RE: LAGEOS reentry

From: Ted Molczan (
Date: Tue Aug 31 2010 - 15:50:07 UTC

  • Next message: Dale Ireland: "RE: LAGEOS reentry"

    Derek Breit wrote:
    > Ummmm... Don't they mean the Uranium will decay in 8 million 
    > years, and are not referring to the satellite??
    Eight million years refers to the time to decay:
    Also, the LAGEOS satellites (76039A / 08820, 92070B / 22195) do not contain uranium; it is the
    conceptually similar satellites, Starlette (75010A / 07646) and Stella (93061B / 22824), that
    contain depleted uranium (at their core). 
    LAGEOS and Starlette/Stella were required to achieve highly stable orbits, which in turn required
    minimizing the effects of non-gravitational forces, mainly atmospheric drag and solar radiation
    pressure. Since those forces are proportional to cross-sectional area divided by mass, both
    satellites were made small and dense. 
    The LAGEOS satellites were placed in high orbits (~5900 km), far above the effects of the
    atmosphere; Starlette and Stella are in much lower orbits (800-1000 km). It might be supposed that
    the use of uranium in the latter was necessitated by the greater drag they experience, but I suspect
    it is more complicated than that. Data on shape, size and mass is available here:
    The relevant values, with calculated A/m are:
                        Dia    m    A/m
                         m    kg   m^2/kg
    Lageos     sphere  0.60  411  0.00069
    Starlette  sphere  0.24   47  0.00096
    So the areal density of the two types of satellites are the same order of magnitude. Had Starlette
    been made of solid uranium, its areal density would have been less than half that of Lageos, so,
    altitude may not have been the primary factor in deciding on a uranium core. At the much greater
    altitude of Lageos, perturbations due to solar radiation pressure are similar in magnitude to those
    due to atmospheric drag in LEO, which may explain the similarity in areal density of the two
    For a sphere, A/m = 1.5 / (D * d); where D is volumetric density, and d is diameter. That formula
    may be convenient for assessing trade offs between size and density.
    Ted Molczan
    Seesat-l mailing list

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