Kepler's constant and average radius of orbit

  • Thread starter Tyyoung
  • Start date
  • #1
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Homework Statement


Given that Jupiter has a mass of 1.9x10^27 kg, and the sun has a mass of 1.99x10^30 kg:

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

b) If Jupiter's orbital period is 11.89 Earth years, calculate the average radius of its orbit.


Homework Equations



K = Gmp/4pi^2 = r^3/T^2
11.89x365x24x3600 = 374963040

The Attempt at a Solution



For a) Kjupiter = Gmp/4pi^2
(6.6x10^-11)(1.9x10^27)/4pi^2

= 3.21x10^15 m^3/s^2

b) Ksun = Gmp/4pi^2
ksun = (6.67x10^-11)(1.99x10^30)/4pi^2

3.36x10^18 m^3/s^2

Ksun=r^3/T^2
Therefore: r^3 = Ksun*T^2
r = cubed rt. of Ksun*T^2
r = cubed rt. of (3.36x10^18)(374963040)^2
r = cubed rt. of 4.724068653x10^35
r = 7.8x10^11m

I am really unconfident about these answers, can someone tell me if I am one the right track or should give up on Physics?
Thank you in advance.
 

Answers and Replies

  • #2
25
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I obtained similar results. Wondering if anyone can verify if they are correct.
 
  • #3
1,137
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well for (b) i guess its pretty correct

i even confirmed it using kepler's 3rd law in its pure form

I dont know what's kepler's constant but if you are given a straight formula so it has to be correct
 

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