# Keplers Constant Question

1. Dec 5, 2006

### quickslant

I am given
K= 5.045 x 10^28
and T = 11.89 Earth years

I know that K = r^3/T^2

what i dont know is what should T be calculated in ? earth years/earth days/ seconds? etc? can someone tell me what T shoudl be calculated in?

2. Dec 5, 2006

I think $$K = \frac{T^{2}}{r^{3}}$$

$$T$$ is the period in any unit.

3. Dec 5, 2006

### quickslant

in my text book it says the other way around im not sure though..
so it doesnt matter what units i use? wont i get a larger or smaller number depending on units?

4. Dec 6, 2006

### OlderDan

You already have T in earth years. That's as good a unit as any, unless of course you have been asked to do something with that information that you haven't bothered to write down. What is the problem you are trying to solve and what are the units of K?

5. Dec 6, 2006

K = m^3/T^2

6. Dec 6, 2006

### vanesch

Staff Emeritus
7. Dec 6, 2006

### OlderDan

You still have not told us the problem you are trying to solve, and you have not given the units specified for K. T is not a unit. I know what they probably are, but I could tell you that

K = 5.045 x 10^28 Parsecs³/century²

and from what you have told us so far, I might be right

8. Dec 6, 2006

### quickslant

sorry i was trying to calculate the average radius of jupiters orbital

9. Dec 6, 2006

### quickslant

i believe what htey wanted is m^3 / Days^2

10. Dec 6, 2006

### quickslant

good lord i wish i never went to that site... way to many equations.. it look like graffiti for a second.. lol :surprised

11. Jan 20, 2007

### MattsVai

the textbook clearly mentions that T is to be calculated in seconds :)

12. Jan 20, 2007

### HallsofIvy

Staff Emeritus
Which textbook is that? Quickslant, go back to where ever you got that K= 5.045 x 10^28 figure. Surely, it gives the units also.