# Homework Help: Using Kepler's constant for Jupiter

1. Aug 3, 2009

1. The problem statement, all variables and given/known data
Given that Jupiter has a mass of 1.9*10^27kg, and the sun has a mass of 1.99*10^30kg:

a)Calculate the value of Kepler's constant for Jupiter

2. Relevant equations

k = r^3/T^2
K = G(M+m)/4pi^2

3. The attempt at a solution

Well since the question only gave me the mass of Jupiter and the Sun I'm assuming that I should use Gm/4pi^2 instead even though this formula should only be used when the masses of the two objects (Jupiter and Sun) are close to each other......right?

Let G = Universal gravitational constant
Let M = mass of the Sun
Let m = mass of Jupiter

K = G(M+m)/4pi^2
k = (6.67*10^-11)(1.99*10^30 + 1.9*10^27)/4pi^2
=3.365*10^18, I'm not sure about the units for this formula

I'm pretty sure I can't use k = r^3/T^2 because the question didn't provide me with enough information to do so, but out of curiosity I googled the radius of Jupiter along with its period and plugged it in the formula:

radius (r) is in meters and period (T) is in days (should it be in days or years?)
k = r^3/T^2
= 71492000^3/4331.57^2
= 1.9475*10^16 m/d

If T was in years (11.89):
k = r^3/T^2
= 71492000^3/11.89^2
= 2.584*10^21 m/y

Ok, so clearly these numbers aren't close to the one above. Can you guys point out what I'm doing wrong?

also another confusion that I have, according to other sources(Wikipedia) Kepler's constant should be k = T^2/r^3 rather than(my textbook) k = r^3/T^2.

and if I use K = T^2/r^3 (for T is in days)
K = 5.134*10^-17 d/m

for T in years:
k = 3.868*10^-22 y/m

2. Aug 3, 2009

### kuruman

Your dimensions need fixing. The dimensions of K (or k) are [Length]3/[Time]2. If you choose the formula that has G in it, it is less confusing to do all calculations expressing the period in seconds, then convert to m3/yr2 for K if you wish. That's because G has "seconds" buried in it as its units are N.m2/kg2 and N = kg.m/s2.

3. Aug 3, 2009

according to what you said, in this case, would it be better to use the formula with G in it?

also, Should the calculation be done in years or days?

4. Aug 3, 2009

### kuruman

Either way should give you the same number. However, if you use the formula with G, the period must be expressed in seconds.

5. Aug 3, 2009