# 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?

Mar 12, 2016

Semi-Major Axis Distance:
$a = {d}_{p} / \left(1 - e\right) = \frac{0.19 \setminus \quad A U}{1 - 0.83} = 1.12 \setminus \quad A U$
Aphelion Distance:
${d}_{a} = a + c = a + e a = \left(1 + e\right) a$
$\setminus q \quad = \left(1 + 0.83\right) \setminus \times 1.12 \setminus \quad A U = 2.04 \setminus \quad A U$

#### Explanation:

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

${d}_{p}$ - Perihelion distance; $\setminus \quad {d}_{a}$ - Aphelion distance;

Eccentricity - Focal Point Relation: $e = \frac{c}{a}$ ...... (1)
Perihelion distance - Eccentricity Relation:
${d}_{p} = a - c = a - e a = \left(1 - e\right) a$ ...... (2)

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

Semi-Major Axis Distance:
$a = {d}_{p} / \left(1 - e\right) = \frac{0.19 \setminus \quad A U}{1 - 0.83} = 1.12 \setminus \quad A U$
Aphelion Distance:
${d}_{a} = a + c = a + e a = \left(1 + e\right) a$
$\setminus q \quad = \left(1 + 0.83\right) \setminus \times 1.12 \setminus \quad A U = 2.04 \setminus \quad A U$