What is the Orbital Period of Eris, the Tenth Planet?

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SUMMARY

The orbital period of Eris, designated as 2003 UB313, can be calculated using Kepler's Third Law, which states that the square of the orbital period (P) is proportional to the cube of the semimajor axis (a). For Eris, with a semimajor axis of 68.048 AU, the formula P^2 = (4π^2/G(m1+m2))(a1+a2)^3 applies. The eccentricity of 0.4336 does not influence the calculation of the orbital period, confirming that it is only relevant for other aspects of orbital mechanics.

PREREQUISITES
  • Understanding of Kepler's Third Law of planetary motion
  • Familiarity with astronomical units (AU)
  • Basic knowledge of gravitational constants (G)
  • Ability to manipulate algebraic equations
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  • Research the application of Kepler's Third Law in different celestial bodies
  • Explore the significance of eccentricity in orbital mechanics
  • Learn about the gravitational constant (G) and its role in astrophysics
  • Investigate the characteristics of other trans-Neptunian objects
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Astronomy students, astrophysicists, and educators involved in teaching celestial mechanics and orbital dynamics.

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Homework Statement


We will compute the orbit of Eris, the infamous \tenth planet" that caused Pluto
to be demoted to \dwarf planet" status. Eris (also called 2003 UB313 in the
textbook) has a semimajor axis a = 68:048 AU and eccentricity e = 0:4336 based
on the best current observations. Calculate the orbital period P of Eris (in years).


Homework Equations


P^2=4pie^2)/G(m1+m2)) (a1+a2)^3


The Attempt at a Solution


would i just use this formula? I know it would work but not sure if i am suppose to use something else cause i was given eccentricity, but could not find any formula that uses it to find orbital period
 
Physics news on Phys.org
The period does not depend on the eccentricity (kepler's third law).
 
yea i figured that out thxs, eccentricity was for the other part of the problem
 

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