The question is the following:
For any object that orbits the sun, Kepler’s Third Law relates the period — the time needed for one orbit — and the mean distance from the sun — the average of the least and greatest distances (recall that the sun is at a focus). Halley’s comet has a period of 76 Earth years (it next returns in 2061), and the least distance from the comet to the sun is 0.59 astronomical unit (one a.u. is about 93 million miles). Use Kepler’s Third Law to calculate the greatest distance from the comet to the sun.
I was just able to draw an ellipse showing that the distance from $F_1$ (point at which the Sun is located) to the comet is $0.59$ a.u. which, when converted to miles, is $54,870,000$. I know that I would have been able to calculate the greatest distance if I knew the eccentricity or at least the mean distance. I don't know how to start answering this question. Any help will be greatly appreciated.
Best Answer
So, when you look at an body's orbit around the sun, you'll end up with an aphelion and a perihelion. These form the major axis, or the longest axis of the ellipse.
We know that $a^3/T^2 = 7.495\times10^{-6}$ from the Third Law.
$a$ represents the semi-major axis, and is half of the sum of aphelion distance (furthest) and perihelion distance (closest).
Given that $T = 27740\ \text{days}$ from 76 years, we can figure out that...
$a^3=(7.495\times10^{-6})\times(27740^2)$ --> $a \approx 17.933354$.
If aphelion $d_1$ + perihelion $d_2$ = twice the semimajor axis $2a$, and the shortest distance is $.59\ \text{a.u.}$, then we can solve for $a$:
$d_1\ \text{(far)} + d_2\ \text{(near)} = 2a\ \text{(long axis)}$
$d_1 = 2a - d_2 = 35.86 - .59 = 35.27\ \text{a.u.}$