Kepler's 3rd Law Question finding how long a year is for Pluto

  • Thread starter Thread starter Jon Bori
  • Start date Start date
  • Tags Tags
    Law Pluto Year
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
Jon Bori
Messages
2
Reaction score
1

Homework Statement



Kepler's Constant for any object or planet orbiting the sun is 3.36x1018m3/s2. Calculate how long a year is for Pluto given the "radius of object" is 3.0x106m and the "mean radius of orbit" is 5.9x1012

Answer: 7.82x109s

Homework Equations


K = T2/R3

The Attempt at a Solution


I tried plugging the values into the equation

T2 = sqrt (5.9x1012m)+(3x106)/3.36x1018m)

= 6.1x1019[/B]

 
Physics news on Phys.org
Jon Bori said:

Homework Statement



Kepler's Constant for any object or planet orbiting the sun is 3.36x1018m3/s2. Calculate how long a year is for Pluto given the "radius of object" is 3.0x106m and the "mean radius of orbit" is 5.9x1012

Answer: 7.82x109s

Homework Equations


K = T2/R3

The Attempt at a Solution


I tried plugging the values into the equation

T2 = sqrt (5.9x1012m)+(3x106)/3.36x1018m)

= 6.1x1019[/B]
What equation is that? The one you listed in the Relevant Equations only involved one distance value, but you seem to have used two ("radius of object" and "mean radius of orbit"). The listed equation also cubed the distance, but you've used a square root to resolve it?

How does the given value of Kepler's constant relate to your Relevant equation? Do the units match?
 
I got it. You only use the mean radius of orbit and I forgot to cube the distance.. lol
 
  • Like
Likes   Reactions: gneill