the sun. The bacon is in orbit already because the bacon is on the earth and the earth is orbiting the sun. When it gets to the asteroid belt, it will still be orbiting though at a different orbital speed. So to calculate the total energy change you have to take into account the orbital kinetic energy.

That’s not all. The earth is itself rotating, and thus you can consider the bacon as starting off with that much speed. This is why the Space Shuttle is launched from Florida: the rotation of the earth is fastest near the equator and so NASA takes advantage of this free kinetic energy. I doubt an unmodified Shuttle could reach orbit from Anchorage, Alaska even if NASA wanted to.

That latter complication involving the rotation of the earth is dependent on both the latitude of the launch location and the direction of the orbit. As such I think we’ll ignore it. It’s a fairly good average assumption anyway, since some launch directions will help you out and some will make the launch more difficult (this is why no one launches their rockets westward – the earth is rotating in the other direction).

Now I’m also going to ignore the first complication, the orbital kinetic energy. This is because we’re not requiring the final orbit to be a circular orbit or any other particular kind of orbit. We don’t care what the bacon is doing when it gets to the asteroid belt. We only want it to get there, and if we add in energy equivalent to the change in potential, we can do it.

So what’s the equation for change in gravitational potential energy? It’s this:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

I’ve finagled an overall minus sign so we can keep everything in positive territory. G is the universal gravitational constant, M is the mass of the earth (or sun when we get to that step), and m is the mass of the bacon.

For finding the potential energy due to the earth, we can drop the second term entirely. The distance to the asteroid belt will be so large compared to the radius of the earth that it might as well be zero. All told we end up with 26.4 MJ needed to get the bacon off the earth. Now repeat the process, replacing M with the mass of the sun, the initial r with the orbital distance of the earth, and the final r with a representative value for the orbital distance of the asteroid belt (say, 2.5 AU)*. I come up with about 241 MJ.

This gives us a total energy of 267 MJ. The energy content of the combustion of gasoline gives about 32 MJ per liter. All total, the energy we’re talking about is equivalent to… about seven liters of gasoline. Not so much. My car has enough energy in its tank to launch a pound of bacon into the asteroid belt, and my car is a little Nissan crackerbox. It’s entirely plausible for Hiro’s car batteries to have that much energy with no sweat.

Now in practice you can’t send pork products on interplanetary voyages with 2 gallons of gas. Energy efficiency is bad enough in a car engine. Energy efficiency in rockets is horrible. But strictly as a matter of energy with no reference to efficiency, the bacon delivery is not actually all that impressive.

*Though this happens to correspond to a thin gap in the belt due to a Kirkwood resonance.

Comments

  1. #1 steve s
    June 17, 2009

    Energy efficiency is bad enough in a car engine. Energy efficiency in rockets is horrible. But strictly as a matter of energy with no reference to efficiency, the bacon delivery is not actually all that impressive

    Ah but look again at what Stephenson said.

    The Deliverator’s car has enough potential energy packed into its batteries to fire a pound of bacon into the Asteroid Belt.

    he did not say

    The Deliverator’s car has enough potential energy packed into its batteries to fire a pound of bacon into the Asteroid Belt, assuming a total BS scenario where you ignore poor efficiency and air resistance etc like Matt Springer does.

    The Deliverator’s batteries have enough power to launch the actual bacon into the actual asteroid belt. So the energy is impressive.

  2. #2 Eric Lund
    June 17, 2009

    I doubt an unmodified Shuttle could reach orbit from Anchorage, Alaska even if NASA wanted to.

    Yes, it could. NASA has launched the Shuttle into polar orbit from Vandenberg AFB (which they use because there is nothing but a handful of small, mostly uninhabited islands south of there until you reach Antarctica). Rotational speed is irrelevant, if not counterproductive, for achieving polar orbit. They can’t get it as high as they can launching from Cape Canaveral into a prograde orbit, and they have to be stricter with payload weight, but it can be done.

  3. #3 Matt Springer
    June 17, 2009

    Steve, I’m not sure that makes sense. The car can’t launch anything, much less with the specific method of a rocket through an atmosphere. The Stephenson quote leaves a method entirely unspecified, and so a purely theoretical approach is not excluded. Pigeonholing the phrase into a description of rocketry specifically is not so much as hinted by the text. There are, after all, theoretical methods of achieving space flight which are very efficient and don’t involve appreciable air resistance – the space elevator being the most obvious.

    But sure, if the energy is going to be expended at the efficiency of current rocket technology it would in fact take a lot more energy.

    Eric: thank you, I did not know that. I stand corrected!

  4. #4 Uncle Al
    June 17, 2009

    Nissan crackerbox “Nissan rice burner” The politically correct phrase would be “pound of eggplant,” thereby not offending Semites (kashrut, halal), PETA, vegans, Canukistanis (unpalatably perverse bacon), Italians (demanding prosciutto di San Daniele or culatello di Zibello), and the Eggplant Active Media Workers’ Collective (eggplant.coop/services/hosting/tos).

    If NASA had any brains (already sci-fi) it would launch from the planarized summit of Mt. Chimborazo, 6384.4 km from the Earth’s center, 6.2682 km altitude, and 01°28′09″S latitude. The difference in energy to orbit versus Cape Canaveral (6375.5 km from Earth’s center, 0.003 km altitude, and 28°27′20″N) is almost two Space Scuttle SSB’s outputs for a fully laden and fueled Space Scuttle, plus air resistance difference. This derives from increased altitude above the geoid (gravitational potential energy, mgh) and increased net throw velocity (mv^2) vs. the fixed stars (given the local point’s circumference at its radius from Earth’s center, 23.934470 hours sideral day, then net (speed)[cosine(latitude]).

    Slightly lessened gee at the summit of Mt. Chimborazo for altitude is offset by the extra rock beneath.

  5. #5 CS
    June 17, 2009

    Slight correction to Eric Lund’s comment:

    NASA planned to launch into polar orbits from Vandenburg, but the program was scrapped after Challenger. They never actually launched from there (but the shuttle was capable of doing so).

  6. #6 abb3w
    June 17, 2009

    Given the tech level of Snow Crash, wrapping the pound of bacon in foil, dropping the package into a large railgun, and hooking the railgun up to the car batteries would appear the most Reason-able method. =)

    Random poking at the internet suggests railguns run about 50% efficient in conversion of electricity to kinetic energy. So, 534 MJ is probably realistic. Still only about 14 gallons of gasoline. On the other hand, this IS a battery we’re talking about. That’s on the order of 2000 times as much as your typical laptop battery holds, and on the order of 100 times the battery in a Prius.

  7. #7 Left_Wing_Fox
    June 17, 2009

    The Deliverator’s car has enough potential energy packed into its batteries to fire a pound of bacon into the Asteroid Belt.

    A wonderfully evocative metaphor, even if the sentence reads as if it was scribed by StrongBad.

  8. #8 Tim Gaede
    June 18, 2009

    Hold on a second there, Matt. To move from a circular orbit of 1AU to an elliptical orbit of 1AU perihelion and 2.5AU aphelion (eccentricity 0.42857), you would need to boost your speed by 19.52%, or 5820 m/s. That means 7.7 MJ for a pound of bacon. So you would need only 34.1 MJ total. Of course, the bacon traded away some kinetic energy to gain that potential energy. The bacon would be moving slowly at its aphelion and begin its decent toward the sun unless…it smacks into an asteroid (although it should do so at a high speed).

  9. #9 Tim Gaede
    June 18, 2009

    To clarify things just a little bit, the energy required may be a lot less than you may initially expect because the bacon “borrows” momentum from the Earth. Furthermore, it trades its kinetic energy for potential energy.

    Carefully taking into account the rotation of the Earth, I see we can get to 2.5 AU from the sun with just 33.42 Megajoules. One pound (0.4536 kg) of bacon with 33.42 Megajoules of kinetic energy would be moving at 12139 meters per second. Since the Earth rotates at 464 m/s at the equator a horizontal launch would be 12603 m/s with respect to (w.r.t.) the center of the earth. The escape speed of the Earth from the equator is 11180 m/s so the bacon’s speed w.r.t. the planet will decrease asymptotically to (12603^2 – 11180^2)^0.5 = 5817 m/s (NOT 1423 m/s as the bacon is essentially given limited time to slow down and momentum is “borrowed” from the Earth), which translates into a 19.53% speed boost w.r.t. the sun. The eccentricity of the orbit would then be (1.1953)^2 -1 = 0.4287, which means an aphelion to perihelion ratio of 1.4287 / (1-0.4287) = 2.50.

  10. #10 Greg
    June 19, 2009

    It’s very cool that you wrote something up on this. In my amateur way, here are a few further thoughts:

    Given that a one pound sphere (!) of bacon would have a radius of about 10 centimeters, and the velocity calculated in comment #9 is about 12.1 km/s, such that the bacon must be moving at at least that speed, how might one calculate a rough estimate for the drag force that must be counteracted to get the bacon out of Earth’s atmosphere? I tried plugging some numbers into some equations, but this only gives me the drag force at an instant (which is on the order of 14MJ*), which will be acting to reduce the velocity at the next instant, etc. If I knew enough calculus, I could guess how to construct an integral that would help, but I don’t, especially given both the change in velocity and the change in the density of the air as the fiery ball of bacon hurtles towards freedom.

    Any (further) thoughts?

    * (If the bacon is a sphere, its drag coefficient is .47, which is supposedly equal to the quotient of the drag force, and 1/2 rho v^2 A, where rho is the density of air (at STP, 1.3 kg/m^3), v is the speed in m/s (12,600), and A the area 0.314 m^2]).

  11. #11 Greg
    June 19, 2009

    Let’s say a rough linear estimate for the density of the air is that it falls off continuously from 1.3 kg/m^2 to 0 in 150km.

    Also, I’m off by a factor of 10 in the area calculation. :(

    I’ll just go crawl under my humanities rock again.

  12. #12 Tim Gaede
    June 19, 2009

    Well Greg, I think calculus by hand would be insufficient here. I ran a quick stepwise computer simulation that assumed air resistance to be proportional the density of the air multiplied by the square of the speed of the object. This is probably taking drag estimates to an extreme because turbulence, compression, and thermal effects would surely factor in. I assumed a streamlined massless container with a drag coefficient of 0.04 and a radius of 2 centimeters. I also assumed air density decays exponentially by a factor of e (2.71828…) every 7.5 km.

    Launching straight upward with a speed of 21635 m/s did the trick. That’s 106 MJ of kinetic energy.

  13. #13 Greg
    June 19, 2009

    Thanks for looking at the problem, Tim. If I’ve learned a couple of things in this, it’s all good. One thing to learn about is the kind of problem that might require approximative (numerical) solutions rather than symbolic ones. One might not know about these if one only does assigned homework problems. Second, I learned that the straight-line graph I looked at for air density as a function of altitude was in fact logarithmic, as anyone who looks at these things enough might well expect.

    Thanks to you and to Matt for considering this silly, but fun question.

The site is currently under maintenance and will be back shortly. New comments have been disabled during this time, please check back soon.