I've had a lot of response to the Shuttle launch tracks I submitted
yesterday.
One question that keeps popping up is:
"How far from a ground track can the Shuttle (or any launch vehicle) be
seen?"
The formula for maximum surface distance on the Earth (in kilometers) an
object at HEIGHT (also in kilometers) can be seen is given by:
DIST = 6371 * {arccos[6371/(6371 + HEIGHT)] / 57.29577}
Formula from "The Satellite Experimenter's Handbook" published by the
American Radio Relay League and uses average Earth radius (ignores
oblateness). Book is kinda old (1985), but formula and
other Tracking Basics described in it still apply!
Check values:
If HEIGHT = 106km (like Shuttle 9 minutes into launch), DIST = 1,154km.
If HEIGHT = 1,460km, DIST = 3,953km.
If HEIGHT = 35,500km (Geosync band), DIST = 9,034km
You can alter the formula to give DIST in statute miles for HEIGHT input
(also in statute miles) by changing "6371" to 3959 in the three places. The
"6371" of course is the Earth's average radius in kilometers.
Of course, at the maximum distance given by the formula, the vehicle would
appear on the horizon! You have to be quite a bit closer to the ground
track to see the vehicle at a good altitude above the horizon.
For the Shuttle, the altitude in km is given below for Launch + (time
given). Data is from STS-88 flight, which is a bit different now that the
Shuttle uses OMS-assist thrusting, but these data should be close enough.
30 sec 1.218km
1 min 8.04
2 min 38.09
3 min 73.22
4 min 95.83
5 min 107.02
6 min 109.07
7 min 105.80
8 min 102.72
9 min 106.88
Hope this helps.
RICK BALDRIDGE
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This archive was generated by hypermail 2b29 : Mon Jan 15 2001 - 09:40:55 PST