Kepler's third law constant

[Mary1910]

Homework Statement

Determine Kepler's third-law constant Ku for Uranus using the data for the first four moons.

Homework Equations

[/B]
(r)^3/(T)^2 = K

Data:
Moon #1 (Ophelia)
r=5.38 x 10^4 km
T=0.375 Earth Days

The Attempt at a Solution

[/B]
I know the question asks for the K constant for the first four moons, but I'm only posting my attempt for the first moon, just so I can confirm that I am solving them correctly.

K=(r)^3/(T)^2

first convert km to m
(5.38 x 10^4km)(1000m)
=5.38 x 10^7 m

and then Earth days to seconds
(0.375 Earth days)(86400s)
=32400 s

K=(5.38 x 10^7)^3 / (32400)^2
K=1.48 x 10^14 m3/s2

Any help would be appreciated. Thank you

 

[mfb]

Calculating the other moons is a great opportunity to test the results - the numbers should be similar for all four moons. If you get different results something is wrong, if you get the same result the answer is right.

 

[Mary1910]

Calculating the other moons is a great opportunity to test the results - the numbers should be similar for all four moons. If you get different results something is wrong, if you get the same result the answer is right.

Thanks, and yes the Kepler constants for the other four moons were very close to the result I have for the first moon.

Question b) In this problem asked me to find the average K value for all four moons. I then added all four together and then divided by four to have 1.47 x 10^14 m3/s2. In question c) we are asked to complete the missing information for the rest of the moons listed.

Ex. Moon #5(Rosalind)
r=6.99 x 10^4 km
T=?
K=?

Since the K value is missing, would it make sense to sub the average K value from part b (1.47 x 10^14 m3/s2) into K=(r)^3/(T)^2 and then solve for T?

Thanks for your help.

 

[mfb]

Since the K value is missing, would it make sense to sub the average K value from part b (1.47 x 10^14 m3/s2) into K=(r)^3/(T)^2 and then solve for T?

Sure.

 

[Mary1910]

Sure.

Thanks