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January 26, 2015

January 26, 2015

Posted by **John** on Saturday, March 17, 2012 at 10:59pm.

- trig -
**Steve**, Sunday, March 18, 2012 at 3:14pmspeed is distance/time

for a circle, C=2pi*r

r = 3960+104 = 4064

C = 25534

speed = 25534mi/10.5day = 2432mi/day

- trig -
**tchrwill**, Monday, March 19, 2012 at 1:48pm/r] sqrt(sqrt[(1.407974x10^16)/(3960+218.4)5280]

= 25,262 feet per second.

The orbital period is

T = 2(3.14)sqrt[22,061,952^3/1.407974x10^16] = 5487 seconds or 91.45 minutes.

- trig -
**tchrwill**, Monday, March 19, 2012 at 2:04pmPart of my previous reply was lost in the posting process.

The orbital radius is 3960 + 2.1(104) = 4178.4 miles or 22,061,952 feet.

The alleged time to complete one orbit is 10.5(24)3600 = 907,200 seconds making the derived orbital velocity

Vc = 152.7fps.

Unfortunately, the real orbital velocity required to remain in a circular orbit derives from Vc=sqrt(µ/r)=

where µ = the earth's gravitational constant and r = the orbital radius in feet =

sqrt[(1.407974x10^16)/(3960+218.4)5280]

= 25,262 feet per second.

The orbital period is

T = 2(Pi)sqrt[r^3/µ]=

2(3.14)sqrt[22,061,952^3/1.407974x10^16] = 5487 seconds or 91.45 minutes.

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