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”##
Save your time - order a paper!
Get your paper written from scratch within the tight deadline. Our service is a reliable solution to all your troubles. Place an order on any task and we will take care of it. You won’t have to worry about the quality and deadlines
Order Paper NowEccentricity 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.