##a=4.0AU##

##e = .5##

##T=2years##

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 = (6.0AU+2.0AU)/2 = 4.0AU##

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) = (6.0AU – 2.0AU)/(6.0AU+2.0AU)=.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=4.0AU##.

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

Taking the square root we get a period of ##2years## for the asteroid.