6
$\begingroup$

Quite a bit has been written about railguns launching stuff to the earth. But has any work been done on using the moon as a pit stop between the earth and mars.

$\endgroup$
8
$\begingroup$

It would be a bad idea but on two technicalities. Railguns produce huge accelerations in a short distance, what's needed for a mobile weapons platform.

For a launcher what you want is a linear motor, not a railgun. Think of a maglev train, but without a speed limit.

It would be a major engineering project that would certainly have to be built almost entirely from lunar or asteroid resources. The track would be hundreds of miles long but you have the advantage that the forces on it are much more predictable than here on Earth.

The longer the track and the higher acceleration you will tolerate on the track the farther you can go. If you take this to its logical conclusion you get a doubled track wrapped around the lunar equator. You build a second track upside-down and overhead and the cars that ride on it are built to ride either track. For 5G on that track you can get an ejection velocity sufficient to take you anywhere from practically grazing the sun to an escape orbit. You still need to bring along some way to stop but your main boost is done.

Clayton brought up transfer windows. I see this as a non-issue. Yes, the launcher will only be pointed in the right direction once a month--but there's no deep space target with a launch window that comes that frequently anyway. It might not be a perfect launch window but such windows aren't a binary state, a sightly mistimed launch just costs you a bit more fuel--but your "fuel" is electricity, dirt cheap compared to rocket fuel you need to haul along.

Clayton also brought up the issue of accuracy. Yes, a perfect ejection isn't going to happen. You're not going to be able to fire artillery and hit Mars. Rocket launches suffer the same limit, the reality is midcourse corrections are needed.

LocalFluff (in a comment to Clayton's post) also addressed the issue of plane changes. Do them in a flyby, it's still a lot cheaper than the whole burn.

Finally, the second technicality: It's cheaper to go to Mars than to go to the Moon, even if you figure that your facility on the moon can pluck a craft out of a very low orbit and land it. (You have a car riding on a track, it shoots cables up as the spacecraft flies overhead. Get it right, they connect and the spacecraft is hauled down and decelerated. Miss and you try again next orbit.) The only reason to use a lunar facility to go to Mars would be if you wanted to do it when you didn't have a decent launch window.

As for the "it's cheaper to go to Mars than the moon" bit: Solar System ΔV map

Start at Earth (bottom left) and work your way up: 9.4 km/sec to reach low orbit Another 2.44 km/sec to reach geosynchronous orbit Another .68 km/sec to reach the moon

Now, from here there are two paths: Going to the moon: .14 km/sec for capture .68 km/sec for low orbit 1.73 km/sec for landing. Note that I was only counting the first two numbers in talking about capturing a vessel in orbit. In reality the cost would be higher as you're not going to get a capture in a 100 km orbit. 1 km periapsis would be more like it. I was simply counting .82 km/sec as the cost of the Moon, but if you have to do it with rockets all the way it's 2.55 km/sec.

Going to Mars: .09 km/sec to leave Earth's SOI .39 km/sec for a Martian transfer orbit.

Note that the total cost is .48 km/sec, well below the Moon.

What's that I hear from the peanut gallery about the .67 km/sec for capture at Mars? Note that red arrow on that line. You get that .67 km/sec from a dip in the Martian atmosphere, not from your engine. If your objective is landing you can go all the way down without burning, if you want to orbit Mars you will need a small burn to circularize once you have used the atmosphere all you can. Even with that circularization burn it's still going to be less than the cost of going to the moon.

One point--you would never actually move around like that diagram shows and many of those numbers aren't actually accurate for the shown maneuver. In practice you always burn as much as you can as close to as massive a body as possible. Thus if your objective is a low orbit about the moon you don't do the ~10 burns (5 arrows + 5 circularization burns) that chart shows, but rather: 1) Launch burn 2) Low orbit circularization burn. (Note: Some rockets are able to do one continuous burn for these two steps. This saves the need to use a restartable engine for this purpose.) 3) Lunar transfer burn. This will take you to the desired altitude above the moon. 4) Circularization burn.

In theory burns 1, 2 and 3 can be combined into a single burn but that gives you an extremely narrow launch window and doesn't actually gain you much of anything as you need to get above the atmosphere before piling on too much velocity anyway.

Likewise, you would not find that .39 km/sec would suffice to get you from Earth orbit to Mars, that number is assuming you do the burn in low Earth orbit.

If you really want to learn about orbital mechanics I strongly suggest the game Kerbal Space Program. While it definitely makes some simplifications the orbital mechanics are pretty accurate (although if you want real numbers you will have to download a mod that gives you our solar system rather than the game one. The stock game uses a more compact solar system to make things happen faster--transfer time to the most distant planet in the game is under 2 years.

$\endgroup$
  • $\begingroup$ Thanks for pointing out that "railgun" is the wrong term to use here. Indeed it should be a very long linear motor. I don't quite understand the phrase "It's cheaper to go to Mars than to go to the Moon..." can you elaborate? $\endgroup$ – uhoh Dec 28 '16 at 2:20
  • $\begingroup$ Nice graphic but the text size is too small for easy reading. The font size should be doubled and the names of the planets should be larger than the other text $\endgroup$ – Uwe Dec 28 '16 at 11:01
  • $\begingroup$ @Uwe Not my graphic, just what Google came up with. It's quite readable if you blow it up to full size. $\endgroup$ – Loren Pechtel Dec 28 '16 at 19:46
  • $\begingroup$ The math on this is quite a bit off. You're comparing surface-of-earth-to-the-moon cost to leo-to-mars cost. Leo-to-moon is far less than leo-to-mars. $\endgroup$ – Stephan Feb 8 '18 at 0:09
  • $\begingroup$ @Stephan No, I'm comparing leo-to-Moon to leo-to-Mars. Both Mars and Venus are closer in delta-v than the Moon because you can aerocapture while you have to do it with rockets at the moon. (A flyby of the Moon is cheaper than a flyby of the other planets.) $\endgroup$ – Loren Pechtel Feb 8 '18 at 3:01
6
$\begingroup$

Yes, but it would be extremely hard, you would need to accelerate it fast enough to escape the moon's gravity, Earth's gravity, then raise your orbit of the sun to achieve a Mars transfer.

But that's not all!

First, the railgun would need to be aimed perfectly: being one centimeter off on the moon will make you thousands of kilometers off at mars.

Second, you would need to be in a specific place on the moon. This would be a problem, since the moon's orbit takes 27 days, and since the moon is tidally locked, it may take longer than the transfer window to be in the correct position to fire the cannon.

Third, the amount of energy required. From low earth orbit, to a Mars transfer orbit, you need 4.3 kilometers/second of delta v. Back to the energy, the US navy's railgun requires 25 megawatts of energy. For comparison, the ISS's acre of solar panels produce 120 kilowatts at max. And the gun only accelerates a 2.5 kilogram projectile to 2.4 kilometers a second.

In conclusion, it's possible, but not feasible with modern technology. It would require too much energy, have so little times it could be used it would be almost useless, and the tiniest mistake could make it fail.

$\endgroup$
  • $\begingroup$ The points seem valid, but a little technical note: according to en.wikipedia.org/wiki/… moving from LEO to Mars transfer (3.8km/s) is more expensive than Moon surface to Mars transfer (2.9km/s). $\endgroup$ – Martin Modrák Dec 27 '16 at 10:33
  • 2
    $\begingroup$ Concerning the "need to be in a specific place on the moon", what about aiming the Lunar rail gun towards a close Earth flyby, to take advantage of an Earth gravity assist towards Mars. Wouldn't that geometrically offer a launch window to Mars every month? (The 2 degrees inclination of Mars' orbit, and the 5 degrees inclination of the Moon's orbit, could maybe be adjusted for by a modest reaction impulse at perigee to take advantage of the Oberth effect). $\endgroup$ – LocalFluff Dec 27 '16 at 13:21
  • 4
    $\begingroup$ Talking about the power draw of a railgun when it's fired is remarkably pointless. There are such things as batteries and capacitors, you know. What's actually relevant is not the power draw for the milliseconds of firing, but the total energy, and you can't measure energy in watts! $\endgroup$ – Nathan Tuggy Dec 27 '16 at 20:14
  • $\begingroup$ The possible acceleration is limited by the loads structural strength. A lot of extra weigth is necessary for the necessary stability, but all that extra weigth is useless when the load has left the railgun. $\endgroup$ – Uwe Dec 28 '16 at 11:09
0
$\begingroup$

yes but you would need to make course corrections on the way. using a maglev directed towards earth you could use the earths gravitational field to magnify any small changes in direction allowing you to shoot in almost any direction you want with only a course correction of 1 to 6 degrees. I could easily see us having little ports we land in and use regenerative breaking on the other side to slow the cargo down. we might also decide to ship things to planets with stronger gravitational pull to generate energy and ship the empty cargo ship back with surplus energy or lighter resources.

$\endgroup$
0
$\begingroup$

This answer only address the navigation aspects of this.

The arc-size of Mars at opposition from Earth is 22 arc-seconds. https://in-the-sky.org/news.php?id=20201013_12_100

The very best current technology can do at least 8 milli-arc-seconds. https://jwst-docs.stsci.edu/display/JTI/JWST+Pointing+Performance

There will be perturbation forces on any projectile from the moon, but assuming those are modeled reasonably well (drag is the most difficult to model which is not an issue for the moon) and accounted for in aiming it should be possible to pull this off with some mid-course corrections.

$\endgroup$
  • $\begingroup$ I don't think the JWST's no-thrust, high-altitude Lagrangian pseudo-orbit environment makes a good comparison for railgun accuracy. (There are also quite a lot of other factors involved, as some of the other answers have mentioned, and an answer that only addresses one aspect of the question is less than ideal.) $\endgroup$ – Nathan Tuggy Mar 13 at 1:46
  • $\begingroup$ I meant to put this as a comment under your point about pointing accuracy being a challenge. We absolutely have the technology for pointing accuracy for this sort of system as JWST can achieve an order of magnitude better. Less extreme control systems can also meet the requirements. $\endgroup$ – Knudsen Number Mar 13 at 1:54
  • $\begingroup$ JWST uses low-acceleration, ultra-low-vibration reaction wheels and thrusters integrated over long periods of time in a nano-gravity environment. None of that works on a railgun round being fired. That's fine, since we don't need those extra 3+ orders of magnitude anyway: mid-course corrections are good enough. $\endgroup$ – Nathan Tuggy Mar 13 at 2:48
  • $\begingroup$ Just figured out what you might have meant by "meant to put this as a comment". Are you talking about space.stackexchange.com/a/19578? Because if so, I didn't post that answer, and I downvoted it (several years ago). $\endgroup$ – Nathan Tuggy Mar 13 at 8:57
  • $\begingroup$ Yes, that is what I was referring to. $\endgroup$ – Knudsen Number Mar 15 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.