Question

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?

Answer 1

Semi major axis a = 3.2 AU. Eccentricity e = 0.4375. Period`= 3.2sqrt 3.2` years = 5.7 years, nearly

Explanation:

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

Perihelion = `a(1-e)` = 1.8 AU.

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

`e = (4.6-1.8)/(2a)` = 2.8/6.4 = 0.4375.

Period = `((2pi)/sqrt mu)asqrt a`

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

So, `sqrt mu = 2pi`, with unit of distance as AU and unit of time as year.

Now, the period of the asteroid `= a sqrt a years = 3.2sqrt 3.2` years=