Question

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?

Answer 1

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`