Orbit to the Moon: nodes position

[andreajf89]

Hi everybody, I'm planning a 2016 mission to the moon similar to Apollo 8 (Orbital mechanics homework). I decided to start from a equatorial parking orbit, and then to perform a plane change. The problem is that I obviously need to know the position of the moon's orbit plane, which, as you know, changes in time. Where can I find the nodes line position in 2016?
Also, the teacher asked us to choose the 2016 day in which the propellant requirement for the mission is the less. How should I proceed?

 

[D H]

http://ssd.jpl.nasa.gov/?horizons

 

[andreajf89]

Ok but is there a way to get all at once all the year node line position?

 

[tony873004]

Are you required to start from an equatorial parking orbit? Why not start from within the Moon's orbital plane? From the ground you can launch directly into the Moon's orbital plane from any location between 28 degrees South to 28 degrees North of the equator. That would save you a lot of fuel, as you wouldn't be burning any to make a plane-change.

A mathless way to figure out when your launch station is in the Moon's orbital plane is to simply wait for the Moon to be directly overhead. But this puts a restriction on the day you must launch. With a little more effort, you can figure out when your launch site intersects the Moon's orbit on any given day. It should happen twice a day. A program like Orbitron, or perhaps Celestia or Stellarium can probably tell you that. The program Orbiter will actually give you a map of Earth, with the Moon's orbital plane superimposed. Just wait until this line intersects your launch pad and you're ready to launch directly into the Moon's orbital plane.

Low-Earth orbit is a very expensive place to be making plane adjustments to your orbit. You can reach the Moon without traveling in its orbital plane, provided that you intersect its orbital plane as you arrive. Being further from Earth, the amount of fuel spent matching planes will be less than if you attempt to match planes in low-Earth orbit. But if I'm not mistaken, the least expensive way would be to launch directly into the plane to begin with. Then when you arrive, you've already got one of the components of your velocity vector wrt the Moon zeroed-out.

It's probably best to launch at a time when the Moon will be at perigee when you arrive, although I might be wrong. The Moon is traveling faster at perigee, requiring a heavier lunar orbit insertion burn. Perhaps its cheaper to spend a little more fuel on your TLI burn to travel to apogee where your orbit insertion burn will less expensive. You could also save a little fuel by timing your launch to coincide with a new or full Moon. It's a little easier to travel away from the Earth in the direction of its L1 or L2 points (wrt the Sun). But the Moon is deep enough in Earth's Hill Sphere that I doubt it would make a huge difference.

If this is orbital mechanics homework, the prof is probably going to want to see some calculations. Look up the formulas to find the line of intersection between 2 planes.

 

[andreajf89]

Dear Tony, thanks for your answer. I began to develope a Matlab code in order to compute all the orbital elements of the mission. Answering your first question, we're required to start from a circular equatorial orbit (H=300 km for example).
Right now this is the orbit, following the patched conics approach:

-300 km circular equat
-burn in order to change plane (I've used a rotational matrix in order to know the position of the moon's orbital plane wrt the equatorial plane)
-elliptic orbit to the moon

As soon as the spacecraft enters the moon's sphere of influence, the reference is the moon, and the spacecraft is in a (relative) hyperbolic orbit.
-burn at periselenium in orbit to acquire a circular moon parking orbit
Then the return is symmetrical.

The script is actually able to compute the Delta v, the time, the periselenium altitude as a function of the injection speed from the parking orbit (earth) to the elliptical transfer orbit.

I've have still two more questions:

1) How can I get a free return trajectory with no moon circularization?

2) Is it possible to arrive in the proximity of the moon taking advantage of the L1 libration point? I know that this is theorically possible if the Jacobi energy constant is set at a proper value, but how can I plan an extremely basic orbit to the moon using the L1 equilibrium point?
Thanks for your time!

 

[Travis_King]

1) You want to come back without using the moon as a pivot? Why? It's extremely time and energy effecient way of doing things...I guess it's a little extra work because in planning your optimal departure day, you'll have to also plan a trajectory that will give you the correct return angle (thus you'll need the proper approach angle). These aren't incredibly difficult, and it'd be hard to justify not using the Moon, IMO.

2) You don't...It's not going to be cost effective to do so. L1 is out past the moon's orbit, in fact it's almost 4 times further out.

 

[D H]

1) You want to come back without using the moon as a pivot? Why? It's extremely time and energy effecient way of doing things...I guess it's a little extra work because in planning your optimal departure day, you'll have to also plan a trajectory that will give you the correct return angle (thus you'll need the proper approach angle). These aren't incredibly difficult, and it'd be hard to justify not using the Moon, IMO.

Wrong. Most translunar trajectories result in the vehicle being ejected from the Earth-Moon system or in the vehicle being in an elliptical orbit about the Earth with a rather high perigee. A free return trajectory flies by the Moon in a special way so that in the case of no lunar injection burn, the return trajectory naturally comes back toward the Earth.

Many of the Apollo missions used a free return trajectory in case of some en route emergency. Apollo 13 was not on a free return trajectory. One of the first things done after the explosion was to put that vehicle onto such a trajectory.

2) You don't...It's not going to be cost effective to do so. L1 is out past the moon's orbit, in fact it's almost 4 times further out.

Wrong again. The Earth-Moon L1 point is between the Earth and the Moon. You are apparently thinking of the Sun-Earth L1 point.

Using the Earth-Moon L1 point as a rendezvous point is a bit more expensive than a direct mission. The concept keeps coming up within NASA and other space agencies because of the are considerable advantages that accrue from L1 rendezvous.

 

[Travis_King]

1) I was operating under the assumption that this was a moon mission aimed at visiting the moon, not flying by it. I suppose I misread the question.


2) Yep. I was. Good call.

 

[andreajf89]

So what burns should I apply in order to be stationary at the L1? And what are the conditions in order to get a free return trajectory?