5
$\begingroup$

Reading this question, what is the best way to make sure most of ISS (or any other large object planned to be destroyed in the atmosphere) will burn and break into relatively small pieces in the atmosphere?

For instance, will it be a retrograde burn? Or will there be some radial composant in this burn, in order to raise apogee, and make a more brutal/steeper entry? For a given amount of de-orbit fuel, what's the best de-orbit burn profile?

Edit: It may be completely irrelevant in comparison with reality, but:

I tested something using stock KSP. Initial conditions (using orbit cheat) is perfectly circular orbit at 86750m Spacecraft setup is one large command pod (white apollo like) one heatshield, two sepratrons (simulating identical de-orbit delta V)

First de-orbit burn was retrograde. resulting perigee was 52966m, ablative material used during re-entry : 153

Second de-orbit burn was 45 deg. between retrograde and radial. Resulting perigee was 60355m, ablative material used : 161

Which would mean, for identical de-orbit delta V, burning with radial composant raises apogee immediately after burn, but leads to higher perigee too (less steep entry) More heat is produced, and landing site is less predictable.

$\endgroup$
  • $\begingroup$ There is no ablative material and no heatshield designed for a reentry of the ISS.The most important goal is to hit the surface of earth in a region where nobody could be injured or killed and nothing valuable could be destroyed. Saving fuel should not be top priority. $\endgroup$ – Uwe Jan 15 '18 at 17:59
5
$\begingroup$

There is no need to have a steeper angle, and in fact, that's probably counter-productive. As explained in this thorough answer; the quickest deorbit would be to have a roughly circular orbit that is stable for a few months (270 km was proposed), and then lower the perigee to below 100 km, the lower the better. This lowering should happen such that the station won't reach below, say, 180 km or so until it reaches the start of the safe path. This will have the lowest energy, and the most time, to allow for deorbiting the station.

If one makes the path steeper, it increases the likelihood of some part of it making it through the entire process.

Bottom line, wait until the station will only be stable for a month or so, and then burn as hard as you can at the apogee, until you are out of fuel. That will give you the best deorbiting capability for something like the space station.

$\endgroup$
  • $\begingroup$ Where do these values (270 km, 100 km, 180 km...) come from? Are you calculating them in your head or are they from sources? If sourced, could you please cite the sources? If you are calculating them, please explain how. Right now this could be totally wrong, there is no way to tell because there are no sources cited. How can future readers know if you are just making this up randomly, or if it is correct? $\endgroup$ – uhoh Jan 12 '18 at 3:18
  • $\begingroup$ These are based off of my experience, and from observing other missions. 100 km is pretty much guaranteed to reenter, especially for something as non-aerodynamic as the ISS. In fact, the most extreme numbers, 270 km and 100 km, came from the linked question as well. $\endgroup$ – PearsonArtPhoto Jan 12 '18 at 3:36
  • $\begingroup$ Could you edit your answer and note where the values are coming from then? Comments are considered temporary and once clarifications are established, they should be incorporated in the post itself. $\endgroup$ – uhoh Jan 12 '18 at 3:44
  • $\begingroup$ The question was would a steeper trajectory matter. While I did include some numbers in there, where the numbers came from doesn't matter so much as answering the question, which was would a steeper trajectory help things in any way. $\endgroup$ – PearsonArtPhoto Jan 12 '18 at 3:48
  • $\begingroup$ I've made an edit, attributing the source of the explanation and values. Linking to better and more thorough answers improves the quality of both answers as well as the site in general. $\endgroup$ – uhoh Jan 12 '18 at 3:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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