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Mass of Jupiter using its moon Sinope

  • Thread starter johnq1
  • Start date
  • #1
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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!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
993
13
Let us use F = ma for the moon of Jupiter.

What is F?
 
  • #3
1,943
241
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)
 
  • #4
993
13
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]
 
  • #5
3
0
The mass of the Juipter moon Sinope is 8*10^16 but I looked that up, didn't figure it from what was given.
 
  • #6
3
0
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:
  • #7
1,943
241
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
 
  • #8
sophiecentaur
Science Advisor
Gold Member
24,288
4,314
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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