If we orbit a non-spherical asteroid or moon at a sufficient distance I believe that we can consider it to be a point mass. Therefore we can take up a conventional orbit.

Assuming our lander craft is of insignificant mass compared with the body we wish to orbit and land on, how do we cope when we get near to the surface?

Presumably an orbit will become more and more chaotic the nearer we get. As we try to land, unless we somehow synchronise with the body's rotation things will be equally tricky.

What implications did this have for Rosetta's Philae and what implications will it have for much larger and maybe more irregular bodies?

Is there some rule-of-thumb method to get an approximate answer or do spacecraft simply have to make minute by minute adjustments?


I understand mathematics to a reasonable level but really I'm more interested in the practicalities of such landings. To what extent can they be pre-calculated and to what extent must they be adjusted on the fly.


For bodies small enough to be non-spherical, the mass is relatively low (as an example, Rosetta's bounces took a long time - it touched the surface at 15:34, 17:25 and 17:32 GMT comet time) but the centre of mass is still the point the lander will orbit.

So a chaotic orbit is not a problem here

What will be problems are:

  1. trying to arrange for the lander's velocity to match that of the landing site. For example if the body is spinning rapidly and an outcrop spins round into the path of the lander that could end badly
  2. staying on the surface - Rosetta tried to use harpoons. Another proposed solution was a low powered rocket pointing upwards
  3. while small bodies won't have an atmosphere as such, outgassing and dust can cause problems

I'm assuming 1 is the one you are most interested in here. The solution Rosetta favoured was to pick a site that was on one of the outer surfaces, rather than on the 'neck' of the comet, as this offered the cleanest route to landing. A landing on the neck would have necessitated flying into a channel on a spinning body, and that would make the calculations much harder.

| improve this answer | |
  • $\begingroup$ Thanks - Would your first paragraph also apply to Hyperion - moon of Saturn? I believe that has a mass of about 100,000 times that of 67P/Churyumov-Gerasimenko. $\endgroup$ – chasly - reinstate Monica Jul 30 '15 at 8:08
  • $\begingroup$ I would suggest so, however, read astronomy.stackexchange.com/q/2092/43 - the answer is not well defined. $\endgroup$ – Rory Alsop Jul 30 '15 at 8:17

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.