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.

Solution:

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.

Solution:

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.
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Solution:

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.
E.g.

(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

Hence:

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:

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


Solution:

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)' )

Solution

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!