a)The recently discovered Eris which is slightly larger than pluto orbits the su

Question

a)The recently discovered Eris which is slightly larger than pluto orbits the sun every 557 years what is its avearge distance ( semimajor axis) from the sun? How does its average distance compare to pluto?

b) A newly discovered planet orbits a distant star with the same mass as the sun at an average distance of 112 million kilometers its orbital eccentricity is .3. Find the planet's orbital period and its nearest and farthest orbital distances from the star.

Possible info needed Eris Period=557yr e=.442

a^3/P^2=Msun

Explanation / Answer

a) Eris’s period is 557 years

a^3 = P^2

a = P^(2/3) = (557)^(2/3) = 67.7 AU

Eris has an average distance (semimajor axis) of 67.7 AU

This is further from the Sun than Pluto, which has a semimajor axis of ~40AU.

b) The orbital period, T, is related to the mean distance (semi-major axis), a, for different orbiting bodies about the same central object, by the proportionality

a³ ~ T²

This is Kepler's Third Law of planetary motion. Knowing T and a for one of the Sun's planets (Earth!), you can derive either a or T from the other, for any body orbiting any 1-solar-mass star. For the Earth,
a[E] = 149,600,000 km
T[E] = 365.26 days (sidereal period!)

So for the mystery planet,
a = 112,000,000 km
T = T[E]·(a/a[E])³ = (112/149.6)^(3/2) · 365.26 days

Now Periastron (nearest to the Star) = Ave.distance (1 - eccentricity) = 0.7487 AU(1 -0.3) = 0.52389 AU
= 102,682,440 km
Apastron (farthest to the Star) = Ave.distance (1+eccentricity) = 0.7487 AU(1+0.3) = 0.97331 AU
= 145,607,176 km

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