# Why would a launch request for a gravitational assist at Mars specify a DLA (Declination of Launch Asymptote) of 30 to 32 degrees?

The recent intriguing question Why is NASA proposing an EMEJ (MEGA) trajectory for Europa Clipper instead of EEJ? quotes "A launch service request... recently put out for Europa Clipper to do a flyby of Mars on its way to Jupiter"

Launch Vehicle Performance: The launch vehicle shall deliver a minimum 6065 kg Europa Clipper spacecraft (SC) with Mars-Earth-Gravity-Assist (MEGA) trajectory characteristics as follows: C3 value of 41.69km2/sec2 and a DLA range of 30-32 degrees.

Launch Period: Europa Clipper will be launched during a 21-day launch period beginning October 10, 2024 and ending on October 30, 2024.

My understanding is that this launch should have an asymptotic velocity relative to Earth of about 6.5 km/s (i.e. above escape velocity) in magnitude and in a direction defined by a Right Ascension and a Declination of Launch Asymptote (RLA and DLA). This direction is a point on the celestial sphere and so is relative to Earth's poles and equator.

Where will this declination of 30 to 32 degrees point? It seems like it's "straight up" if the launch was from a latitude of 30 to 32 degrees, is that right? Is this just a fancy way of specifying "where on Earth" the launch will be, or is it a strong constraint on the launch in other ways as well?

• I don't have time for a full answer, but the DLA for a launch opportunity is a complex interaction of the destination's ecliptic inclination, Earth's obliquity, the latitude of the launch site (call it L), and the details of the heliocentric departure asymptote vector with respect to Earth's orbital motion vector. Notably, a departure from Earth with a DLA of, say, 30°, does not yield a transfer orbit with an ecliptic declination of 30°, or 30°+L, or 30°+23.4°+L, or anything like any of those. It will be much smaller, the vector resultant of 6.5 km/s with Earth's 29.8 km/s. – Tom Spilker Feb 8 at 17:51
• ... The direction of that 6.5 km/s vector (in the case of the specific example you cite) will be whatever gives you the needed ecliptic inclination needed to reach Mars. That might be zero inclination, if it'll reach Mars at a node, i.e. as it crosses the ecliptic plane, but that isn't always the case. – Tom Spilker Feb 8 at 17:55
• @TomSpilker oh I see I think, thanks! If we could continue to see it from Earth (and as DSN needed to), might it at least appear to remain somewhere near +30° declination for a few days from Earth? – uhoh Feb 8 at 22:03