Kepler's Laws-What am I doing wrong here?

1. Jan 5, 2007

AznBoi

Kepler's Laws--What am I doing wrong here?

1. The problem statement, all variables and given/known data
Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22 x 10^5km. From these data, determine the mass of Jupiter.

2. Relevant equations
T^2=(4(pi)^2/GM)r^3

3. The attempt at a solution
I converted the radius from km to meters and substituted all of the numbers to their variables. However, I still came up with the wrong answer. Please help me check my work, thanks!

Divided r^3 from both sides:
(1.77)^2/(4.22x10^8m)^3= 4(pi)^2/(6.673 x 10^-11)M

Multiplied universal gravitation constant with M(mass) on both sides:
(4.1688x10^-26)(6.673x10^-11)M=4(pi)^2

solved for M: M=4(pi)^2/ (2.78184 x 10^-36)

I came up with: 1.4192 x 10^37

The correct answer is: 1.90 x 10^27 kg

What am I doing wrong? Thanks!!

2. Jan 5, 2007

chroot

Staff Emeritus
Don't plug in numbers until you're done with the algebra.

Solve for M:

$T^2 = \frac{4 \pi^2 r^3}{G M}$

$M = \frac{4 \pi^2 r^3}{G T^2}$

$r = 4.22 \cdot 10^8\,m$

$T = 152928\,s$

Plug into the equation, and you're done. You appear to have left your T in units of days, which is inconsistent with the SI mks system.

- Warren

Last edited: Jan 5, 2007
3. Jan 5, 2007

AznBoi

Ah I see.. So is every single variable in Physics in the SI unit? Thanks a lot!

4. Jan 5, 2007

chroot

Staff Emeritus
You're free to use any unit system you want, but you must be consistent. If you want to use G = 6.673 x 10^-11, you must understand its units:

$G = 6.67300 \cdot 10^{-11} \,\textrm{m}^{3} \,\textrm{kg}^{-1} \,\textrm{s}^{-2}$

That means that everything else in your equation must in terms of meters, kilograms, and seconds: the standard mks SI system.

If you really wanted to use days as your fundamental unit of time, you could, but G would be:

$G = 0.498136781 \,\textrm{m}^{3} \,\textrm{kg}^{-1} \,\textrm{day}^{-2}$

- Warren

5. Jan 5, 2007

AznBoi

Wow, thanks for all the superfluous replies! Okay, I understand it clearly now. Thanks to you!

6. Jan 5, 2007

chroot

Staff Emeritus
No problem. Try plugging in your values for r in meters and T in days, using the "version" of G that I provided.

- Warren

7. Jan 5, 2007

8. Jan 5, 2007

AznBoi

Yeah, sometimes I don't know if the vocabulary I've learned makes sense in what I say/write. Here is the definition: Being beyond what is required or sufficient. How come it doesn't work with chroot's replies?

9. Jan 5, 2007

chroot

Staff Emeritus
Well, the word "superfluous" usually means that something was pointless or needless, not that it was unexpectedly generous.

You might say "Man, this textbook is horrible! It's full of superfluous sidebars," in order to disparage it. You probably wouldn't tell your mom "Thanks for all these superfluous Christmas gifts!"

- Warren

10. Jan 5, 2007

AznBoi

Oh I see why berkeman said that to me then. Sorry if you thought of it as the other way. My vocabulary isn't that great and it's still developing. I mean I know the definitions to some words but I just haven't had much experience with them. That's why some of my sentences are weird lol. Alright, thanks for the vocab. lesson xD I'll try not to use the word superfluous as something that is too super. Thanks to both of you.