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OpenStudy (anonymous):

Keplers Laws

OpenStudy (shane_b):

What's the question?

OpenStudy (anonymous):

Intuitive....insight.... \[T^2=Cr^3\]

OpenStudy (anonymous):

Oh yeah, Proportions.....

OpenStudy (turingtest):

uh oh, math processing error

OpenStudy (anonymous):

what happened? LOL

OpenStudy (turingtest):

oh there we go, intuitive insight? Not sure what to say about it, I've just about never used that equation.... rather self explanatory, no?

OpenStudy (anonymous):

oh semi major axis....

OpenStudy (anonymous):

|dw:1362612752315:dw| what is the relationship b/w Kepler's and the semi major axis?

OpenStudy (anonymous):

ok found it....sorry LOL

OpenStudy (turingtest):

lol so... any more misgivings?

OpenStudy (anonymous):

question. why is this true?

OpenStudy (turingtest):

|dw:1362613146649:dw|you mislabeled the semi-major axis as well, I think

OpenStudy (anonymous):

|dw:1362613168947:dw| is this their setup?

OpenStudy (turingtest):

right, so in your problem, the sun is here|dw:1362613265550:dw|

OpenStudy (turingtest):

the closes they are is 0.1AU, so we have|dw:1362613346232:dw|

OpenStudy (anonymous):

oh yeah....that makes more sense

OpenStudy (turingtest):

the whole distance across is 2 times the semi-major axis, or 2R, and you can see that the farthest the plant can be is on the other side of the ellipse|dw:1362613607318:dw|

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