Calculating Mass-Kepler's 3rd Law

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SUMMARY

The discussion centers on the application of Kepler's 3rd Law, specifically the formula a³/p² = M, where 'a' represents the semi-major axis in astronomical units and 'p' represents the orbital period in years. The user calculates (6x10^9)³ / (7.8x10^8)² and arrives at 3.6x10^11 MSun, but expresses confusion regarding the conversion of the result and the units involved. The community clarifies that the result is indeed in solar masses (MSun) and confirms that the calculations are approximately correct, indicating a misunderstanding in the division step.

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  • Understanding of Kepler's Laws of planetary motion
  • Familiarity with astronomical units and solar masses (MSun)
  • Basic algebraic manipulation skills
  • Knowledge of scientific notation and unit conversion
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winterrose
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Okay, so I understand that a3/p2=M

I must be missing something though.

I know that (6x10^9)3 / (7.8x10^8)2 = 3.6x10^11 MSun

but I don't understand what step I'm missing. When I divide, I end up with 0.3550^12

what am I doing wrong?

Is my answer in kilograms, and do I have to convert it to MSun?
 
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welcome to pf!

hi winterrose! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
winterrose said:
… = 3.6x10^11 MSun

but I don't understand what step I'm missing. When I divide, I end up with 0.3550^12

they're (aproximately) the same, aren't they? :wink:
 

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