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Go for Launch

In previous articles, I talked about optimal solutions for a transfer from a given non-nominal GTO to the geostationary orbit, basing my musings on a hypothetical example case. Now I present some results on what I think can be achieved in this case with a conventional propulsion system providing a delta-v of 1.5 km/s, which is the nominal transfer cost from GTO to GEO: Sort of like an industry standard. Many launch vehicles have a defined GTO that is specific to this launch vehicle and requires 1500 m/s of delta-v to reach GEO.

(Author’s Note: All I am providing here is my personal opinion and some results of mathematical calculations I have performed myself based on publicly available information. These results could have been obtained by anyone with a background in orbital dynamics. Nothing I write here shall be construed as reflecting the official position of my employer. Essentially, what we have here is a mathematical problem: A spacecraft is in a given initial orbit and shall reach a certain target orbit B with a certain amount of delta-v. What I am presenting is a mathematical solution to a given mathematical problem. Now, back to the blog article …)

In the example scenario I am regarding, the direct transfer from the achieved launcher insertion orbit to the geostationary orbit would cost around 2.2 km/s. This assumed that there no waiting period was applied during which orbital perturbations would cause some of the orbital parameters to drift to an overall orbital geometry that makes the insertion less costly. In the considered scenario, using natural drift would imply a waiting period until mid-September and lower the insertion cost to 1627 m/s.  This is less than 10% above the nominal value of 1500 m/s, and there are normally some margins on the propellant load, so it might be within the capabilities of the regarded hypothetical satellite.

If the propulsion system can’t deliver 2200 m/s and waiting until September is not an option, then the only way out is to accept a final orbit that is different from the geostationary orbit. In my example scenario, I am assuming an initial orbit that is inclined by 20.6 deg, with an argument of perigee of 232.9 deg and an apogee radius of 49,543 km. I am also assuming 1500 m/s delta-v capability, not one iota more. Anything that the system can deliver on top of tat is a welcome bonus.

I am not considering splitting the manoeuvres into multiple parts and using even limited waiting periods and the associated drift on the elements. That certainly may have some effect but I am just doing a math exercise here and I won’t go to such level of detail.

Right. Based on the above assumptions, one can certainly lower the eccentricity and one can lower the inclination, but not both to zero. Not with only 1500 m/s of available delta-v capability. There is an infinite amount of possible solutions. First of all, I just forget about the semi-major axis and the orbital period for the time being. Sure, the target orbit shall be geosynchronous, but that will just have to be fixed somehow as part of the final orbit insertion using the spacecraft’s ion propulsion system that is foreseen for station keeping on the operational orbit.

In order to handle this problem of an infinite number of options, what I do is as follows:

  • Choose four fixed target inclination values: 0, 2, 4 and 6 deg with respect to the Earth equator.
  • Compute the lowest possible eccentricity that can be achieved with up to three impulsive manoeuvres, which may add up to no more than a total delta-v of 1500 m/s. As I said, I’m not worrying about the semi-major axis right now.

So here is what I get:

Credit: Michael Khan: Inclination/eccentricity combinations achievable with  1500 m/s of delta-v in the given example scenario
Final inclination [deg] 0 2 4 6
Final eccentricity 0.5 0.4 0.346 0.292

Unsurprisingly, the closer to zero the target inclination, the larger the residual eccentricity. Obviously, the more delta-v is invested to reduce the inclination, the less is available to correct the eccentricity.

So, those are the results of my calculations.

Now, let’s see. If I remember correctly, a geostationary telecommunications satellite recently was launched into an initial orbit that is about as the one I assumed; the orbit was quite different from the targeted launch orbit. The current orbital elements in form of TLEs can be obtained here. At the time of writing this article, I see the data set from 2018/2/22 16:50:37 UTC. There, I read the following:

  • Eccentricity: 0.2883266
  • Inclination: 5.5445 deg
  • Argument of perigee: 273.839 deg
  • Apogee height: 47429 km (with means the apogee radius is 53807 km)

Taking the inclination and eccentricity values and comparing them to my table above, it looks as if the current orbital state of that real-world satellite happens to be right on the curve of the solutions I found for my example scenario. This would also indicate that for the transfer of that satellite, close to 1500 m/s of delta-v must already have been spent.

Recent announcements by the satellite operator on the web appear to indicate that the satellite will reach its final orbit by June 2018, which is indicative of an extended phase of ion propulsion use to reduce the inclination to zero and to circularise the orbit at the geostationary radius.



Ich bin Luft- und Raumfahrtingenieur und arbeite bei einer Raumfahrtagentur als Missionsanalytiker. Alle in meinen Artikeln geäußerten Meinungen sind aber meine eigenen und geben nicht notwendigerweise die Sichtweise meines Arbeitgebers wieder.

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