What’s up with Rosetta? Part 2

Today’s update continues our 18 August post, What’s Up with Rosetta.

Here are some interesting numbers related to Rosetta’s path around the comet. Note that, in this post, we’re citing numbers as planned; reconstructed values are not available until after the flight dynamics teams do their detailed orbit determinations.

All times shown in UTC = CEST - 2

From 6 to 17 August, 09:00 UTC (during the first 100km pyramid orbit, also called ‘Big CAT’ in our last post) the minimum distance from the comet achieved by Rosetta was ca. 90 km on 11 August around 21:00 UTC.

On 17 August, starting at 09:00 UTC, we started the descent toward the next triangular orbit (‘LittleCAT’) as follows:

  • During the orbital arc from Sunday, 17 August to Wednesday, 20 August at 09:00 UTC, the minimum distance achieved was ca. 79 km on 19 August around 07:00 UTC.
  • On 20 August at 09:00 UTC, Rosetta conducted a manoeuvre to go further ‘down’; the distance at the time of the burn was ca. 92km.
  • Rosetta crossed the 79 km distance between 02:00 and 03:00 UTC on 21 August; and will be on a continuous decay until late in the evening of 22 August, when we reach ca. 60 km
  • On Sunday, 24 August at 09:00 UTC, we will do the manoeuvre to start the first arc of the second pyramid (‘Little CAT’); we will be at about 72 km.
  • At the crossing (point of closest approach) between 24 and 25 August, we will go below 60km

During the second pyramid (24 August until 3 September 09:00 UTC), the minimum distance will be about 52-53 km and will be reached on 25 August and 29 August (at ca. 09:00 UTC) and on 1 September at ca. 22:00 UTC, i.e. the mid-points of the three arcs.

On 3 September at 09:00 UTC (the end of the second pyramid), Rosetta will be again at about 72 km; from there onwards will be on an (almost) continuous decay toward the 30 km orbit. For this next leg we do not yet have the operational orbit – it is being prepared right now!

Comments

15 Comments

  • Simon Frederik says:

    Question : How much will the cenrtifugal force (due to the rotation of 67P) be at the ecuator of 67P and with the weak gravity there how will it balance these opposite forces?

    • Gerald says:

      Roughly calculated, the velocity at the comet's equator at a distance 2km from the rotation axis is about 0.28 m/s. The escape velocity is about 0.82 m/s for a simplified point mass with the current estimate of 1e13 kg. Hence we get a net attraction. The previous mass estimate of 3e12 kg has been too low by a factor of about 3, and led to an escape velocity of 0.46 m/s.

    • Paul Stewart says:

      Hi Simon,

      I do not know by exactly how much the rotation of 67P would reduce the effect of gravity for objects on its surface, but the centrifugal force would be small even in comparison to its gravity.

      Consider the same effect taking place on Earth, a much larger body that experiences a full rotation every 24 hours, compared to 67P's 12.7 hours. According to this article, the effect of Earth's rotation on gravity reduces the pull on objects at its equator by only 0.3%.

      Since Earth's gravity is ten thousand times that of 67P, and it looks from my ballpark estimate that the centrifugal force would be about 2,000-4,000 times less than that of Earth's, the proportional force (I could be way off here). This means that the total reduction of 67P's gravity at its equator would be around 1% or so. I'd be interested to see the actual numbers, but I think I have the scale about right!

      http://en.wikipedia.org/wiki/Earth%27s_gravity#Latitude

      • Gerald says:

        I'm getting up to 23%: Escape velocity 0.82 m/s divided by sqrt(2) returns velocity 0.58 m/s for a circular orbit. Velocity due to rotation is 0.28 m/s. That's 0.48 times the velocity of the circular orbit. This quotient needs to be sequared to 0.23, since the centrifugal force is proportinal to the square of the velocity.
        Calculation basis are: a rotation period of 12.4 h, distance from the rotation axis 2 km, point mass of the comet 1e13 kg.

  • TESTARD says:

    Triangular orbits of Rosetta on 67P comet approach.
    How had been forecasted these orbits in the software of Rosetta before launching as we didn't know, seen in details from Earth, the shape of 67P nucleus ?
    Or these erratic orbits had them been estimated after approach ? Keeping in mind the necessary minutes for an order to go and return from such a long distance ?
    Thanks for your reply.

    • Gerald says:

      It hasn't been forcasted in detail. The necessary burns are calculated on the basis of the measuremets.
      But at a distance of 100 km errors due to uncertain gravity are low. More precise data become more relevant for closer orbits.

    • Daavs says:

      Since the triangular trajectories are still pretty far away from the comet, its weird shape currently does not have a strong effect on how its gravity pulls on Rosetta. Recall that gravity is proportional to mass divided by distance squared: if the distance to the closest part of the comet is barely different from the distance to the furthest part, then its shape won't cause much inconsistency in the gravity force while Rosetta moves and the comet spins.

      The irregular shape will begin to have a much stronger effect on the orbits during the close characterization phase and lander deployment..

  • Donald Q says:

    What are the forces of these thrust maneuvers, in KG or Newtons, when doing course corrections?

  • Marco says:

    Thanks for the very detailed report!

  • Roland says:

    Why does Rosetta use so complicated triangular flybys? Why not head for a circular orbit right away?

    • Brian Deschene says:

      Since 67P's exact shape, density, gravity field was not known before arrival, it was not possible to plan the specifics on entering orbit beforehand. Also, not until close observation was it possible to pick where Philae would land which the eventual orbit has to accommodate. This triangular fly-around is being done to only use fuel in three short bursts during each circuit to go around 67P to survey for landing sites and simultaneously determine 67P's gravity field for eventual traditional orbit.
      See separate post...
      Determining The Mass of Comet 67P/C-G

  • Alte kalt says:

    Use

    http://www.wsanford.com/~wsanford/calculators/gravity-calculator.html

    To calculate, set the comet mass to 1E 13 kg/m3
    The radius is about 2000 m give and take a bit depending where you are on the surface.
    The rotation period is 12.7 h
    The the circle is 6.28* r and out of this you ca calculate the orbit velocity. Then use

    http://www.calctool.org/CALC/phys/newtonian/centrifugal

    To calculate the centripetal force and, heureka keep hanging on with fingers and toes, also try to calculate the geostationary orbit radius.

    With this animation you can estimate the best and worst case situation and their properties
    http://www.esa.int/spaceinimages/Images/2014/07/Rotating_view_of_comet_on_14_July_2014

    Got surprised?

    • Alte kalt says:

      Ah yes, i just forgot the hard part and that is the effect of the shape and density profile of this body, still once you are settled on the comet and don't tremor too much you will stay put even in the worst case position. It still would be a free fall feeling.

  • Alte kalt says:

    The circular geostationary orbit is about 3280 m radius plus minus the errors of mass, so lets say about 3 km. but the rosetta will, as i can see, go in the opposite direction as the comet spin direction is making it a fraction harder to pinpoint the landing spot of Philae. Still this is a low speed mission in every aspect.

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