Calculating orbital speed/period

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The discussion focuses on calculating Jupiter's orbital speed and period relative to the Sun using the given mass of the Sun and Jupiter's average orbital distance. The calculations yield an orbital speed of approximately 98,200 mph and an orbital period of about 11.88 Earth years. The formulas used include Vorb = sqrt(GM/r) for speed and T = (2pi*r)/sqrt(GM/r) for the period, with the distance converted from AU to meters. Participants are encouraged to verify the results by comparing them to known data about Jupiter's actual orbital period. The calculations appear to be correct based on the provided parameters.
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Homework Statement


The mass of the sun is 2x10^30kg and the planet Jupiter averages an orbital distanceof 5.2 AU. Calculate the orbital speed (in mph) and orbital period (in Earth years) of Jupiter relative to the Sun: One AU is one Astronomical Unit which is the average earth-sun distance... 1 AU = 1.5x10^11m.

Homework Equations



Vorb= sqrt(GM/r)

T=(2pi*r)/sqrt(GM/r)

r = (1.5x10^11m)(5.2) = 7.8x10^11m
G = 6.67x10^-11 Nm^2/kg^2
M= 2x10^20kg

The Attempt at a Solution



I plugged in the numbers and ended up with...

Vorb = 4.39 x 10^4 m/s = 98199.91 mph
T= 374751912s = 11.88 years

Does this look right?
 
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Why don't you look up the orbital period of Jupiter and see for yourself?
 

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