Planet is 2.00 a.u. when closest to sun. eccentricty is .500. find period

AI Thread Summary
To find the orbital period of a planet with a perihelion distance of 2.00 astronomical units (a.u.) and an eccentricity of 0.500, the semi-major axis (A) is calculated to be 4.00 a.u. Using Kepler's third law, the relationship T^2 = a^3 is applied, leading to T = 8 years. The calculations confirm that the period is indeed expressed in years. The discussion highlights the importance of correctly applying Kepler's laws to determine orbital characteristics.
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Homework Statement



a planet is 2.00 a.u. when it is closest to the sun. its eccentricity is .5000. Find its period in years.

Homework Equations





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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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?
 
anyone i got a 7 15 dead line
 
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