Re: EUVE reentry probabilities

From: Bjorn Gimle (
Date: Wed Jan 30 2002 - 13:06:28 EST

  • Next message: Harro Zimmer: "EUVE Decay Forecast 02"

    The area of a band around a latitude is R*R*width * 2 * pi * cos(lat), not
    considering oblateness.
    So between two latitudes, it is the integral, or 2*pi*(sin(lat1)-sin(lat2)),
    or for your example 4*pi*0.3987. For the whole sphere it is 4*pi, so 40% is
    your answer.
    The uninhabited space rarely comes into play - even if it decays W-E across
    Sahara, it would usually be visible from one of the coasts or central
    states, and just two or three scattered nomads would be sufficient to cover
    the area - but they would probably not post a message on SeeSat-L.
    If a decay is visible over a rectangle (2*500) * (2000+2*500), that
    is 0.6% of the Earth's surface.
    Seen on a time scale, if EUVE has a mean motion of 16.55, it is above 10
    degrees from Honolulu on 9 passes/day, totalling 12 minutes/day. Adding 5
    minutes/pass of decay duration, increases its chances to about 4%, not
    considering daylight.
    This is because about 50% of Earth never saw EUVE, and Hawaii can see up to
    northern apex, ie a lot more passes than areas on lower latitudes.
    ----- Original Message ----- >
    > Two of the obvious factors are 50% daylight and 70%
    > ocean (and nn% of land uninhabited), and there's the
    > Earth's equatorial bulge (and maybe its oblateness),
    > and of course the percentage of cloud cover (which I
    > don't know).  What is the percentage of the Earth's
    > surface within the tropics (+/- 23.5 degrees)?  I
    > don't know how to do that math.  So I would like to
    > learn more of the technicalities on this topic.
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