Determine the mass of Jupiter using data about a moon

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Homework Help Overview

The original poster attempts to determine the mass of Jupiter using data from its moon Io, including its mass, orbital period, and mean distance from Jupiter. The problem involves applying concepts of centripetal acceleration and gravitational forces.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between centripetal acceleration and gravitational force, with the original poster attempting to derive the mass of Jupiter through calculations involving orbital parameters. There is a focus on the correct interpretation of velocity in the context of the problem.

Discussion Status

Some participants provide feedback on the calculations, pointing out errors in the original poster's approach. The original poster acknowledges a mistake in calculating velocity and mentions correcting it, leading to a successful result. However, there is no explicit consensus on the overall approach or methodology.

Contextual Notes

The discussion includes a note on the use of LaTeX formatting for clarity in mathematical expressions. There may be constraints related to the accuracy of the calculations and the assumptions made about the orbital mechanics involved.

Calpalned
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Homework Statement


Determine the mass of Jupiter using the data for the moon Io.
Mass of Io: ##8.9*10^{22}##
Period: ##1.77## Earth days
Mean distance from Jupiter in km: ##422*10^3##

Homework Equations


Centripetal acceleration ##a = \frac{v^2}{R} ##
Universal law: ##\frac{GMm}{R^2} ##

The Attempt at a Solution


##\frac{m_Iv^2}{R} = \frac{m_IM_JG}{R^2} ##
## v^2 = \frac{M_JG}{R} ##
##M_J = \frac{v^2R}{G} ##
##\frac{(2\pi R)^2R}{G} = 1.419*10^{37} ## kg. My answer = wrong
Just like in the post about binary star systems, I see no error in my calculations, yet I am ten magnitudes too great...
Correct answer is ##1.9*10^{27}##
 
Last edited:
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##2 \pi R## is not a velocity.
 
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gneill said:
##2 \pi R## is not a velocity.
Thanks Gniell. Once again I forgot to divide ## 2\pi R ## by T. I solved the problem again, corrected my mistake, and got the right answer.
 
Just a little ##\LaTeX## note: you can say ##8.9 \times 10^{22}## (8.9 \times 10^{22}) instead of using an asterisk.
 
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