# What is the Mass of Jupiter?

1. Nov 15, 2007

### Eternalmetal

1. Europa was one of the moons discovered by Galileo. It has a mass of 4.9E22 kg and a period of 3.6 days and distance of 6.7E8 m. What is the mass of Jupiter?

2.
3.6 days = 311040 sec
F = G m m / r^2
F = Mv^2 / r
T^2 = (4pi^2/Gm) R^3

3. I threw the numbers into Kepler's formula and get a negative exponent answer, and at this point im sort of lost. Somebody please put me in the right direction. And no I dont know exactly what the distance is from, but assume the obvious (whatever it may be) for the time being. Distance from the surface or from the center of Jupiter I dont know.

I spent more time dwelling over this, and used 3 formulas: v = 2pi r/t to find the velocity of the moon's orbit. Then I used Kepler's formula T^2 = 3E-19 R^3 to find the radius of Jupiter (plugging 311040 sec into T). I think used the velocity and radius I calculated to find the mass of Jupiter using the formula v = sqrt(G m/r). I derived this formula by comparing centripetal force equation with Newtons law of universal gravitation (probably a memorized formula in most cases I would assume). Doing all of this, I got a value of 1.88E28. I double-checked with wikipedia and I am off by a decimal point (according to them). I assume I am now on the right track, but could someone just assure me that I used the right logic to come to my answer?

Last edited: Nov 15, 2007
2. Nov 15, 2007

### mgb_phys

I think you just did the arithmatic wrong
Kepler's law (t/2pi)^2 = a^3 /G(M+m)

G(M+m) = 4pi^2 a^3 / t^2, where G=6.6E-11, I get the right answer (1.86E+27kg)

3. Nov 15, 2007

### Eternalmetal

What did you use/how did you calculate the "a" value? (I am assuming you mean radius, at least thats how we learned it).

4. Nov 15, 2007

### mgb_phys

Yes 'a' is radius ( strictly semi-major axis of the ellipse, hence a )
You are given it as 6.7E8 m in the question!

5. Nov 15, 2007

### Eternalmetal

Interesting, I rechecked my work plugging in 311040 for T, 4.9E22 for one of the masses, and 6.7E8 m for the radius, and did indeed get it wrong, again. I tried this time, following your formula and got it right. Sorry for my stupidity, ive just been extremely stressed out lately. I hope I dont screw up my math on my test tomorrow, although I usually am pretty good with that part of physics. Thanks for your help.

But would the method I used earlier (and got the answer slightly wrong) still work for this? Considering 4pi^2/Gm is a constant, I am assuming my logic is not flawed. However, I guess I should just stick with using one formula to find the answer, now that I can.

Last edited: Nov 15, 2007
6. Nov 15, 2007

### mgb_phys

Your formula is for the case for the mass of the moon being negligible, it's what I did first because I misread the question as Io - which is pretty small.

Remember to check units and do a quick order of magnitude estimate from the exponnets.
It's easy to miss a +/- in an exponent on a calculator.

Good luck.