Question
Two satellites are in circular orbits around Jupiter. One, with orbital radius r, makes one revolution every 17 h. The other satellite has orbital radius 4.2r. How long does the second satellite take to make one revolution around Jupiter?
Answers
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.
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).
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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