24
$\begingroup$

Lot of questions involve shooting things into the Sun. But there are no aliens on the Sun: they are on the Moon. I want to drop things on them, but since there's lot of them, I can just drop a tungsten rod anywhere on the Moon and that's good enough. But how much delta-v does one need to just hit the Moon (as opposed to orbiting or landing).

The Wikipedia page on delta-v budgets indicates that the delta-v to reach Earth-Moon L1 from LEO is modest: about 0.77km/s*

enter image description here

Now, from my understanding of the Langrange points, all you need to do is ''just'' tip over the "saddle" of the L1 point and you'll fall into the Moon (and if you fall just short, it's a slow spiral back to Earth).

enter image description here     The Lagrange points for the Earth-moon system. Credit: David A. Kring, LPI-JSC Center for Lunar Science and Exploration

Is it that simple: the minimum delta-v to hit the Moon is to get to L1 and then just tip over the lip? And does that apply to any two-body system?

* Let's assume it's right despite the [Citation needed]

$\endgroup$
6
  • 2
    $\begingroup$ I like your question very much! +1 and often wonder about a puzzler or space-golf tag $\endgroup$
    – uhoh
    Jan 27 at 21:45
  • $\begingroup$ "I can just drop a tungsten rod anywhere on the Moon and that's good enough" – do you mean "The likelyhood to hit someone there is great enough", or do you expect some other consequences from dropping the rod, affecting the whole moon? $\endgroup$ Jan 28 at 22:56
  • 2
    $\begingroup$ How soon do you need the tungsten rod to get there? It's likely possible to save a small amount of delta-v by waiting several months to get multiple gravity assists from the Moon before the actual impact. (You'll probably also want some small maneuvering thrusters on your tungsten rod for mid-course corrections, because these kinds of orbits tend to be quite sensitive to small perturbations in the initial trajectory, and you don't want to miss the Moon — or, worse yet, accidentally hit the Earth instead.) $\endgroup$ Jan 29 at 8:14
  • $\begingroup$ I don't think your understanding of Lagrange points is correct. If you're just inside the L1 point you won't be spiralling back to Earth, because that would require you to be losing kinetic energy, which won't happen that far outside the atmosphere. Likewise I don't think there's any guarantee that you'll hit the moon if you're just on the other side of L1. It's a saddle point in gravitational potential, not in a Lyapunov function. Others might know better but I'd guess in either case you'd orbit chaotically around the Earth-Moon system until you eventually collide with one body or the other. $\endgroup$
    – N. Virgo
    Jan 29 at 14:11
  • 1
    $\begingroup$ Note that for shooting at the moon there's no reason for tungsten. Tungsten rod kinetic attacks are for attacking planets with atmospheres where you need something that is dense (to minimize drag losses) and refractory (so it can take more before melting.) Furthermore, there's no reason for a rod--once again, that's for attacking through atmosphere. For shooting at airless worlds you simply want a guidance unit and it's fuel tank--better to hit with less mass than to miss. $\endgroup$ Jan 31 at 2:10

4 Answers 4

45
$\begingroup$

This is an excellent and fun question, bravo.

First, the table is being read incorrectly, here's how to properly read it:

proper reading

For example, to go from LEO-Ken to EML1 is given as 3.77 km/s, not 0.77 km/s.

Second, the Earth-Moon L1 point 'orbits' Earth with the Moon. It has significant tangential velocity. Getting to L1 doesn't do much as you have to stay with L1:

L1 flyby animation (Personal work)

where the red dot is L1. The Earth and Moon are to scale, the dashed line is geostationary orbit.

Another clue comes from reading the from LEO to C3=0 at 3.22 km/s (this value checks out, re: citation needed). C3=0 is an Earth escape orbit so it certainly takes less $\Delta V$ to simply crash into the Moon.

I built a quick and dirty 2D sim to find the minimum $\Delta V$ from a 185 km (100 nm) low Earth orbit to lunar impact (same sim used in above animation). It assumes a circular orbit for the Moon so the real Moon's eccentricity will affect these results slightly:

wide dV span (Personal work)

$\theta_0$ is the initial phase angle of our tungsten rod spacecraft and the Moon. Zooming in on the light, boomerang shaped section (where trajectories fly closest to the Moon):

min dV closeup (Personal work)

A "Lunar Close Approach" of one lunar radius indicates impact with the lunar surface. The minimum $\Delta V$ appears to be 3135 m/s (at a phase angle of 2.045 radians). This trajectory looks like so:

animation 2 (Personal work)

It impacts the lunar surface at over 2500 m/s. Here is a closer look at the (explosive?) impact:

impact close-up (Personal work)

It even impacts on the Earth facing side!

$\endgroup$
7
  • 5
    $\begingroup$ Nice answer! (Doesn't the 3.77 km/s assume you want to orbit at L1? i.e. the L1 velocity is already included) $\endgroup$
    – Roger Wood
    Jan 28 at 6:45
  • 1
    $\begingroup$ @RogerWood I think it does mean that $\endgroup$ Jan 28 at 12:53
  • 2
    $\begingroup$ "It even impacts on the Earth facing side!" My intuition says that's because your trajectory probably doesn't have exactly the minimal delta-v, and that the optimal trajectory would probably lead to a grazing impact near the "trailing pole" of the Moon, slightly on the far side — basically where your perilune would be if you just barely missed the Moon. (I wonder where exactly that point would be. I feel like it would be a lot easier to eyeball if I could see your trajectory from a frame centered on the Moon.) Of course, my intuition could also just be wrong. $\endgroup$ Jan 29 at 8:01
  • 3
    $\begingroup$ @IlmariKaronen I can see some logic in that idea; however, the system is quite sensitive. I tried just 1 m/s less $\Delta V$ (3134 m/s) and could not get any lunar impacts with milliradian spacing for the phase angle. $\endgroup$ Jan 29 at 14:50
  • 1
    $\begingroup$ According to this answer by Mark Adler, you could potentially save around 100 m/s by skipping LEO altogether (leaving your perigee deep under ground). $\endgroup$
    – TooTea
    Jan 30 at 13:30
9
$\begingroup$

The answer from @BrendanLuke15 is excellent. A simple insight is that basically you're just throwing the tungsten rods into a high elliptical orbit and waiting for the Moon to run into them. This elliptical orbit has a semi-major axis of roughy half that of the Moon's orbit and will therefore have twice the specific orbital energy. This also follows from the fact that the kinetic energy is close to zero at the far end or the elliptical orbit and only the (negative) potential energy remains.

Orbital energy is proportional to minus the square of orbital velocity for circular orbits. The orbital velocity in LEO is 7.8 km/s and the orbital velocity of the Moon is 1.022 km/s. The specific orbital energy at the Moon's orbit is $(1.022/7.8)^2 = 1.72\%$ of that in LEO and the specific orbital energy of our highly elliptical tungsten orbit is thus twice that number or $3.43\%$.

Adding $C_3=3.22$ km/s to LEO velocity gives the escape velocity of 11.02 km/s (which, not coincidentally, corresponds to twice the LEO orbital energy). Taking $11.02^2$ as our zero potential energy (at infinity) reference, we now have LEO orbital energy as $7.8^2 - 11.02^2 = -60.6$ and the tungsten orbital energy as $3.43\% \times -60.6 = -2.08$ on this weird energy scale. So we're going to need a $\Delta V$ to get from -60.6 to -2.08 which is a difference of 58.52. So we have $(7.8 + \Delta V)^2 - 7.8^2 = 58.52$ giving $\Delta V = 3.125$ km/s - which just a little bit off from BrendanLuke's fine answer.

$\endgroup$
8
$\begingroup$

It is possible to save a bit of fuel, according to this paper. The problem space is quite complex. The best that this paper was able to figure out was a theoretical 3100 m/s to the L1 and a 627 m/s to the Moon (Orbit) from there, to give 3727 as the minimal delta v to orbit. Just to crash one doesn't actually need the last bit. The best orbit in practice they were able to find was 3265 m/s, taking 100 days to reach the Moon as such.

$\endgroup$
5
$\begingroup$

First off, the table is misleading: for every transfer to an equatorial or KSC-inclination low-Earth orbit, it assumes aerobraking. This makes the table asymmetric, and the delta-V for going from EML-1 to LEO isn't equal to the delta-V for going the other direction. The actual cost of going to EML-1 from LEO is 3.77 km/s.

Second, no, reaching EML-1 is not enough. From EML-1, a slight nudge will put you into an unstable, very high Moon orbit that will quickly turn into an unstable, very high Earth orbit. To hit the lunar farside, you need about another 1 km/s of delta-V to lower your periapsis to just barely graze the surface; to hit some other point, you'll need more.

$\endgroup$

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.