Saturday, August 6, 2011

Solution to Simple Astronomy Problems (6)

To recap, the problems are given first in order, then their solutions:

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.


Based on the graph shown with scale in cm, 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

This could also have been found directly from the graph (measuring the distance of f1 or f2 from 0) and thereby obtaining: c = 5.6 cm

So: e = c/ a = 5.6 cm / 8 cm = 0.7

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.


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 a (and remembering the units used):

a^3 = P^2

But we know a = 8 AU (subst. AU for cm in the graph). Solving for P:

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

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.


The key to the solution is to insert 'Earth" into the solar system and thereby normalize Kepler's law so the constant of proportionality, k = 1.

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

Thus, with appropriate substitutions, we can let P be the period for Planet Y and let a be the distance for Planet X.

To solve for part (a) then, we have: P(X) = 8 yrs, we need a(X):

a(X) = {[P(X)]^2}^1/3 = [{8 yr}^2] ^1/3

a(X) = [64]^1/3 = 4 AU

(b) Since: (P(Y) / 1 yr)^2 = k(a(Y)/ 1 AU)^3

where a(Y) = 2.8 AU


P(Y) = [a(Y)^3]^½ = {[2.8]^3}^½ = 4.7 yrs.

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.


This is based on noting that the gravitational field intensity g ~ 1/r^2

That is, it is inversely proportional to the distance from the center squared. Hence, the remaining entries in the table as as follow, in ascending order:

g/9, 4g/9, g/16, 4g/ 25 and g/25

Given this relationship (g ~ 1/r^2) then at a distance 10r we have: g/100,

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)' )


Given the longitude circle is 1 deg 46’ west of the standard longitude this means the time difference is about: 4 min. + 3 mins. = 7 mins. earlier or 8:39 a.m. LMT. Then the Equation of Time for the particular date is approximately E = - 10 min. so:

App. Solar time = 8:39 a.m. – 10 mins. = 8:29 a.m.

Since the hour angle of the Sun (HA ☉) is referenced with respect to local noon then this means we have:

(HA ☉) = 90 o – ( 45 deg + 7 ¼ deg ) = 90 deg – 52 ¼ deg

(HA ☉) = 38 deg E., approximately!

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