SpaceX launched two GEO comsats, and is about to launch two more, that use all-electric ion propulsion.

Now, I'm wondering, how much time would it take to actually make a satellite like that operational. After all, while ion propulsion is cool and lightweight, it has a very, very low thrust. (Eutelsat 117 West B apparently has only about 200 mN — that's millinewtons — of thrust)

Obviously I can just look it up for an operational satellite, but would be interested in calculating the answer for any arbitrary satellite.


I'm taking Eutelsat 117 W B as an example.

The simple, obvious answer is "about half a year" — this satellite was launched in March and entered service in October 2015.

But I wanted to figure out the answer with physics and come up with roughly the same answer:

The satellite has a mass of m = 2200 kg and thrust of T = 0.2N. It takes about dv = 1.6 km/s to circularize the orbit from GTO to GEO.

Therefore, we need a change of momentum of:

dp = m * dv = 1.6 km/s * 2200kg = 3.52 * 10^6 kg m/s

to achieve our final orbit. Since change in momentum is equivalent to impulse:

dp = J = T * t

we can calculate the time it takes, with our tiny thrust, to achieve that impulse:

t = dp / T = 3.52 * 10^6 / 0.2 = 1.76 * 10^7 s = 204 days

204 days is about 6 months and 3 weeks.

  • $\begingroup$ At least that seems to make sense to me — so I wanted to share my numbers, but I don't know if I'm not oversimplifying something. So if my physics is wrong, I'd love to hear how to properly calculate that for any satellite. $\endgroup$
    – radex
    Jun 6 '16 at 9:44
  • 2
    $\begingroup$ Possible oversimplifications: Eutelsat 117 West B is not actually in GEO (its orbit is not circular), so your dv might be off. And of course m is not constant, but decreasing through the “burn”, so you would have to figure out how much reaction mass is used, then solve a differential equation to get an even-more-accurate value. $\endgroup$ Jun 6 '16 at 12:14
  • $\begingroup$ I've added a related question. $\endgroup$
    – uhoh
    Jun 7 '16 at 11:48
  • $\begingroup$ It's always okay to accept your own answer if you are satisfied with it. $\endgroup$
    – uhoh
    Feb 26 '19 at 0:09
  • $\begingroup$ is that assuming it can burn the whole time instead of just when it's near the highest point? $\endgroup$
    – xaxxon
    Jan 7 at 22:59

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