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If an L1 space station was used as a staging area for a mission to Mars, it looks like we would use the Oberth effect in a close flyby to Earth to make the burn.

But I just realized - the idea, as stated, has a free degree of symmetry. I imagine that we need to leave (v_infinity) with a velocity vector in the direction of Earth's orbit (as per the Hohmann transfer orbit). We have to stay in the elliptic plane. That means there are 2 well-timed passes that could accomplish this in a month (neglecting the position of Mars). Standing on the moon's north pole, looking at the Earth, the trajectory could either fly on the left of right of Earth. Put another way, the flyby could be prograde or retrograde.

One trajectory would obviously be better than the other, but which one?

Obviously one factor would be the velocity of the moon itself. But maybe the gravitational potential relative to the sun would factor in as well? Initially thinking about it, the retrograde option would seem to start at a higher solar gravitational potential.

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EML1 is moving in a prograde orbit at about .86 km/s. A .7 km/s braking burn will drop it into a prograde orbit with a perigee deep in earth's gravity well (about 300 km altitude).

To go to a retrograde orbit, you would need to completely kill your .86 km/s velocity. But at 0 km/s, you'd fall straight to earth for an impact. To raise perigee to above earth's atmosphere, you'd need another .17 km/s. So the retrograde burn would require a little more than 1 km/s.

So the burn dropping you into a prograde orbit is preferable as it takes less delta V.

This is only one time during the moon's orbit when a braking burn would drop a ship to a perigee at the correct longitude. But during the launch window it's possible EML1 could be as much as two weeks away from the place where you want to do the burn.

This can be mitigated by letting your payload orbit more than once. For example, if EML1 reaches the place where you need to do the burn 10 days early, go ahead and do the burn. It will take 5 days to reach perigee. Then another 10 days it will complete another orbit and it's back at the perigee location you want, but now it's the right time as well as the right place. And you can make the orbit more or less than 10 days with small perigee or apogee burns.

Another option is to drop from EML2, do a small burn at perilune to get to deep perigee. It takes a little more than 8 days to reach perigee. And it takes less delta v than dropping from EML1 -- about .4 km/s.

Farquhar diagram

After the perilune burn drops the ship to a deep perigee, it too is in about a 10 day orbit.

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The Oberth effect isn't the only reason for flying past another celestial body during an interplanetary flight. There is also the gravitational slingshot effect which increases the speed of a vessel without using any fuel.

To perform a gravitational slingshot around a celestial body which results in an increase in relative velocity, one needs to pass behind that body on its orbital path. Passing in front of it reduces the relative velocity.

Gravitational slingshot with speed-gain Gravitational slingshot with speed loss

Such a fly-by will most certainly be on the "right" side as seen from moon's north pole. The reason is that the moon and thus also the EML-1 orbit the earth in a counter-clockwise direction (seen from north). Passing earth "left" would require to change from a prograde to a retrograde orbit, which requires more fuel than just going from a circular to a highly eccentric orbit.

Image source: Wikimedia Commons

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    $\begingroup$ The orbits shown are hyperbolic. The fly by can be positioned so the hyperbola's turning angle can either subtract or add to the heliocentric orbit velocity. However this isn't applicable to a drop from EML1 into an elliptical orbit about the earth. $\endgroup$ – HopDavid Mar 30 '14 at 0:24

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