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December 18, 2014

December 18, 2014

Posted by **Aaron** on Sunday, October 2, 2011 at 9:34pm.

- earth -
**SraJMcGin**, Sunday, October 2, 2011 at 9:58pmTry some of the following links for information:

http://www.google.com/search?q=how+to+find+the+orbit%2C+radius%2C+speed+%26+weight+of+a+satellite+moving+around+our+earth+4+times+a+day+&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

Sra

- physics -
**Aaron**, Sunday, October 2, 2011 at 10:33pmA 10.0 kg mass is 1.00m to the left of the origin, and a 20.0 kg mass is 3.00 m to the right of the origin, both on x-axis. Find the gravitational potential energy and the gravitational field on x-axis at a) x =-2.0m b.) the origin, c.) x = 4.00 (set Zero P.E @ infinity in terms of G.

- earth -
**tchrwill**, Tuesday, October 4, 2011 at 11:17amThe orbital period of any satellite derives from

T = 2(Pi)sqrt(r^3/µ) where T = the period in seconds, 6(3600) = 21,600, r = the orbit radius in seconds and µ = the earth;s gravitational constant.

Therefore,

21600 = 2(3.14)sqrt(r^3/1.407974x10^21)

from which r = 55,002,293 ft. = 10,417 miles.

The velocity derives from V = sqrt(µ/r)

V = sqrt(1.407974x10^21/55,002,293) =

The orbital height and period are totally independant or the weight.

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