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

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SUMMARY

The discussion focuses on calculating the orbital period of a planet with a perihelion distance of 2.00 astronomical units (a.u.) and an eccentricity of 0.5000. The user applies Kepler's Third Law, stating that the square of the period (T) is proportional to the cube of the semi-major axis (a). After deriving the semi-major axis as 4.00 a.u., the user calculates the period T as 8 years, confirming the application of the formula T² = a³ is correct. The final conclusion is that the orbital period of the planet is indeed 8 years.

PREREQUISITES
  • Understanding of Kepler's Laws of Planetary Motion
  • Familiarity with astronomical units (a.u.)
  • Basic algebra for manipulating equations
  • Knowledge of eccentricity in orbital mechanics
NEXT STEPS
  • Study Kepler's Third Law in detail
  • Explore the concept of eccentricity and its implications on orbits
  • Learn about the calculation of orbital parameters in celestial mechanics
  • Investigate the significance of perihelion and aphelion distances
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Astronomy students, physics enthusiasts, and anyone interested in celestial mechanics and orbital dynamics will benefit from this discussion.

c-murda
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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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