## Friday, August 5, 2011

### Tackling Simple Astronomy Problems (6)

IN this instalment we examine a number of short answer questions of the type that were tested regularly by many of us during the CXC Astronomy course in the late 70s. Mostly, we used short answer format questions to obtain a quick grasp of the extent to which students mastered the principles and how well they could apply them.

As an example consider the appended diagram which shows an ellipse said to be representative of an asteroid orbit. The sketch shows a point P(x,y) on the orbit and also the two foci, f1 and f2. An example problem is to find the eccentricity (e) of the orbit, given only the information evident in the graph.

How to proceed?

First one must be aware of the assorted parameters for the ellipse. There are basically three: a (the semi-major axis), b (the semi-minor axis) and c, the distance between the foci or c = f1f2.

What is c, mathematically?

It is the square root of the difference between the semi-major axis squared, and the semi-minor axis squared:

c = (a^2 - b^2)^½

Note that when c = 0 we have a = b. (One constant fixed radius: e.g. r = a = b)

This means the figure is a perfect circle.

When, however, a > b then the shape alters to elliptical.

Again, in general, since the shape of an ellipse requires TWO descriptors, or parameters- one needs to know BOTH a and c to calculate the eccentricity, e, viz.:

e = c/ a = (a^2 - b^2)^½/ a

Solution:

Now, from the graph, one obtains a = 4 cm and b = 2.8 cm, so c = 2.85 cm. Then to obtain e:

e = (4^2 - (2.8)^2)^½/ 4 = 0.71, which is highly elliptical as an asteroid orbit would be.

Then, there is Kepler's Third Law of Planetary Motion' or 'The Harmonic law,' can be written out in terms of:

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

That is, the squares of the periods (P1, P2) of revolution of two planets (being compared) equal k times the cubes of their semi-major axes (a1, a2). The last is also known as the 'mean distance from the Sun'. Usually - to make things simpler- we adopt Earth values for P2 and a2, and convert the other planet's (that we wish to find) for P1 and a1. In the case of the (other) planet's distance, we therefore want to solve for a1.

In the original preparatory workshops I usually laid out the problem as follows:

We take P2 = 1 year (the period for Earth to make one revolution). We take a2 = 1 AU or astronomical unit (= 93 million miles). Then the equation simplifies to:

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

with k a constant of proportionality that can be set equal to 1. Or, on solving for a1 (and remembering the units used):

a1 = {[P1]^2}^1/3

Example Problem:

If for Mars, P1 = 687 days, find its mean distance from the Sun (note mean distance taken to be equal to the semi-major axis, a).

Solution:

First convert Mars' period P1 to years, given one Earth year = 365 days. Therefore: P1 = (687 d/ 365 d/yr) = 1.88 years.

Then, solving for a1:

a1 = {[1.88]^2}^1/3 = {3.53}^1/3 = 1.52

I left out units, but we understand that a1 is in AU so

a1 = 1.52 AU

This can also be converted into kilometers:

a1 (km) = 1.52 A.U. x (1.49 x 10^8 km/AU) =

2.26 x 10^8 km

1) A CXC student draws a sketch (Fig. 1) of what he asserts is an asteroid orbit. Using the graph shown, find:

a) the semi-major axis a

b) the semi-minor axis b

c) the eccentricity e.

2) 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.

3) A solar system 100 light years from Earth has two planets, X and Y. Planet X has a period of 8 Earth years, and Planet Y has a distance from its Sun of 2.8 astronomical units.

a) Using Kepler’s 3rd law of planetary motion, find the distance of Planet X in astronomical units and kilometers.

b)Using the same law, find the PERIOD of Planet Y in Earth years.

4) Consider the table shown below, applied to circular satellite orbits around the Earth:

------------------------------------------------------------------------------
Distance from Center of Earth: 2 r --3 r ---3r/ 2 ---4 r ---5 r/2 --- 5 r
------------------------------------------------------------------------------
Acceleration of gravity--- ----g/4
------------------------------------------------------------------------------

Complete the Table above given that r = 6.4 x 10 ^6 m. Hence or otherwise, deduce the value of the acceleration of gravity g at distance 10r.

5) You are required to calculate the hour angle of the Sun (HA ☉) from a place with longitude 163 deg 14’ E. The observation is made at a standard time of 8:46 a.m. on March 10, the standard time being referred to the meridian of longitude 165 deg E. (The Equation of Time for the particular date is approximately E = - 10 min. Note that Equation of Time is just the difference (Apparent time – Mean time). If you need to refer to: Tackling Simple Astronomy Problems (1) )