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Let's say I launched something into lunar orbit with minimal of propellant - just enough for trajectory corrections and then a final push to de-orbit.

What is the slowest crash landing speed (both tangential and vertical)?

Sources?

Edit 1: I would like a crash trajectory be at the most 30 degrees from vertical.

Edit 2: Wikipedia describes "Hohmann transfer orbit" :

The orbital maneuver to perform the Hohmann transfer uses two engine impulses, one to move a spacecraft onto the transfer orbit and a second to move off it.

Any way to end up on the moon without a 2nd burn? If yes, what would be the collision speed.

BTW, If this is really too vague of a question I have no problem deleting it.

gene
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    What do you mean by deorbit? A burn that completely cancels the horizontal component of velocity, a burn that changes a circular orbit into an elliptical one whose perilune just barely intersects the Moon's surface, or something else? – David Hammen Sep 02 '19 at 20:23
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    Is there a way to just "shoot for the moon" without planning for the de-orbit burn? Or setting up orbit's apogee so that the Moon crashes into the launched object? – gene Sep 02 '19 at 20:38
  • @gene What do you mean by "slowest crash landing speed"? That entirely depends on the spacecraft itself. A spacecraft might be able to survive a 200 m/s impact and another spacecraft might break at a speed of 2 m/s. It depends on the space craft. – Star Man Sep 02 '19 at 21:05
  • I did not say "survivable" speed... :-). Just looking for numbers. Is there a orbital calculation tool online/downlodable I can play with? – gene Sep 02 '19 at 21:08
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    @gene I think you have to put more detail in your question. It is unclear what you are asking. Using the example in my comment above, a spacecraft might impact at 200 m/s and another at 20 m/s. It depends on the space craft (eg, amount of fuel, thrust of thrusters, specific impulse, etc). I suggest editing it. Thanks :) – Star Man Sep 02 '19 at 21:21
  • @gene Unless you're asking for the slowest speed of a spacecraft in history to impact the moon which is answerable. But asking "what's the slowest crash landing speed in general is vague and unclear. – Star Man Sep 02 '19 at 21:28
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    The slowest collision speed is just shy of zero. There are lots of ways to accomplish this. – David Hammen Sep 02 '19 at 21:37
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    Related: https://space.stackexchange.com/q/2103/32284 – Star Man Sep 02 '19 at 23:22
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    @DavidHammen I just wrote this, so now I'm curious what those ways might be! – uhoh Sep 02 '19 at 23:24
  • "just enough for trajectory corrections and then a final push to de-orbit." You do not need a final push to de-orbit if you eject from LEO on a collision course. – Polygnome Sep 03 '19 at 08:28
  • [This suggestion devoid of any sort of maths or knowledge ;-)] Could you construct your ship in two parts? If so, you could construct your lander to be lighter than the other part of your ship. You pick a decent path you like the look of, and then jettison the lander from the ballast (no fuel required). The lander is now travelling slightly slower than the ballast and may arrive at the slower speed you require. – Ralph Bolton Sep 03 '19 at 09:33
  • A "relatively slow" crash just happened – Magic Octopus Urn Sep 07 '19 at 19:14

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tl;dr: I don't think there is any scenario where you can strike the Moon with low velocity by using a small impulse to leave orbit. You can hit sideways with an orbital velocity of about 1680 m/s, or vertically with escape velocity the square root of 2 larger at 2376 m/s.

Let's say I launched something into lunar orbit with minimal of propellant - just enough for trajectory corrections and then a final push to de-orbit.

From low lunar orbit

When in orbit around Earth, say at 400 km, "a final push to de-orbit" would be a small impulse to lower the perigee to about 100 or a little higher. Then each time the spacecraft passed near perigee it would loose a little more velocity due to drag, slowly circularizing near perigee. After that, it would spiral due to drag and eventually reenter the main part of the atmosphere and quickly either burn up, or fall to the ground if it had proper heat shielding and aerodynamics.

But the Moon is tricky. If it were a nearly perfect gravitational sphere, then your burn would lower the perilune to just above the average lunar surface where it would strike whatever boulder or crater rim might be sticking up. This would happen at the lunar orbital velocity given by the vis-viva equation

$$v= \sqrt{GM/a}.$$

The standard gravitational parameter of the Moon $GM$ is 4.905E+12 m^3/s^2 and the semimajor axis $a$ would be the lunar radius 1.737E+06 meters. That puts the velocity at about 1680 m/s.

Since the Moon has quite a lumpy gravity field all you need to do is to bring the spacecraft to a very low orbit and just wait. Due to gravitational perturbations, or those from the Earth and Sun, eventually its constantly changing orbit will bring it into contact with the surface.

There are no small orbital corrections from a low lunar orbit that can bring it down within 30 degrees of vertical. You'd have to do a major burn to loose most of that 1680 m/s of orbital velocity very quickly, so that it would just "fall straight down".

From high lunar orbit

If the Moon were all alone in space, you could put yourself in an absurdly high lunar orbit, let's say 1 million kilometers. At that altitude your orbital velocity would be only 70 m/s and a delta-v equal to that would stop you in your tracks. However, then you'd fall towards the Moon and accelerate.

Your velocity at impact dropping from an altitude $a$ to the lunar radius $R$ would then be

$$v= \sqrt{2 GM\left(\frac{1}{R} - \frac{1}{a}\right)}.$$

If you plot those versus the starting semi-major axis, you can see that the delta-v you'd need to fall out of orbit, which is the orbital velocity, drops with increasing altitude, but the resulting impact velocity due to acceleration towards the Moon rapidly rises.

There's no gentle delta-v followed by a gentle impact.

enter image description here

What about a clever 3-body orbit?

But what if I know about the chaotic 3-body orbits of minimoons that uses both the gravity of the Earth and the Moon, and I wanted to look for a crazy orbit that starts near a stable orbit, but "goes chaotic" and eventually touches down on the surface of the Moon, or slows very close to it?

This doesn't happen. I think there is a good Stack Exchange Q&A on this somewhere in Space Exploration, Astronomy, or Physics, but I can't find it.

The argument goes like this: orbits work just as well forwards and backwards in time. So if such an orbit existed, then the backwards scenario would also have to be possible; you'd be able to hold a rock near the surface of the moon, give it only a slight nudge, and it would mysteriously start flying away from the Moon and end up in a high orbit.

That doesn't happen, it just falls to the surface with a silent but none-the-less perceived thud.

uhoh
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    I really like the "orbits work just as well forwards and backwards in time." logic. – gene Sep 03 '19 at 00:54
  • @gene thanks, but I can't take credit for thinking of it first; I read it elsewhere in SE. I'll keep looking for the original post. Great question by the way! – uhoh Sep 03 '19 at 00:55
  • what if your spacecraft was in a low lunar orbit? If you were just a short distance above the surface when you did your delta-v maneouvre, you would not accelerate much due to gravity as you fall. Taken to the limits, I think I would call this a "landing". – craq Sep 03 '19 at 04:57
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    @craq yes, but that falls under the "no small burns available" part of the answer. If you're in a super-low orbit, you're going very fast, so you need to make a very large burn to null your horizontal velocity. And if you can manage that, you could just do a proper landing. – hobbs Sep 03 '19 at 06:41
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    tldr : You can't stop without slowing down first. If you don't do it yourself, the moon will do it for you. – J... Sep 03 '19 at 17:19
  • About the last point with the 3-body orbit, and the counterargument with the "reverse orbit" : it could be possible to hold a rock near the surface of the moon, give it only a slight nudge, and have it float away. But in this case the foreign body has to be so massive and so close that the Moon would be within its Roche limit, and be itself torn apart. – vsz Sep 03 '19 at 17:28
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    @Quietghost Thank you for the edit to the equation but I've rejected it for the following reasons; 1) I've also used the equation to make that plot, so the better procedure would be to simply leave a comment that you suspect the equation was wrong. That way I can check, and if it was I can re-plot as well. 2) That's not the vis-viva equation, that's the velocity at impact when dropped from the height of the original orbit. You can see that if the original orbit were at infinity, then $1/a$ goes to zero and the impact velocity would be $$\sqrt{2\frac{GM}{R}}$$ – uhoh Sep 03 '19 at 22:34
  • @Quietghost and that's where "the square root of 2 larger" comes from in my first sentence. I think you made the edit too quickly, without reading the answer carefully to see what the equation represents, and without considering the impact on the rest of the question. – uhoh Sep 03 '19 at 22:35
  • @vsz the question is about the actual real world, not some end-of-the-world scenario, so the answer is as well. – uhoh Sep 03 '19 at 22:42
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    @J... lithobraking is a time-tested way of slowing down for landings, I think most space programs have used it either intentionally or unintentionally. – Peteris Sep 04 '19 at 00:31
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    @Peteris True, and OP didn't actually say anything about the number of pieces the ship was required to land in. – J... Sep 04 '19 at 00:45
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    @uhoh I see it now, you are correct. I saw the $R$ and $a$ in the equation, and the first part of your sentence above didn't register, but the second part did. Hence I thought it looked like the vis-viva as $a$ is common for SMA. – Quietghost Sep 04 '19 at 13:42
  • @Quietghost no problem. Ya it does look a lot like the vis-viva equation because that's derived from kinetic vs potential energy as well, but for a circular orbit rather than a "linear orbit". Actually, the idea of a linear orbit is so interesting that I've just asked What's the eccentricity of an orbit (trajectory) falling straight down towards the center? – uhoh Sep 04 '19 at 14:13
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The orbital problem can be unintuitive and tricky if you're not used to thinking about such things, so it may help to simplify the problem to a suborbital one which doesn't really require mathematics for our intuition to come up with the right answer.

Consider being on top of a very tall building and ask the question of how you can drop or throw a bowling ball so that you achieve a soft landing without engines or parachutes. If you just drop it without giving it any other velocity it will fall to the ground at incredible speed. If you throw it with any horizontal velocity, it will fall to the ground at incredible speed, but will hit the ground further away.

Landing on a celestial body is effectively the same problem - no matter which way you approach it, you're falling into a gravity well. All free-fall trajectories are accelerating trajectories, and that acceleration is always in the direction of the surface you're trying to land on. Without an atmosphere to slow you down (to at least terminal velocity) you'll always be speeding up on your way to the ground. There's no free lunch - if you want to stop slowly you have to expend energy to do so. Given that constraint, the most efficient way to make such a controlled landing has popularly (thank KSP) gained the term Suicide Burn.

J...
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Yes, you can hit the Moon with a spacecraft much in the way you can hit somebody in the head with a snowball even when they're running, and it's been done a couple of times. Some of the earliest US and Soviet lunar missions were effectively snowballs that we threw at the Moon, of which the first was the Soviet Luna 2 probe, which you can read about here.

Happy Koala
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  • My question is "what is the slowest collision speed" - whatever the variables are? The very nice link you provided mentions "impacting the moon at about 3.3 km/s ". Nice to know they left some titanium and aluminum that no longer needs to be refined. :-) – gene Sep 02 '19 at 21:02
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    @gene Without a dedicated landing engine, there’s no way to reach the moon at less than 2.3km/s no matter what initial trajectory you approach on. – Russell Borogove Sep 02 '19 at 21:47
  • If you can provide a reference - I'll accept that as an answer. – gene Sep 02 '19 at 21:52
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    wait, you hit somebody in the head with a snowball "a couple of times"? – uhoh Sep 02 '19 at 22:32
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    Oh more than a couple... – Happy Koala Sep 03 '19 at 09:10