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:

------------------------------------------------------------------------------
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
------------------------------------------------------------------------------

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!

Solutions to Simple Astronomy Problems (5)

We proceed now to the solutions for the problems given in instalment (5). Recapping them, with the respective solutions to follow:

(1) The Right Ascension of Dubhe in Ursa Major is 11h 02m. Its declination is 61 deg 56 minutes. From a ship at sea it's observed to transit 42 degrees 30 minutes north of the zenith on April 8, at 1 a.m. according to the ship's chronometer (which reads Greenwich Mean time or G.M.T.)

Find the position of the ship.

Solution

If Dubhe transits 42 deg 30 min north of the zenith then its meridian zenith distance (MZD) is 42 deg 30 min.

The ship's latitude may be found from:

Lat. = Declin. - MZD = 61 deg 56 min - 42 deg 30 min = 19 deg 26 min North

To obtain the longitude we require:

Greenwich time of transit - local time of transit, or GTT - LTT for short.

Where: LTT = Star's RA - Sun's RA

Now, on April 8, the Sun's RA = 18 days x (4 min/day) = 72 mins. = 1 h 12 min

(Note: this is 18 days elapsed after March 21 when the Sun's RA = 0h)

Then:

LTT = 11 h 02 min - 1 h 12 min = 9h 50 min

According to the ship's chronometer, the star transits at 1 a.m. (13 h) therefore GTT = 13h 00m

Then the ship's longitude is:

L = GTT - LTT = 13 h 00 min - 9h 50 min = 3 h 10 min

Converting this to degrees (1 h = 15 deg), we get 47½ degrees. This must be WEST longitude since the GTT is later than LTT.

Then the coordinates are:

Lat. = 19 deg 26 min N, Long. = 47½ deg W



(2)(a) Using a clear diagram, show that the hour angle of a star is related to the Star's RA, the Sun's RA and L.M.T. (local mean time) by:

RA (star) = Sun's RA - (HA) + L.M.T.

Solution

This is shown in Fig. 1



(b)Shaula (RA = 17h 31m) transits at 10 p.m. on June 29th. Find the hour angle it exhibits from your location (local hour angle).

Solution

This can be done from either the diagram or the expression (RA (star) = Sun's RA - (HA) + L.M.T.) which really amount to the same thing.

A simpler tack is to use: LST= HA + RA of object

and it can be ascertained the LST for the date is 15h 31 m

SO: HA = LST - RA = 15h 31m - 17h 31m = -2h 00m, or - 30 deg


(3) The star Canopus (RA = 6h 20m) is observed to have a local hour angle = 45 deg on Feb. 10th for a given location.

(a) What is the local sidereal time.

(b) At what local mean time and standard time would Canopus transit?

(c) What is the approximate LST at noon on the same date?


Solutions

(a)Again: HA = LST - RA

So: LST = HA + RA = 3h 00m + 6h 20m = 9h 20m


(b) We have:

LTT = Star's RA - Sun's RA

Feb. 10th is 39 days before March 21 (assume non-leap year) so:

Sun's RA = 0h - 39 days x (4 min/day) = 0h - 156 mins = 0 h - 2h 36 m

Sun's RA = 21h 24m

LTT = 6h 20m - 21h 24m = -17h 36m

Or: 24 h 00m - 17h 36m = 6h 24m

(c) From (a) the LST on the date is 9h 20m, and this is 9h 20m past noon. Since noon is 9h 20m earlier, then LST(noon) = 6h 20m - 9h 20m = -3h 00m, or:

24h 00m - 3h 00m = 21h 00m.


(4) Study the diagram for Fig. 1. Using the diagram and inferences regarding time - including Sun's changes in Right Ascension, give the date for which the diagram is referred. If the star in Fig. 1 has an RA = 7h 10m then find its hour angle for the observer (Obs). Hence, or otherwise, obtain the longitude of the observer.


Solution

The diagram is shown again in Fig. 2.

Since the local sidereal time (LST) at the observer location is 6h then this is the RA on the meridian. But the Sun is on the meridian at the antipode or at 18h 00m. Hence, the date must be Dec. 23rd or the winter solstice, since the same RA coincides with the Sun.

If the star's RA = 7h 10m we can find the HA from the angular relationship as:

HA = RA - LST = 7h 10m - 6h 00m = 1h 10m = 17 .5 deg

The longitude of the observer is: L = GST - LST = 9h 00m - 6h 00m = 3h 00m

Converting to degrees: L = 45 deg (W) (since GST > LST)


(5)(a) Regulus (10h 07m) is observed to transit on a given date. Given the same date, what would be the local hour angle for: (i) Denebola (RA = 11h 47m), and (ii) Arcturus (RA = 14h 14m)?

(b) Find the LST one half way between the transits of Denebola and Arcturus. How much later would these two stars set than Regulus?

(c) What RA circle must be rising when Arcturus is setting?

Solutions

(a) We use: LHA (local hour angle) = RA on meridian (LST) - RA(star)

Here, RA meridian = RA (Regulus) = 10h 07m

Therefore:

LHA (Denebola) = 10h 07m - 11h 47m = -1h 40m (or 22h 20m)

LHA (Arcturus) = 10h 07m - 14h 14m = -4h 07 m or 19h 53m

(b) Find (RA(Arcturus) - RA(Denebola) = 14h 14m - 11h 47 m = 2h 27m

Then the LST midway between their transits is found by interpolating:

11h 47m + ½(2h 27m)= 13h 00.5m

Thus, Denebola would set about 1h 40m after Regulus. (E.g. 10h 07m + 1h 40m = 11h47m)

(c) When Arcturus is setting, the RA circle rising is:

14h 14m + 12h 00m = 26h 14m

or:

26h 14m - 24h 00m = 2h 14m