# What would your altitude be after you had achieved escape velocity from the moon?

I have been trying to work out how much ΔV would be required to deorbit a spacecraft after it had achieved escape velocity from the moon but I don't know how to work out what the radius of the orbit would be after it had escaped the moons gravity.

Say you were in a lunar orbit at 100km, and you made you burn at your lowest altitude (relative to earth) e.g. moons orbit minus 100km, when you got out of the moons gravity well and into the earths, how big would your orbit around the earth be?

Would I just have to rearrange the vis-viva equation to calculate my radius with the speed that I had after the burn?

I know thats not very well put but I hope you understand.

• Check the answers to How do orbital elements change when force is applied orthogonal to the velocity vector? But your question isn't really clear, for example in which direction are you launching and where from? If it's from the surface of the Moon and towards the Earth-Moon barycenter, then it's a duplicate of the mentioned question. Also, escape velocity is in $v_e = \sqrt{\frac{2GM}{r}}$ relation with the radius to center of mass $r$, so that also makes your title unclear. You'd be at altitude at which you achieved $v_e$. Commented Apr 8, 2014 at 15:30
• I have tried to clarify my question a bit, as it isn't the same as the question you referred to. Commented Apr 8, 2014 at 17:26
• I'm still not sure I follow you. Are you asking what's the orbital period and speed of the Earth-Moon L1 Lagrange point (EML1)? The period is same as Moon's orbital period, and I'd have to dig for what's its velocity, but should be in some answer or comment of mine... somewhere. Or I could just calculate it again, if that's what you're asking? BTW Lunar orbit isn't exactly circular either. Commented Apr 8, 2014 at 18:11
• My guess is your burn would actually happen on the far side of the moon (furthest from Earth). However this is informed by familiarity with patched conics, which may be entirely irrelevant with regards to a transfer orbit in a restricted three body system.
– ben
Commented Feb 8, 2019 at 5:13

There isn't a single altitude associated with escape velocity. You could use a Jules Verne-style gun and achieve escape velocity at altitude 0. From the orbit you describe, it depends on how much thrust you use.
You may be able to escape from the moon without reaching escape velocity, if you can get to a point where Earth's gravity is stronger than the Moon's.

• What I'm trying to say is after I move out of the moon's sphere of influence, how can I calculate the radius of my orbit around the earth? I have already calculated the amount of delta v needed to achieve escape velocity and I want to calculate the amount of delta v needed to bring my periaps down inside the atmosphere after the spacecraft has left the moons sphere of influence. To do that though I need to know the radius of my orbit around the earth, which I don't know how to calculate Commented Apr 8, 2014 at 20:47
• The point where the earth's gravity is stronger than the moon's is only about 37,000 km above the moon's surface. But this isn't sufficient. What's needed is a point where earth's gravity is cancels centrifugal force + lunar gravity. Also known as EML1. This is about 56,000 km above the moon's surface. Commented Apr 8, 2014 at 23:14

A parabolic departure from the moon isn't needed. A ellipse with an apolune near EML1 suffices.

When first messing around with 3 body scenarios I tried using the vis-viva equation and other tools from two-body mechanics. It bit me, especially if I had trajectories passing through EML1 or EML2. For such a trajectory, a tiny change in initial velocity can make a big change later on.

To get an apolune near EML1, you need a perilune near the far side. Here is an orbital sim firing from the moon's far point:

Pale blue dot in the middle is earth. White circular arc is moon. The slowest red pellet was fired at 2.315 km/s and fell back towards the moon. The fastest pellet (blue) was fired at 2.355 km/s. Most these pellets sailed past EML1 and into approximately 100,000 km x 300,000 km earth orbits.

Once in a 100,000x300,000 km orbit, a .6 km/s apogee burn can drop you to an atmosphere grazing perigee.

However there is a direct route that takes less delta V:

All these pellets depart from the moon's surface at the far point. Launches are westward. The earth grazing blue pellet was fired off at 2.6 km/s

• What software did you use to make this? Commented Apr 8, 2014 at 20:50
• I used Bob Jenkins' Java app for orbital sims: burtleburtle.net/bob/java/orbit/index.html With his permission I used his Java to make my own sims based on solar, moon, and earth masses as well as position and velocity vectors. From various locations a user can fire 11 pellets moving a range velocities. I call it EML shotgun. clowder.net/hop/railroad/emlShotGun.php Smaller time increments make the model more accurate I set increments at 60 seconds Commented Apr 8, 2014 at 21:21
• May I assume the green arc is EML-1? Commented Nov 16, 2014 at 15:42
• Each colored arc represents a pellet with a slightly difference velocity with the red and blue pellets representing fastests and slowest. The green arc is a pellet having a velocity somewhere between the etremes I set. EML1 is indicated as a darker arc that runs along side the moon. It's barely visible. Commented Nov 17, 2014 at 20:34

Escape velocity for 2 body systems is not a super clean issue, and I would avoid using it as it clouds the question a little.

The dv to change between an Earth-Moon transfer orbit and a 100km lunar orbit is of the order of $$700\text{m}\text{s}^{-1}$$ (directly, in one orbit). There is lots of info on the options and compromises of the moon-earth transfer burns (or free returns etc) from the apollo missions as you would expect.