# 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!?

##a=4.0″AU”##, ##e = .5##, ##T=2″years”##

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 = (R_a – R_p)/(R_a + R_p) = (6.0″AU” – 2.0″AU”)/(6.0″AU”+2.0″AU”)=.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 ;

##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″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″years”## for the asteroid.