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The Phys.org article Bound for Mars—countdown to first interplanetary launch from California says:

In the early morning hours of May 5, millions of Californians will have an opportunity to witness a sight they have never seen before - the historic first interplanetary launch from America's West Coast. On board the 189-foot-tall (57.3-meter) United Launch Alliance Atlas V rocket will be NASA's InSight spacecraft, destined for the Elysium Planitia region located in Mars' northern hemisphere. The May 5 launch window for the InSight mission opens at 4:05 am PDT (7:05 EDT, 11:05 UTC) and remains open for two hours. (emphasis added)

[...]

The United Launch Alliance two-stage Atlas V 401 launch vehicle will produce 860,200 pounds (3.8 million newtons) of thrust as it climbs away from its launch pad at Vandenberg Air Force Base, near Lompoc, California. During the first 17 seconds of powered flight, the Atlas V will climb vertically above its launch pad. Then it will begin a pitch and yaw maneuver that will place it on a trajectory towards Earth's south pole.

[...]

Mach One occurs 1 minute and 18 seconds into the Atlas V's powered flight. At that time the vehicle will be about 30,000 feet (9 kilometers) in altitude and 1 mile (1.75 kilometers) down range. Two minutes and 36 seconds later, the Atlas first stage will shut down at an altitude of about 66 miles (106 kilometers) and 184 miles (296 kilometers) down range. The Centaur second stage (carrying InSight inside a 40-foot-long payload fairing) separates from the now-dead first stage six seconds later. Ten seconds later, the Centaur's engine kicks in with its 22,890 pounds (101,820 newtons) of thrust, which will carry it and InSight into its 115-mile-high (185-kilometer) parking orbit 13 minutes and 16 seconds after launch. This parking orbit will last 59 to 66 minutes, depending on the date and time of the launch. The Centaur will then re-ignite for one last burn at one hour and 19 minutes after launch, placing InSight into a Mars-bound interplanetary trajectory. Spacecraft separation from the Centaur will occur about 93 minutes after liftoff for the first May 5 launch opportunity as the spacecraft is approximately over the Alaska-Yukon region.

US polar orbit Launches are usually from California due to geography. See How does one dogleg from Florida to a sun-synchronous orbit? for more on that.

But high energy interplanetary launches usually take advantage of the ~0.4 km/sec delta-v "kick" from the rotation of the Earth by launching Eastward.

Why will this launch use a polar parking orbit?

below: "NASA's InSight to Mars undergoes final preparations at Vandenberg AFB, Calif., ahead of its May 5 launch date." Credit: NASA/JPL-Caltech

NASA InSight Spacecraft

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  • $\begingroup$ Slightly related: Why would InSight's arrival date at Mars be fixed, and independent of the launch date? $\endgroup$
    – uhoh
    Apr 18, 2018 at 12:53
  • $\begingroup$ Also related: space.stackexchange.com/questions/25831/… $\endgroup$
    – Russell Borogove
    Apr 18, 2018 at 15:21
  • $\begingroup$ @RussellBorogove Thanks for the link. It looks like the polar orbit is just a result of the choice of coasts, rather than anything related to the orbit or trajectory to Mars. And that is just is a result of availability. I don't see a way to make this question any different. So just close as duplicate? $\endgroup$
    – uhoh
    Apr 18, 2018 at 15:33
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    $\begingroup$ Well, if someone can come up with more information about the trajectory, the answer might be interesting. If this is just a pop sci article describing a 160º azimuth flight (forced by Vandenberg's geography) as "toward the south pole", then the other QA covers it, but if the launch is going closer to polar than that, there might be a reason. $\endgroup$
    – Russell Borogove
    Apr 18, 2018 at 15:36
  • $\begingroup$ @PearsonArtPhoto I'll rollback and post this as a new question. I'd thought the comments above made it clear about it being a duplicate. $\endgroup$
    – uhoh
    Apr 18, 2018 at 17:05

1 Answer 1

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Launching East to take advantage of the rotation of the Earth would violate the safety requirements for a launch (Casualty statistics). Basically, if the rocket exploded it could injure too many people, or the rocket stage might drop on someone. The options then become retrograde or polar. At least if they do an initial polar launch they aren't going opposite of the Earth's rotation.

As to why the East coast, the general thought is they had plenty of extra capacity, and didn't want to fight the schedule in Cape Canaveral. Plus they could have a "first" and get a bit more attention.

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  • $\begingroup$ This is of course what I meant by "geography" in the question. I haven't asked why US launches don't fly over the continental US. The rest of this answer is explained in this answer as noted in comments. I'm in the process of revising the question since as it stands it's a duplicate of that one. $\endgroup$
    – uhoh
    Apr 18, 2018 at 16:11
  • $\begingroup$ The "first" does sound plausible, I wonder if you can add this as a second answer to Why does InSight plan to launch from Vandenberg Note also, I've just finished the edit to the question. $\endgroup$
    – uhoh
    Apr 18, 2018 at 16:20
  • $\begingroup$ You stated "But high energy interplanetary launches usually take advantage of the ~0.4 km/sec delta-v "kick" from the rotation of the Earth by launching Eastward.". I answered why it launched polar, and not east or west, which was what you initially asked. $\endgroup$
    – PearsonArtPhoto ♦
    Apr 18, 2018 at 16:28
  • $\begingroup$ I assumed that interplanetary launches would want the kick, and so would launch from Florida, and the fact that this was launching from California meant somehow that a polar orbit was necessary for some strange reason; otherwise, why launch from California? Since that time, the comments pointed out that this has been answered already, and that the reason is not related to the orbital mechanics of polar orbits, but instead, scheduling. $\endgroup$
    – uhoh
    Apr 18, 2018 at 16:45
  • $\begingroup$ I've rolled back to the original question. $\endgroup$
    – uhoh
    Apr 18, 2018 at 17:08

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