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To: Da Bilge Troll
I just can't believe that orbital period. It would have to be so close to the star that friction would quckly bring it down.

The square of the orbital period is proportional to the cube of the orbital radius and inversely proportional to the mass of the central body, in this case the star.

Plugging in the 1/3 the mass and the 1.94 day period we get an orbital radius of 0.04406 * the earth's orbital radius about Sol, or about 4,083,172 miles. That seems plausible, but it is close. Mercury is just over 1/3 the distance from Sol that Earth is.

51 posted on 06/13/2005 1:57:32 PM PDT by El Gato
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To: El Gato

Thanks for the calculations. That IS close! I wonder if they have enough data to determine if the orbit is stable?


55 posted on 06/13/2005 2:25:25 PM PDT by Da Bilge Troll (Defeatism is not a winning strategy!)
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