Expert: Philip A. Stahl - 1/31/2012

Question
QUESTION: 1. An asteroids closest approach to the Sun (perihelion) is 2AU and farthest distance from the sun (aphelion) is 4AU. What is the semi major axis of its orbit?

2. What is the period of its orbit?

3. What is its eccentricity?

ANSWER: Hello,

This sounds suspiciously like a homework problem which I don't do. However, I can provide some assistance to enable you to complete the solution on your own.

The semi-major axis, a, which is being sought, is just the *mean distance* of the object or planet from the Sun. You ought to be able to work this out from the given information.  Once this is found, which we symbolize as 'a', then the period can be found using Kepler's 3rd or harmonic law, e.g.

(P1/ P2)^2 = k(a1/ a2)^3

Thus, the period P1 of planet (1) is to the period of planet (2) as the semi-major axis cubed of planet (1) is to the semi-major axis cubed of planet (2).

Generally, to simplify the treatment we take "planet 2" to be Earth. Then P2 = 1 year (the period) and  a2 = 1 AU or astronomical unit.

So this simplifies to:

(P1/ 1 yr)^2 = k(a1/ 1 AU)^3

Let k = 1 be a proportionality constant, and then you just solve for the period P1, since you already know a1.

The eccentricity e can be found from the semi-major axis a and the semi-minor axis b. (You will have to work out what b is, and a good exercise here would be to try to sketch the ellipse made by the asteroid, with the data you're given - labelling perihelion, aphelion, semi-major axis etc.)

Then: e = c/ a

where: c = (a^2 - b^2)^½

Therefore:

e = (a^2 - b^2)^½ / a

Hope this helps! The solutions now ought to be nearly automatic!

---------- FOLLOW-UP ----------

QUESTION: Can you do an example with some other numbers? I learn better if I can follow the steps and figure out why the answer is so. Thanks!

Answer
Hello,

Okay, let's make it more concrete using the problem below - with the accompanying diagram as a reference.

Problem:

A student draws a sketch with square units in centimeters (Fig. 1) of what he asserts is an asteroid orbit.

a)Using the graph shown, find:

i) the semi-major axis a

ii) the semi-minor axis b

iii) the eccentricity e.


b) Given the same student orbit sketch, if we let the dimensions be astronomical units (AU) instead of centimeters, then use the graph to estimate the period of the asteroid.

Solutions:

We tackle Part (a) first.

Based on the graph shown (note the semi-major axis a and semi-minor axis b are labeled and the Sun is situated at one focus with the perihelion distance marked at ‘x’ below it, and the aphelion at the other end with another ‘x’)  we have:

a = 8 cm, b = 5.7 cm

The eccentricity e = c/ a = (a^2 - b^2)^½/ a

So: e = [(8)^2 - (5.7)^2]^½/ 8

e = [(64 - 32.5)]^½/ 8 = 5.6 cm/ 8 cm = 0.7


Part (b)

Here, Kepler's 3rd law is required.

(P/ P')^2 = k(a/ a')^3

We take P' = 1 year (the period for Earth to make one revolution). We take a' = 1 AU or astronomical unit (= 93 million miles, or Earth's semi-major axis). Then the equation simplifies to:

(P/ 1 yr)^2 = k(a/ 1 AU)^3

with k a constant of proportionality that can be set equal to 1, provided a is in AU and P in yrs. Or, on solving for P (and remembering the units used):

P = [a^3]^½ = {[8]^3}^½ = (512)^½ = 22.6 yrs.


Hope this concrete illustration helps!