If an asteroid has a perihelion distance of 2.0 A.U. and an aphelion distance of 6.0 A.U., what is its orbital semi-major axis, eccentricity, and period? Thanks!?

Dec 5, 2015

$a = 4.0 \text{AU}$,
$e = .5$,
$T = 2 \text{years}$

Explanation:

The semi-major axis, $a$ of and ellipse is half of the major axis, the total distance between perihelion and aphelion. The semi-major axis is therefore equal to;

a = (R_a + R_p)/2 = (6.0"AU"+2.0"AU")/2 = 4.0"AU"

Eccentricity is defined as the ratio of the distance between two focus of the ellipse, ${R}_{a} - {R}_{p}$, and the length of the major axis, ${R}_{a} + {R}_{p}$. This can be expressed mathematically as;

$e = \frac{{R}_{a} - {R}_{p}}{{R}_{a} + {R}_{p}} = \left(6.0 \text{AU" - 2.0"AU")/(6.0"AU"+2.0"AU}\right) = .5$

The period, $T$, can be found using the Kepler's 3rd law, which states that the ratio of the period squared to the semi-major axis cubed is a constant for all objects orbiting the same body. In other words, for two objects that orbit the sun;

${T}_{1}^{2} / {a}_{1}^{3} = {T}_{2}^{2} / {a}_{2}^{3} = C$

Or, restated;

${T}_{1}^{2} = {a}_{1}^{3} / {a}_{2}^{3} {T}_{2}^{2}$

If we use the Earth as object $2$, then ${T}_{2}$ and ${a}_{2}$ are both $1$. We can calculate the period of the asteroid using its semi-major axis, $a = 4.0 \text{AU}$.

T_1^2 = (4.0"AU")^3/(1"AU")^3 (1"year")^2 = 4.0"years"^2

Taking the square root, we get a period of $2 \text{years}$ for the asteroid.