Since the comet has such low gravity, will Rosetta be able to maintain a conventional orbit around the comet? If so, how long will that orbital period be? I am guesstimating about a day (24 hours). Altitude 3 miles, orbital path 20+ miles, escape velocity 1 MPH. 184.108.40.206 (talk) 04:05, 16 November 2014 (UTC)
- I believe that Rosetta isn't in a 'free' orbit. It does several powered 'turns' in each orbit (I think, every 60 degrees). So, in effect, it's flying a bunch of nearly straight lines at much higher than escape velocity. I read someplace that Rosetta's speed relative to the comet is about 25 meters/second (sorry, I don't remember where I saw that). According to our article, the orbital distance is 29km (roughly) - although at times, they've reduced that to as little as 10km and started out at 100km. If the orbit was circular, then the circumference of that orbit is 2 pi x 29000 meters, which is around 200,000 meters. At 25 m/sec we get an orbital period of around 7000 seconds...around 2 hours. For a more exact answer, we'd need more details about the shape of the orbit and the fuel burns to keep it like that. When it was out further from the comet, I believe the 'natural' orbit was 26 days.
- An orbit that long would be useless while interacting with the lander because there would be periods of many days when they'd be unable to communicate. Hence the powered turns. SteveBaker (talk) 15:45, 16 November 2014 (UTC)