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planet is 2.00 a.u. when closest to sun. eccentricty is .500. find period |
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| Apr28-08, 02:29 PM | #1 |
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planet is 2.00 a.u. when closest to sun. eccentricty is .500. find period
1. The problem statement, all variables and given/known data
a planet is 2.00 a.u. when it is closest to the sun. its eccentricity is .5000. Find its period in years. 2. Relevant equations 3. The attempt at a solution i got this so far... don't know if it is right PD + AD = A -AD - PD = F <SUBTRACTING THESE TWO> 2PD = A - F 2PD = A - (EA) 2PD = A(1-E) 2PD/(1-E) = A A = 2(2)/(1.-.5) A= 8 a.u. then for a a= 8.00/2 a= 4.00 a.u. T^2 = a^3 T = sqrt(a^3) T = 8 im lost T = 8 what? years? |
| Apr28-08, 02:53 PM | #2 |
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kepler's 3rd law says Period( in years) is directly proportional to semi axis (a) cubed. correct?
so was that last line: T^2 = a^3 correct? |
| Apr28-08, 03:57 PM | #3 |
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bump
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| Apr28-08, 04:11 PM | #4 |
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planet is 2.00 a.u. when closest to sun. eccentricty is .500. find period
anyone i got a 7 15 dead line
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