6
$\begingroup$

Space guns have a lot of confounding factors for getting to LEO. The projectile must be a rocket capable of large delta v in order to circularize the orbit. Then, giving more sideways velocity requires a more shallow angle of launch, which makes the (already troublesome) drag even worse. Both of these problems would be mitigated if the destination is switched from LEO to somewhere else, and the gun shoots almost exactly vertical. If you could do this, sending regular supplies to somewhere in Earth-Moon space would have advantages.

As a qualifier, this requires larger velocities than LEO, which couldn't be delivered with a light gas gun. These unprecedented speeds would be very difficult to achieve, but it's still possible in theory with a rail gun or Gauss gun, for instance. Actually, I have something vastly more radical in mind. Agnostic to the gun type, the problem of a low Delta V path to somewhere that won't crash into the Earth is an interesting question.

Let's say, for instance, that we shoot vertically upward from Earth aiming for EML-1. I realized from an orbital simulator that this is the time-reverse of just falling from that point. Here, a test particle starts at what would be EML-1, but is stationary in the Earth-moon CM reference frame. So the test particle is moving $0.86 \text{ km}/\text{s}$ relative to EML-1 reference frame.

simple fall

But I've neglected the rotation of the Earth itself. The equator itself has a velocity of $0.46 \text{ km}/\text{s}$, which will affect the mechanics of approaching EML-1, if launched straight up from the ground. Even a perfectly vertical shot would be following a highly elliptical orbit (although could crash into Earth). I think this would reduce the Delta V to EML-1, but the math isn't obvious.

Question:

  • With a single shot, would it be possible to achieve the EML-1 or 2 position & velocity without using any on-board propellant?
  • If not, what would be the minimum Delta V needed to reach L1,2,3,4,5, or any staging area that is not a chaotic orbit, and will not crash into the Earth or Moon?

In a childish sense of how hills work, I can roll a ball up to the top, and it could balance there, given a perfect shot. This might not be quite as simple, but I'm not sure.

$\endgroup$
  • $\begingroup$ An obligatory flashback to Jules Verne's Columbiad. I'm afraid that you have to add another degree of freedom to the problem, though (inclination of the shaft). $\endgroup$ – Deer Hunter Feb 20 '14 at 23:02
  • $\begingroup$ You still have the problem that unless your projectile is really big you won't get it through the atmosphere, period, no matter what it's launch velocity. $\endgroup$ – Loren Pechtel Feb 22 '14 at 1:37
  • $\begingroup$ @LorenPechtel And the nuclear option would only be used for payloads in the 10s of tons at minimum. I can't even guess what the upper limit would be, if there even is one it would be due to geology. The limiting forces and heating come in the accelerating phase. SHARP and Quicklaunch wanted to send tiny things to orbital velocity. You could easily launch something 100x those masses at sqrt(2) times the velocity, particularly with the lower mass-thickness from angle and altitude. $\endgroup$ – AlanSE Feb 22 '14 at 4:04
  • $\begingroup$ Logically, your upper bound for this problem would be the 11.2 km/sec Earth escape velocity. It also seems that EML-2 is going to be slightly easier to reach than EML-1 in terms of ΔV budget. $\endgroup$ – Jerard Puckett Feb 22 '14 at 16:30
  • $\begingroup$ @JerardPuckett But why? If we only count the Delta V in order to match the tangential velocity, then the EML-2 radius is obviously larger than EML-1, and the angular velocity is the same for both. So if we're just going from stationary to matching the velocity, EML-1 wins. But then there's a gravitational assist effect... $\endgroup$ – AlanSE Feb 22 '14 at 19:00
3
$\begingroup$

I think there's a simplistic way to constrain this problem. Just apply simple conservation of angular momentum. Take the moon out of the picture entirely and just focus on the rise from the surface of Earth to the L1 distance.

With the center of the Earth as the origin, we easily identify a tangential and radial component of the velocity. The tangential component is the only one that counts toward angular velocity. The angular momentum balance is simply:

$$ r_1 v_{t1} = r_2 v_{t2} \\ v_{t2} = \frac{ r_1 v_{t1} } { r_2} $$

These values are all specified by the nature of the problem. Roughly, we find:

$$ v_{t2} \approx \frac{ (6,300 \text{ km}) ( 0.46 \frac{ \text{ km} }{s}) } {326,000 \text{ km} } \approx 5 \frac{m}{s} $$

Basically, any angular momentum you start with will be virtually irrelevant for the purposes for circularization of extremely high orbits.

Out of the orbits that rotate along with the Earth-moon system, it seems that a fancy shot to EML-2 would win with 0.4 km/s, which is better the 0.86 km/s for EML-1.

You could go beyond lunar orbit, but you would need to be over 3 lunar distances in order to beat EML-2, and the stability of such an orbit would get extremely iffy. The Earth-sun L2 point would probably beat that. The Earth-sun L1 and L2 points are in the neighborhood of 0.3 km/s.

Maybe you could just do a very highly elliptical orbit below lunar distance. I went through the calculations for that (system of energy and angular momentum with 2 unknowns). If you do a burn at apogee at 1/2 lunar distance then you need a burn of about 0.36 km/s in order to reduce the eccentricity sufficiently. On the bright side, Earth's rotation is now giving you more nearly 0.01 km/s for an apogee boost as opposed to the 0.005 km/s at L1 distance. Whoo hoo.

I doubt that orbit is very stable, but you could push the tradeoff to either side. So it looks like the Lagrange points are the winners.

$\endgroup$

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.