Mass of Jupiter given a Moon's mass, orbital velocity and distance.

  • #1
Hello all, been meaning to make an account here and participate but haven't been around to it. So sorry that my first post is asking for homework help! :|

Homework Statement


The planet Jupiter has a moon Europa (m = 5x10^22kg) that is orbiting at a velocity of 14,000m/s at a distance of 7x10^8m measured from the center of Jupiter to the center of the moon. What is the mass of Jupiter?



Homework Equations



According to the lecture notes for class, we have the law of universal gravitation [itex]F=\frac{Gm1m2}{r^2}[/itex] and then the formula for "Velocity of a satellite in orbit" [itex]V=\sqrt{\frac{GM}{r}}[/itex]. There's more things like field theory, etc but may not be relevant.


The Attempt at a Solution



I can't seem to get a good result, google says the mass of Jupiter is 1.898x10^27kg although when I plug the numbers into the velocity formula and solve for M, i keep getting a mere 205604.

This exact same question is posted on yahoo answers, but the first guy's solution is something beyond my course, and everyone else just doesn't seem to be right.

I'm stuck, not sure what formula/equation I need to use and what throws me is that with all the given information I'm not sure if it's all needed or if it's there for distraction.

I would really like just a quick hint to get me in the right direction instead of an entire work through :)
 

Answers and Replies

  • #2
6,054
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The first guy uses Kepler's third law. The others use essentially your method. Their results are correct within the accuracy of the original data.
 
  • #3
Bandersnatch
Science Advisor
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contrivance said:
when I plug the numbers into the velocity formula and solve for M, i keep getting a mere 205604.
Here's a hint:
Dividing by a-b means the same as multiplying by ab
 
  • #4
The first guy uses Kepler's third law. The others use essentially your method. Their results are correct within the accuracy of the original data.
I see, this makes sense now. Well after all this stress last night I discovered I was making a calculator error. putting too much information in at once and forgetting to put parenthesis where needed to get the correct result. I see that our result isn't actually that far off from Google's and my professor has been known to round given information up or down significantly.

I solved for [itex]M[/itex] in the Velocity equation and ended up with about 2.0x10^27 as others did.

What's funny about that Yahoo answer post is I just noticed it was posted three days ago so it might have been posted by someone in my class.

Thanks friends!
 

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