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.

edit: While I'd originally asked why this launch will use a polar parking orbit, it's clear to me now from answers to Why does InSight plan to launch from Vandenberg that the polar orbit just follows from geography, and that follows from launch site availability and scheduling.

The 59 to 66 minutes sounds like a 3/4 turn around the Earth, and then the 2nd burn of the second state will put InSight and its other payloads on a trajectory to Mars.

Question: How long will the 2nd burn of the Centaur 2nd stage last? How far from Earth (roughly) will it be when Insight is deployed? And will the Centaur do any further propulsive maneuvering (either main engine, or thrusters) to ensure that it doesn't follow InSight or the other payloads all the way to Mars?

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


I understand I am (3 years) late to the party, but I was recently linked to this question and figured I could determine a pretty accurate way to answer question 2 of 3 (how far from Earth at separation). I know the question asks for "roughly" how far but I have seen many of OP's questions looking for far greater detail on other topics so I am hopeful this will be appreciated.


We can use the $C3 = 8.204 \frac{km^2}{s^2}$ with the 322.14 second escape burn and 540.04 second coast between the end of the second stage burn and spacecraft separation to determine how far from Earth InSight was at separation/deployment.

To model the acceleration provided by the second stage you first need to determine how much $\Delta V$ the second stages gives the spacecraft:

  • $V_{park}=7798 \frac {m}{s}$ (using @Bob Jacobsen's answer of 185 km parking orbit)
  • $V_{PI}=11,393 \frac {m}{s}$ (using the C3 value @ 185 km altitude)
  • $\Delta V = 3596 \frac {m}{s}$ (subtract the two)

Above was done assuming an instantaneous burn (very not true), but we just needed a representative $\Delta V$ value and the errors this brings gets dealt with by distributing this $\Delta V$ over the burn interval of 322.14 seconds in an exponential fashion. For an assumed constant thrust the acceleration from a rocket stage (that is losing mass) is exponential w.r.t. time. The goal is to find a function that satisfies this: $$\int_{0}^{t_{burn}} a(t)dt = \Delta V = 3596 \frac {m}{s}$$ of the form $ a(t)=e^{\frac {t}{B}}$ (edit: this is not correct, but a good approximation that eliminates the need for detailed specifications of the booster stage and mission specific data). $a(t)$ is the acceleration (in $\frac {m}{s^2}$) of the combined upper stage and spacecraft during the escape burn. I used an iterative algorithm to find $B=85.6549$ for this scenario.

We can now simulate both the burn and coast phases of the trajectory with any trajectory sim of your choosing to get the following (or close to) results: InSight Separation

Answer: The separation occurs at an altitude of 1933 km in this simulation.

At the end of the simulation the C3 was recalculated: $C3_{sim}=8.111 \frac {km^2}{s^2}$ or 1.14% different from the actual C3, meaning any errors from prior assumptions is mostly eliminated. One could further refine the value of $B$ to drive the simulated C3 to the actual C3 but I leave this as "an exercise for the reader."

  • 1
    $\begingroup$ Hey that’s an excellent answer. Thanks for sharing. $\endgroup$
    – user39728
    Apr 27 at 2:46
  • 1
    $\begingroup$ Thank you for such a substantial answer! I'm out of my league; what is $B = 85.6549"$ for this scenario"? $\endgroup$
    – uhoh
    Apr 27 at 3:47
  • 1
    $\begingroup$ @uhoh that is the value of $B$ that solves $\int_{0}^{t_{burn}} e^{\frac {t}{B}}dt = 3596 \frac {m}{s}$ $\endgroup$ Apr 27 at 11:03
  • 1
    $\begingroup$ Oh I see it now, thanks! $\endgroup$
    – uhoh
    Apr 27 at 11:13

The NASA press kit has some information. It shows the 2nd and last Centaur burn from second 4736.9 to 5059.8, so a total of 322.9 seconds.

It also has some information on post-separation behavior:

Shortly after the release of InSight, the Centaur will begin an avoidance maneuver taking itself out of the spacecraft's flight path to avoid hitting either the spacecraft or Mars. Shortly after InSight separation, MarCO-A will be released by its CubeSat dispenser, the Centaur will roll 180 degrees, MarCO-B will be released, and then the Centaur will complete its avoidance maneuver.

I think that means there are two additional maneuvers after the 2nd Centaur burn, but it's not clear how large those are i.e. thrusters or main engine.

They're not so clear about altitude. It does say that "the parking orbit is nearly circular at an altitude of 115 miles (185 kilometers)", so the start of the 2nd burn should be about there. Not sure what the end will be.

  • $\begingroup$ 79 minutes after launch the 2nd stage engine lights for a second time while it's in a circular 185 km parking orbit. At 93 minutes after launch is separation. I guess that's still not enough information though. $\endgroup$
    – uhoh
    Nov 23 '18 at 15:03

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