Question #f33c5

1 Answer
Feb 2, 2018

The aphelion of this very eccentric orbit would be 27.793 A.U.

Explanation:

Use Kepler's third law first to find the mean radius of the orbit (actually the semi-major axis of the ellipse). Since T is in (Earth) years and r is in A.U., we can use the simplification that k=1 in the formula:

T^2 = kr^3T2=kr3 becomes T^2 = r^3T2=r3

so, taking the cube root of each side, r = T^(2/3) = 54^(2/3)=14.287 A.U.r=T23=5423=14.287A.U.

This is the mean radius (or semi-major axis), so the diameter (major axis) is twice this value:

d=28.573 A.U.d=28.573A.U.

This value is the sum of the aphelion and perihelion, and so, the aphelion must be

28.573 - 0.78 = 27.793 A.U.28.5730.78=27.793A.U.

(Strange the question would ask for the answer to the nearest thousandth of an A.U. when the given information did not have this precision!)