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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
 

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?
 
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Apr28-08, 02:53 PM   #2
 
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
 
bump
 
Apr28-08, 04:11 PM   #4
 

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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