# Lunar Polar Orbit

With all this talk about lunar bases being at the poles, especially Shackleton Crater, I was thinking about the navigation. I am assuming that the lunar landers would stay there and ferry people & stuff to and from the surface. So the trip from Earth would be to lunar orbit for rendezvous, not to the surface. I am seeing the Earth side being a similar situation. A shuttle would lift off to a space station and a different ship would go orbit to orbit. We would have to do orbital insertion in both directions, and disregard atmospheric breaking or liftoff. Here's my thinking. Since the lunar day is once a month, the rotation advantage of being near the equator is nil. Since the moon is on average about 240,000 miles away and the diameter is about 2K, that would mean, from the earth's point of view, there wouldn't be much of a difference between shooting for the lunar equator or the lunar pole. So I am thinking the delta V requirements of achieving lunar polar orbit wouldn't be much different than the delta V for lunar equilateral orbit. Can anybody do the math on that?

• Since you mention you will be "do(ing) orbital insertion in both directions" do you mean 1) inserting into lunar orbit both from the lunar surface, and from Earth orbit, or do you mean 2) inserting into lunar orbit from Earth, and Earth orbit from the Moon? – uhoh Jan 14 '17 at 9:45
• I believe they did the math for this as far back as the Apollo program, you may be able to find those calculations pretty easily. – John Feltz Jan 14 '17 at 18:31
• Is it possible to get a free return trajectory when aiming for a lunar polar orbit? – Suma Jan 14 '17 at 21:48
• LEO to the North pole is about 6.4 km/s by my calculations, just slightly more than landing in the lower latitudes. Biggest drawback is you don't have any time return, launch windows back to earth open each two weeks. hopsblog-hop.blogspot.com/2013/08/lunar-ice-vs-neo-ice.html – HopDavid Jan 15 '17 at 3:48
• Just as a clearification of the question, I said"orbital insertion" in both directions. Yes I meant from the transfer trajectory. (As oppose to a landing or free return) – Johnny Robinson Jan 15 '17 at 14:10

Ordinary Lunar Landing in Lower Latitudes

A LEO to Moon transfer orbit is a 300 km x 378022 km orbit. These numbers refer to distances above earth's surface, not distance from earth's center.

Perigee speed of this transfer orbit is 10.9 km/s while LEO speed is about 7.7 km/s. TLI (Trans Lunar Injection) is about 3.2 km/s.

At apogee of this transfer orbit the space craft is moving .19 km/s with regard to earth while the moon is moving 1.02 km/s with regard to the earth. If the transfer orbit and moon's orbit are coplanar, the arrival Vinfinity would be (1.02-.19) km/s or about .83 km/s.

When the spacecraft enters the moon's sphere of influence, the path can be modeled as a hyperbolic orbit with regard to the moon. Speed of something on a hyperbolic path is $\sqrt{V_{esc}^2+V_{inf}^2}$. At the moon's surface Vesc is about 2.376 km/s. So speed of a spacecraft near the surface is $\sqrt{.83^2+2.376^2}$ km/s or about 2.516 km/s.

You may notice I'm using three significant figures here. The moon's orbit varies so it's not possible to give a general answer with this precision. However I do have a motive as you'll see later.

Adding the LEO TLI burn of 3.2 km/s to this 2.516 km/s and you get 5.716 km/s. I round up to 6 km/s since there would likely be gravity loss as you land on the moon as well as course corrections enroute.

Lunar Landing At the Poles

You'd still have a LEO TLI burn of about 3.2 km/s.

But with my route to the poles, the spacecraft enters the moon's sphere of influence somewhat above the moon and the Vinf velocity vector lies in the same plane as the line from lunar north to south pole. This can be accomplished if the transfer orbit and and lunar orbit aren't coplanar.

Since the orbits aren't coplanar, the launch needs to be timed so the spaceship arrives at the moon as it's crossing the line of nodes, where the two planes intersect. If departing from a equatorial LEO, launch windows open every two weeks or so.

But now the apogee velocity vector isn't anti-parallel to the moon's velocity vector. So you can't just subtract one from the other to get your Vinf.

Now you need to use vector subtraction. Assuming the two vectors differ by 29º, you get:

So the new Vinf is .86 km/s.

Let's plug that into

$\sqrt{V_{esc}^2+V_{inf}^2}$ km/s =
$\sqrt{.86^2+2.375^2}$ km/s =

2.526 km/s

So now you see why I used more significant figures than I'm comfortable with. Departing from an equatorial orbit and landing on a pole only increases surface hyperbola velocity by .01 km/s. I would still round up to 6 km/s for equatorial LEO to moon's pole.

So in terms of delta V, the hit is insignificant (in my opinion). Take into account that an equatorial LEO is easier to enter from a place like French Guiana's spaceport and it might be even be a little easier.

The big disadvantage is constraints on launch windows. The moon only crosses a line of nodes each two weeks so you don't have any time return. However launch windows each two weeks is still pretty good (in my opinion).

Another disadvantage is the poles lie on the moon's terminator. So if you have sunlight, the shadows are long. This might make landing harder if you're relying on visual input.

Orbits are time reversible so same goes for the trip back. However with the return trip aerobraking can shed delta V when you arrive at perigee.

Irrelevant Note

I'm feeling mortal and have been thinking more about my bucket list. A list item is making a comic book with hard science fiction and some science fact. I'm a big fan of yours and am hoping to get some of your illustrations in my comic book. Especially the polynesians whose population has reached the limits their little island can support.

• Very nice! Is it possible to guestimate how large the delta-v penalty would be if the launch window is opened a bit - say +/- 2 days? – uhoh Jan 16 '17 at 4:55