Delta-v to hit the moon: is reaching Lunar L1 enough?

Question

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*

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).

    ^{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]}

Answer 1

This is an excellent and fun question, bravo.

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

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:

(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:

(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):

(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:

(Personal work)

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

(Personal work)

It even impacts on the Earth facing side!

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

Great answer @BrendanLuke15 (https://space.stackexchange.com/a/58005/54582) Here's a simple answer from a Solar System delta-V map (below).

Add the two delta-Vs from low Earth orbit (250km) to Moon Transfer and you get 2440 + 679 = 3119 m/s. From Moon Transfer it's all downhill. If you get the timing right you will impact the Moon at 145 + 679 + 1721 = 2545 m/s. Get the timing wrong and you will stay in an elongated Earth orbit, which may eventually hit the moon.

These agree remarkably well with @BrendanLuke15, 3135 m/s and 2536 m/s respectively. Reaching L1 is not enough, but reaching "Moon Transfer" is.