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To: driftdiver

“17,000 is a little slow for it to be in orbit.”

Depends on how high it is . . .


20 posted on 06/04/2012 6:23:38 PM PDT by Born to Conserve
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To: Born to Conserve
Anything lower than 90 minutes requires a lot of input power because of the atmosphere and that's... well. I'll let you calc the orbit. A cook can do it.

/johnny

25 posted on 06/04/2012 6:26:37 PM PDT by JRandomFreeper (Gone Galt)
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To: Born to Conserve

This is true!


26 posted on 06/04/2012 6:26:49 PM PDT by coldbluesteel (Endo)
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To: Born to Conserve
Yeah. v2/r = g(R/r)2, so v = sqrt( g R ) sqrt( R/r ).

sqrt( g R ) represents the speed of an orbit at distance R, or about 7901 m/sec = 17674 mph and the speed of higher orbits are lower by the second factor.

Then we require r/R =( 17674/17000 )2 ~= 1.08, giving an altitude of 0.08 R or about 317 miles .

69 posted on 06/04/2012 8:12:02 PM PDT by dr_lew
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