A comit orbits the Sun with a sidereal period of 64.0 years. what is the semi axis of the orbit?

Question

At the Aphelion, the comet is 31.5 AU from the Sun. How far is it from the Sun at Pherihelion?

Answer 1

a = (64 yr)^2/3 = 16 AU (Keplers third law)

We also know:

Pe = a - e and Ap = a + e

(e = linear eccentricity)

So.

e = Ap - a

Pe =  a - Ap + a = 2 * a -  Ap = 0.5 AU

Answer 2

16, since p^2 = a^3.

.5 AU, since (31.5 + .5) / 2 = 16.

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The semi-major axis is 16 AU.  Since orbital period (p)^2 = semi-major axis (a)^3, then a^3  = p^2.  The orbital period is 64 years, so square that and then get the cube root.  The answer is 16.  Since the units are years, which are earth units, the result is in A.U.'s (also earth units).

The semi-major axis is the average distance from the sun.  An average is two numbers added together then divided by two.  If the aphelion is 31.5, and the average is 16, then it follows that that perihelion is .5, since (31.5 + .5) / 2 = 16.