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September 23, 2014

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Posted by **jimmyjonas** on Tuesday, February 22, 2011 at 8:48am.

- physics -
**tchrwill**, Tuesday, February 22, 2011 at 11:11amThe orbital period is defined by

T = 2(Pi)sqrt[r^3/µ)] where T = the period in seconds, r = the orbital radius and µ = the planet's gravitational constant.

17hr = 61,200 sec.

61,200 = 2(Pi)sqrt[r^3/µ] from which µ = r^3/94,873,103

At a radius of 4.2r,

T = 2(Pi)sqrt[74.088r^3(94,873,103/r^3] = 23.288 hours.

- physics -
**tchrwill**, Tuesday, February 22, 2011 at 1:15pmOOPS - the hand was quicker than the eye.

The orbital period is defined by

T = 2(Pi)sqrt[r^3/µ)] where T = the period in seconds, r = the orbital radius and µ = the planet's gravitational constant.

17hr = 61,200 sec.

61,200 = 2(Pi)sqrt[r^3/µ] from which µ = r^3/94,873,103

At a radius of 4.2r,

T = 2(Pi)sqrt[74.088r^3(94,873,103/r^3] = 146.32

hours.

The same result derives from Kepler's third law, T/To = [R/Ro]^(3/2) where

T/17 = [4.2R/R]^(3/2).

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