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

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Kepler's 3rd Law is applied to calculate Pluto's orbital period using the formula K = T²/R³, where K is Kepler's constant. The correct mean radius of orbit for Pluto is 5.9x10^12 m, and the radius of the object is not necessary for this calculation. The initial attempt mistakenly included both distances and misapplied the square root instead of cubing the mean radius. After correcting the approach, the calculated orbital period for Pluto is approximately 7.82x10^9 seconds. Accurate application of Kepler's Law is essential for determining orbital periods.
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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]

 
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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
 
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