# Re: EUVE reentry probabilities

From: Bjorn Gimle (b_gimle@algonet.se)
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

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) sq.km., 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 ----- >
>
> http://wwwvms.utexas.edu/~ecannon/oddsofseeingareentry.htm
>
> 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