Mass of Jupiter using its moon Sinope

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Homework Help Overview

The discussion revolves around calculating the mass of Jupiter using data related to its moon Sinope, specifically employing a modified version of Kepler's third law. Participants are exploring the relationship between the orbital period and distance of Sinope to derive Jupiter's mass.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to apply Kepler's third law and Newton's version to calculate Jupiter's mass. There are questions about the relevance of Sinope's orbital distance and period in the calculations. Some participants suggest using gravitational force equations to derive the mass.

Discussion Status

There is ongoing exploration of different methods to calculate the mass of Jupiter, with some participants questioning the validity of using Kepler's third law in this context. Multiple interpretations of the problem are being discussed, and some guidance has been offered regarding the correct application of formulas.

Contextual Notes

Participants note confusion regarding the wording of the problem and the necessity of knowing Sinope's mass. There is also mention of discrepancies in calculated values, indicating potential misunderstandings or misapplications of the formulas involved.

johnq1
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Homework Statement


Using a modified version of Kepler's third law and data about Sinope calculate the mass of Jupiter.

Keplers Third Law: p^2=a^3
Newton's Version: p^2=a^3 / M + M
p^2 = a^3/M
M=a^3/p^2
Sinope's period of orbit = 2.075 years
Average orbital distance of Sinope is 0.158 AU

Homework Equations



Keplers Third Law: p^2=a^3
Newton's Version: p^2=a^3 / M + M
p^2 = a^3/M
M=a^3/p^2

The Attempt at a Solution



I need Jupiter in solar mass units and then in kg by multiplying it by the mass of the sun (2.00*10^30).

Here is what I have so far but its not making sense.
p=2.075
p^2=4.305625
a^3=4.305625
a=1.62684209

if m=a^3/p^2 and p^2=a^3 I am just going to come up with the same answer of 2.075 which doesn't seem to help me at all. I think the problem is worded bad and I am confused as to where the orbital distance of Sinope comes into play.

Any tips on how to proceed would be appreciated!
 
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Let us use F = ma for the moon of Jupiter.

What is F?
 
Keplers Third Law: p^2=a^3

This is only valid for orbits around the sun (or a star with the same mass),
so you can't use this.

You can find an answer only using p^2 = a^3/M (with M expressed in solar masses)
 
Can we use:

F = ma ( for the moon of Jupiter)

[itex]\frac{GMm}{R^{2}}[/itex] = mR[itex]\omega[/itex][itex]^{2}[/itex]

Gm = [itex]\frac{R^{3}4\pi^{2}}{T^{2}}[/itex]

T[itex]^{2}[/itex] = [itex]\frac{4\pi^{2}R^{3}}{GM}[/itex]
 
The mass of the Juipter moon Sinope is 8*10^16 but I looked that up, didn't figure it from what was given.
 
You can find an answer only using p^2 = a^3/M (with M expressed in solar masses)

so would it be...
M= .158/2.075
M= .076144578313253 AU
.076144578313253(2*10^30) = 1.52289157 × 10^29kg

That's off by over 37% so that doesn't seem right.
 
Last edited:
johnq1 said:
You can find an answer only using p^2 = a^3/M (with M expressed in solar masses)

so would it be...
M= .158/2.075
M= .076144578313253 AU
.076144578313253(2*10^30) = 1.52289157 × 10^29kg

That's off by over 37% so that doesn't seem right.

It's about a hundred times too heavy. You used M = a/p, and not M = a^3/p^2
 
johnq1 said:
The mass of the Juipter moon Sinope is 8*10^16 but I looked that up, didn't figure it from what was given.

Why would you need to know the mass of Sinope?
Which is the relevant mass?
 

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