# An asteroid has an aphelion distance of 4.6 A.U. and a perihelion distance of 1.8 A.U. How do you calculate the orbital semi major axis, eccentricity, and period?

May 14, 2016

Semi major axis a = 3.2 AU. Eccentricity e = 0.4375. Period$= 3.2 \sqrt{3.2}$ years = 5.7 years, nearly

#### Explanation:

Aphelion = a(1+e) = 4.6 AU.

Perihelion = $a \left(1 - e\right)$ = 1.8 AU.

a = (4.6+1.8)/2 AU = 3.2 AU.

$e = \frac{4.6 - 1.8}{2 a}$ = 2.8/6.4 = 0.4375.

Period = $\left(\frac{2 \pi}{\sqrt{\mu}}\right) a \sqrt{a}$

In the case of the Earth, a = 1 AU and period is 1 Earth year.

So, $\sqrt{\mu} = 2 \pi$, with unit of distance as AU and unit of time as year.

Now, the period of the asteroid $= a \sqrt{a} y e a r s = 3.2 \sqrt{3.2}$ years=