In the video of the recent Iridium-6/GRACE-FO launch at about T+ 08:13 (curently at 24:12 in the video) the announcer says:

Now in order to fly an efficient trajectory for both of the payloads, we’re actually now running the second stage engine at lower power. That means the burn will be longer. For folks who used to seeing us shut down the upper stage engine nine to nine and a half minutes in to flight, today we’ll actually be shutting down the engine just past T+10 minutes into flight. As planned for today, the longer burn at lower power obviously takes more time, but gives us a more efficient trajectory.

GRACE-FO was deployed in a ~500 km circular orbit, which was done with a single burn to circular trajectory, rather than using an initial elliptical orbit plus a second burn to circularize.

There was then a roughly 45 minute coast phase in this nearly-circular orbit, followed by a short (~8 second) prograde burn putting the 2nd stage in an elliptical orbit, presumably with an apogee of either 666 km for Iridium Next "storage" or 780 km, the target altitude for operational satellites. (This was then followed by a short, 8 minute additional coast.)

Question(s): Why is the longer, lower power first 2nd stage burn more efficient? And what is the purpose of the ~45 coast phase, why couldn't the 2nd burn and then Iridium deployment start relatively soon after the GRACE-FO deployment? Since the quote includes "...in order to fly an efficient trajectory for both of the payloads..." are these two related somehow?

note: I've asked these as a single question because they may be linked. If they are not, they could be asked separately, but at the moment I think the answers may have some overlap.


1 Answer 1


The coast appears to be to align for a plane change. The Iridium satellites are at an 86.4 degree inclination while GRACE-FO is at 89 degrees.

Inclination changes are expensive, but the F9 2nd stage can provide a lot of $\Delta v$. According to the webcast, that 8 second burn took the speed from 27442 km/hr to 27582 km/hr, about a 40m/sec change in speed or 1/2g. The perpendicular plane change needs about 300 m/sec, or about 3.5g for 8 seconds. Together, that should be within the capability of a F9 second stage with low tanks and a partial payload.

  • $\begingroup$ I see, so if that were true, then the 2nd burn is not completely prograde. It is somewhat, because the altitude begins to rise significantly during deployment. Would 8 seconds really be long enough for a ~2.6 degree plane change? It's a small change, but plane-changing is an expensive delta-v maneuver, and the burn happens over Madagascar (roughly 20 degrees South). $\endgroup$
    – uhoh
    Commented May 23, 2018 at 6:06
  • $\begingroup$ TLEs do show 43476, 7 at 89.0 degrees, and 43478-82 at 86.7 degrees already, so it does make sense. $\endgroup$
    – uhoh
    Commented May 23, 2018 at 6:57
  • 1
    $\begingroup$ @uhoh Added a bit on the numbers needed $\endgroup$ Commented May 23, 2018 at 14:04
  • 2
    $\begingroup$ @uhoh Nothing definite. The GRACE-FO orbit is comparatively low and circular. Perhaps the usual transfer-burn-plus-circularization burn approach would have had two separate burns too close together. Has SpaceX ever done a hot restart of a 2nd stage? $\endgroup$ Commented May 25, 2018 at 4:18
  • 1
    $\begingroup$ That's a fascinating question on its own, and worth asking. Right now you are at Q/A = 0/41, why not give it a try? $\endgroup$
    – uhoh
    Commented May 25, 2018 at 4:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.