Kepler's 3rd law to determine mass of jupiter

AI Thread Summary
The discussion focuses on using Kepler's 3rd law to calculate the mass of Jupiter from a graph of log10(a) vs log10(P). The user has created a graph with a trend line of 3/2 but is confused about how to derive the mass from it. They initially used the equation m1+m2 = (4*P^2*a^3)/(G*P^2), which led to an incorrect large value. Other participants point out potential errors in the user's equation, suggesting a misprint involving the use of Pi instead of P. The conversation emphasizes the importance of verifying equations against textbooks or class notes for accuracy.
bemc
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my problem involves creating a graph of log10(a) vs log10(P) and to calculate the mass of Jupiter from the graph.

I have created the graph and it seems to be right since the of the trend line is 3/2. My problem is that I am unsure how to go about calculating the mass from the graph.

I used the orbital period and length of the semi-major axis of the Galilean satellites and 3 others to get the equation y = 1.5012x - 8.1973

I have tried using m1+m2= (4*P^2*a^3)/(G*P^2), though that produced a rather large value and doesn't utilize the graph at all.

Any help or nudges in the right direction would be greatly appreciated, Thanks!
 
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You might try posting this same question in the Homework Questions section of PF
https://www.physicsforums.com/forumdisplay.php?f=153

They are good at giving nudges without actually doing the problem for you.

It looks to me as if you have a typo or misprint in your equation. You wrote a P instead of a Pi (the number 3.14...)
at one point.

m1+m2= (4*P^2*a^3)/(G*P^2)

Better check in your textbook or your notes from class. A plain letter P would stand for the period.
On the other hand Pi^2 is a number roughly about equal to 10. So 4*Pi^2 is roughly about 40.
Once you check to make sure you don't have any major misprints like that, then if you need help
you could see what you can learn at the PF Homework forum.
 
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thanks for the input - it is actually P, being the period, in my equation, not Pi. I hadn't realized that I had posted in the wrong area.
 
Seriously, you need to check your equation in the textbook or the class notes.
In the way you have written it, you have P^2 both in the numerator and in the denominator.
In one case that is correct, it is supposed to be the period squared.
In another case it is not correct---where you have written P^2 you should have written Pi^2.

The Kepler law equation, as usually written, has a pi^2 in it, and what you have written does not. So it looks to me like you have screwed up the equation. Better check.
 
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