The asteroid Icarus has a perihelion of 0.19 AU and an orbital eccentricity of 0.83. How do you calculate the asteroid's orbital semimajor axis and aphelion distance from the Sun?

1 Answer
Mar 12, 2016

Semi-Major Axis Distance:
#a = d_p/(1-e) = (0.19\quad AU)/(1-0.83) = 1.12\quad AU#
Aphelion Distance:
#d_a = a+c=a+ea=(1+e)a#
#\qquad=(1+0.83)\times1.12\quad AU = 2.04\quad AU#

Explanation:

#a# - Semi-major distance; #\quad b# - Semi-minor distance;
#c# - Centre-to-Focal Point distance; #\quad e# - Eccentricity of the ellipse;

#d_p# - Perihelion distance; #\quad d_a # - Aphelion distance;

Eccentricity - Focal Point Relation: #e = c/a# ...... (1)
Perihelion distance - Eccentricity Relation:
#d_p = a-c = a-ea = (1-e)a# ...... (2)

Using Equation 2, solve for the semi-major axis distance

Semi-Major Axis Distance:
#a = d_p/(1-e) = (0.19\quad AU)/(1-0.83) = 1.12\quad AU#
Aphelion Distance:
#d_a = a+c=a+ea=(1+e)a#
#\qquad=(1+0.83)\times1.12\quad AU = 2.04\quad AU#