AST 3722C  Spring 2008
Homework #2 – Solutions
(a)
[0.5 points]
Suppose we’re in Orlando so
φ
=
φ
0
= 28
.
6
◦
. Find the upper and lower
bounds on
δ
such that
a
can be zero. I.e. what is the range of declinations of objects that
are neither circumpolar nor permanently below the horizon?
This is similar to a problem from homework #1. Remember that you found that the
southernmost declination that you can see from your location is
φ
−
90
◦
. In this case, that’s
−
61
.
4
◦
. See the solution set for that problem if you don’t see why this is.
As for the maximum declination, think about extremes. If you were at the North Pole,
δ
= 0
◦
at your due north horizon. As you walk south, Polaris would drop in the sky toward
the due north horizon. At the Equator, Polaris would have dropped all the way down and
so
δ
= 90
◦
at your due north horizon. You should be able to visualize that therefore the
maximum declination you can see is 90
◦
−
φ
, or in our case
61
.
4
◦
.
The inelegant way to do this is to use the formula on Handout #5.
A
= 0 and
a
= 0, so:
sin
δ
= sin
a
sin
φ
+ cos
a
cos
φ
cos
A
= sin 0
◦
sin
φ
+ cos 0
◦
cos
φ
cos0
◦
= cos
φ
= sin(90
◦
−
φ
)
⇒
δ
= 90
◦
−
φ
= 61
.
4
◦
.
(b)
[0.5 points]
Let
t
up
(
δ
) be the time that an object at declination
δ
is above the horizon
over the course of a day. What is
t
up
(90
◦
) and
t
up
(
−
90
◦
) for us in Orlando?
At
δ
= 90
◦
, we’re looking at the North Celestial Pole, basically Polaris. Polaris is always
up – it’s circumpolar! Therefore
t
up
(90
◦
) must be
24 hours
.
At
δ
=
−
90
◦
, we’re looking at the South Celestial Pole. We can never see that far
down; remember from the last problem that the minimum
δ
we can see is
−
61
.
4
◦
. Therefore
t
up
(
−
90
◦
) must be
0 hours
.
(c)
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 Spring '09
 Fernandez
 Astronomy, Orlando, celestial pole, Polaris, N/A N/A N/A, Celestial coordinate system, Pole star, Circumpolar star

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