Astronomy 204: Homework #4 - Solutions
1.
Astronomers talk of “declination zones” when describing what stars can be seen from
each latitude on the earth. They typically identify three zones:
• Circumpolar – stars that are always visible from a particular latitude.
• Rise & Set – stars that can rise or set from a particular latitude.
• Never Rise – stars that can never be seen from a particular latitude.
(a) Use a meridinal altitude diagram to derive a heuristic for identifying the 3 zones from
any latitude.
(b) Identify any locations on the earth that are exceptions to the rule (i.e. that don’t have all 3
zones).
One method of approaching this is to use a meridinal altitude diagram.
Z
CE
NCP
N
S
SCP
You know that the declination at the NCP is +90° and that the altitude of the NCP is
equal to you latitude. A star that just sweeps down and touches the north point on the
horizon must define the limit of the circumpolar region. Any star with a lower declination
than this would actually set. The declination of this star must be 90-latitude. Thus, the
circumpolar region always goes from 90-latitude to the pole of the hemisphere that you are
in. Similarly, if you picture a star that just skims the south point on the horizon it must
define the limit of the never rise zone which must extend from –(90-latitude) down to -90.
The rise and set region must be what is left -(90-latitude) to +(90-latitude).
Note that if you apply this rule to the equator and poles you do not 3 regions. The
equator has only rise and set stars and the poles have only circumpolar and never rise stars.
2.
Draw the equinox and both solstice paths for the sun in a horizon diagram for an observer
at a latitude of 78°S.
Your drawing should have the SCP 78 degrees above the south point. Thus, the CE is 12 degrees
above the northpoint.
The equinox path stars at the east point, rises to 12° meridinal altitude above the north point, and
sets directly west.
The winter solstice path is in the sky all day long from a meridinal altitude of 11.5° above the
south point to 35.5° above north.
The summer solstice is below the horizon all day long from a meridinal altitude of 11.5° below
the north point to 35.5° below south
Z
35.5
SCP
WS
12 CE
E
11.5
N
-11.5
S
SS
W
-12
-35.5
3.
If the Local Sidereal Time is 8:45, what is the hour angle of Sirius? Draw a horizon
diagram for an observer at 22°N, draw in Sirius, and estimate its azimuth and altitude.
From Appendix 4 in your text, the RA of Sirius is approximately 6h45m.
HA = LST – RA – 8h45m - 6h45m = 2h
Thus, Sirius is 2 hours past the meridian.
This is sufficient to estimate the position of Sirius in that it is 2 hours past the meridian and it has
a declination of -16°.
4.
Proxima Centauri (α Centauri C) is the closest star to the Sun and is a part of a triple star
system. It has the epoch 1950.0 coordinates (α, δ) = (14h26.3m, -62°38’) while the center of the
system is located at (14h36.2m, -60°38’).
(a) What is the angular separation of Proxima Centauri from the center of the triple star system?
(b) If the distance to Proxima Centauri is 4.0 x 1018 cm, how far is the star from the center of the
triple system?
∆α = 9.9m = 2.48°, ∆δ = 2°
using
2
2
2
( ∆θ ) = ( ∆α cos δ ) + ( ∆δ )
= ( 2.48° )( cos 61° )  + ( 2° )
2
2
= 2.33°

 2.33°
s = d ∆θ = ( 4.0 × 10 cm ) 
 57.3 °
rad

18


 1AU 
17
= 10,844 AU
 = (1.63 × 10 cm ) 
13 
 1.5 × 10 


5.
Adapt the script from last week to print out two plots of blackbody curves of different
temperatures one stacked on top of the other. Then apply your script to creating blackbody
curves which depict the sun and Rigel. Turn in hardcopies of your plot and a printout of your mfile.
Hints: One approach would be to place the entire graphics routine from last week in a
for-end loop and let the variable p range from 1 to 2. You should make use of the subplot
command to create the two separate plots -- subplot(m, n, p) breaks the figure window into an
m-by-n matrix of small subplots and selects the pth subplot for the current plot. You should
place your two temperatures into an array and use the p variable to access the appropriate value
for each plot.
% m-file to graph and label 2 blackbody curves
clear
T =[5800,12000];
% Enter the sun’s and Rigel’s temperature in arrays
c = 3.00e8;
h = 6.623e-34;
k = 1.38e-23;
npoints = 1000;
lambda = 200.0:700.0/(npoints-1):900.0; % Wavelength in nm
for p = 1:2,
for i = 1:npoints,
int(i) = ((2*h*c^2)/(1e-9*lambda(i))^5)/(exp(h*c/(1e9*lambda(i)*k*T(p))) - 1);
end;
hold on;
grid on;
title('Blackbody Curve');
xlabel('Wavelength (nm)');
ylabel('Intensity (W/m^2)');
semilogy(lambda,int,'r-')
%Calculate wavelength peak - Apply Wien's Law
lambdamax = 2.9e6/T(p);
% Calculate Energy Flux - Apply Stefan-Boltzmanns' Law
sigma = 5.67e-8;
eflux =sigma * T(p)^4;
% Print values to screen - coordinates in actual axes values
ycoord =((2*h*c^2)/(1e-9*lambdamax)^5)/(exp(h*c/(1e9*lambdamax*k*T(p))) - 1)
print1 = strcat('Peak Wavelength: ',num2str(lambdamax),' nm');
print2 = strcat('Energy Flux: ', num2str(eflux),' W/m^2');
if lambdamax < 500
% Control locations of text printouts
text(lambdamax+60,ycoord, print1, 'color','blue')
text(lambdamax+60,ycoord*0.9, print2, 'color','blue')
else
text(lambdamax-320,ycoord, print1, 'color','blue')
text(lambdamax-320,ycoord*0.9, print2, 'color','blue')
end;
end;