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Orbital velocity of Rosetta

  1. Dec 3, 2015 #1
    Hi All,

    I know that Rosetta was orbiting around 67P/Churyumov–Gerasimenko. This asteroid has extremely low gravitational force acting on Rosetta. I believe that if the celestial body is heavy (mass), then there would be greater gravitational pull on the satellite and the orbital speed of it's satellite should be fast near it's surface so as to avoid collision.

    I have below questions.

    1. Since there is almost zero gravitational pull on Rosetta by 67P/Churyumov–Gerasimenko, does the Rosetta orbit the comet very slowly ?

    2. The comet itself rotates. In order to make observations of the comet (Geo-stationery orbit), the orbital speed of Rosetta should match with the rotational speed of comet. Now assume if rotational speed of comet is very fast. Now in this case, Rosetta should match that speed so that it will be stationery to the comet. But at that speed it will fly away from comet. So, what maneuvers they put in place to achieve this ?

    Thanks...
     
  2. jcsd
  3. Dec 4, 2015 #2

    Simon Bridge

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    1. Fast and slow are relative terms. What are you comparing with?
    2. To orbit faster the spacecraft has to apptoach closer. But you do not need to match orbital velocity to rotational velocity to map the surface... it's actually counter productive. ie satellite mapping of the Earth is easier from a polar orbit, do you see why?
     
  4. Dec 4, 2015 #3
    Thanks Simon. I intend to keep Rosetta stationery relative to comet. Hence I would like to match the orbital and rotational speed.
     
  5. Dec 4, 2015 #4

    davenn

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    Why ?
    it's not a very good idea from an exploration point of view
     
  6. Dec 5, 2015 #5

    Simon Bridge

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    ... what is wrong with getting Rosetta into a stationary orbit then?

    Compare: if you wanted a geostationary orbit, which is a stationary orbit about the Earth, then you need satisfy particular conditions...[/I]
     
  7. Dec 5, 2015 #6

    SteamKing

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    Yes. The escape velocity for 67P is approximately 1 (one) meter per second. Orbital velocity, of course, must be less than this velocity.

    https://en.wikipedia.org/wiki/67P/Churyumov–Gerasimenko
    'stationery' is stuff you write letters on.

    'stationary' is standing still in one place.

    If Rosetta is placed in a stationary orbit w.r.t. 67 P, it can only map the surface directly below its cameras.

    https://en.wikipedia.org/wiki/Rosetta_(spacecraft)
     
  8. Dec 6, 2015 #7

    Janus

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    I don't know if that is possible. Rosetta presently orbits at ~29 km with an orbital speed of ~0.15169 m/s and a period of ~14 days. the rotational period of 67P/Churyumov–Gerasimenko is 12.4 hours. The orbital radius for its mass if you treat the mass as a point source is 3.229 km. However, the body is an irregular shape with some dimensions greater than this distance, so the point source solution is not applicable. There just might not be a stable geosynchronous orbit.
     
  9. Dec 6, 2015 #8

    Simon Bridge

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    @Janus: was hoping to get OP to realise that and write it out.
    The process of thinking about it should (fingers crossed) also suggest a solution...
     
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