Posted by jimmyjonas on Tuesday, February 22, 2011 at 8:48am.
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] = 23.288 hours.
OOPS - 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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