Credit/Permission: For text, © David Jeffery. For figures etc., as specified with the figure etc. / Only for reading and use by the instructors and students of the UNLV astronomy laboratory course.
This is a lab exercise with NO observations.
But for reference, see Sky map: Las Vegas, current time and weather.
Sections
We will learn something about the sky as it is dealt with in astronomy. This is entirely an inside lab. There are NO observations.
We do touch on the following topics:
Some of the
Tasks can be completed ahead of the lab period.
Doing some of them ahead of lab period would be helpful.
However, you can print a copy ahead of time if you like especially if
want to do some parts ahead of time.
You might have to compensate for updates in this case.
The Lab Exercise itself is NOT printed in the lab ever.
That would be killing forests
and the Lab Exercise is designed to be an active web document.
General remarks about quiz prep are given at
Quiz Preparation: General Instructions.
For DavidJ's lab sections, the quiz prep is doing all the items listed here and self-testing with the
Prep Quizzes and Prep Quiz Keys
if they exist.
However to complement and/or supplement the reading, you should at least
read the intro of a sample of the articles
linked
to the following keywords etc.
so that you can define and/or understand some keywords etc. at the level of our class.
The celestial sphere
is an imaginary sphere centered on the
Earth's center that is
infinitely more remote than all astronomical objects.
It does NOT exist.
But despite that detail, you can project all
astronomical objects
onto the celestial sphere from the location of the
Earth
and the locate them on the
celestial sphere.
A 2-dimensional angular coordinate system can be superimposed on the
celestial sphere for
location purposes.
There are several such systems each with its own rationale.
In this lab, we consider
equatorial coordinates
and
horizontal coordinates.
The basic idea of the celestial sphere is
illustrated in the figure below
(local link /
general link: celestial_sphere_000_basic_idea.html).
Under reconstruction.
Sub Tasks:
From the
Earth's perspective, the whole
celestial sphere
and all the astronomical objects
rotate westward
around the Earth once per day as a basic motion.
We call this rotation a geometrical
rotation.
The
eastward of
the Earth on its
axis
is called a physical rotation---though it also
geometrical rotation---but NOT just a geometrical rotation.
In our present context, "geometrical" and "physical" have special meanings which are
shorthands for much more wordy descriptions.
To explicate:
However, when you make a point of saying geometrical motion, you usually mean that
the reference frame
is NOT an
inertial frame.
If you do NOT make a point of saying "geometrical", you often mean that the
motion is relative to an
inertial frame.
So motions relative to
inertial frames
can be explained using
physical laws.
Geometrical motions CANNOT be explained by
physical laws without
making them physical motions by referencing them to an
inertial frame.
For example, you CANNOT explain the rotation of the
whole observable universe
around the Earth (taking the
Earth as
at rest)
by physical laws without
first explaining the Earth's
rotation relative to the observable universe
using physical laws, and then
explaining the first motion as geometrical motion.
In our modern understanding in terms of
general relativity,
they are free-fall frames
in uniform external gravitational fields.
Further explication is given in the very, very long figure below
(local link /
general link: frame_inertial_free_fall.html).
After reading the figure, "geometrical" and "physical" motions are now clear, hopefully.
For observational purposes, we usually take the
Earth's perspective
and think of the
celestial sphere as rotating
westward.
It's just easier.
It goes around once per day.
More precisely, once every
sidereal day = 86164.0905 s
= 1 day - 4 m + 4.0905 s (on average)
which is a bit shorter than the
metric day
=24 h = 86400 s
and the
solar day = current mean value 86400.002 s
(which is relative to the
Sun: i.e.,
solar noon to
solar noon period).
The sidereal day
is the time between transits
of the meridian
of a fixed star.
Note:
These motions are small compared to the daily rotation of the
celestial sphere
for anything much farther away than the
Moon
and for most observational purposes entirely negligible for
astronomical objects
well outside of the Solar System.
Why is the physical rotation period of the
Earth relative to the
inertial frame of the
fixed stars
(i.e., the
sidereal day = 86164.0905 s
= 1 day - 4 m + 4.0905 s (on average))
SHORTER than the
solar day = current mean value 86400.002 s
(i.e., the solar noon to
solar noon period)?
Answer in sentences with reference to the figure below
(local link /
general link: sidereal_solar_time_2.html).
HINT: You have to consider the Earth's
revolution around the Sun.
Answer:
While the Earth rotates eastward on
its axis,
it is also moving eastward along its orbital path.
Diagrammatically
it is clear that this orbital motion means the
Earth must rotate
more than one rotation relative to the fixed stars
in order to complete a solar day.
The extra time to complete the solar day
is about 4 minutes as the task-specified figure shows.
The Earth's axis
extended to the celestial sphere
becomes the
celestial axis
which has a north celestial pole (NCP)
and a south celestial pole (SCP).
The equator
projected onto the celestial sphere
from the Earth's center
is the celestial equator.
The celestial equator
is a great circle.
The astronomical objects
carried around with the daily rotation of the
celestial sphere execute
small circle
rotations
around the celestial axis
or great circle
rotations if they are actually on the celestial equator.
For more explication, see the figure below
(local link /
general link: celestial_sphere_001_features.html).
What is a great circle?
What is a small circle?
Answer in sentences.
Answer:
Both things are circles on a
sphere.
A great circle
divides the sphere in half
and a small circle
divides into unequal parts.
For most purposes, the
Earth is regarded as
a point compared to the
celestial sphere.
But human observers are regarded as
points on the
Earth which
appears as infinite plane
tangent to their location
on Earth.
This plane
divides the celestial sphere
into two parts.
The line of division is the horizon.
The situation is further explicated in the figure below
(local link /
general link: celestial_sphere_002_horizon.html)
The horizon
is a _________________ circle on the celestial sphere.
Answer: "great"
Sub Tasks:
HINT: The figure below
(local link /
general link: declination_altitude.html)
might stimulate your memory.
The equatorial coordinate system
is the main
celestial coordinate system
used for locating
astronomical objects on
the celestial sphere.
The equatorial coordinate system is explicated
and illustrated in the figure below
(local link /
general link: celestial_sphere_003_eqcoord.html).
Fill in the blanks below:
Below is figure
further explicating the
celestial sphere
and equatorial coordinates.
Below is an all-sky sky map
with the equatorial coordinate
lines marked on.
By eye-balling the
celestial globe
(if the instructor
remembered to put it out for you) or the
sky map shown in the figure above
(local link /
general link: sky_map_all_sky.html),
what is the approximate average
declination (Dec or δ) of:
Using
TheSky
find the southernmost
star
(i.e., the star closest to the
SCP) with
TheSky
reference magnitude
(which must be close to
the apparent V magnitude)
brighter (i.e., less) than or equal to 3.00 (i.e., ≤ 3.00).
What is its:
HINTS:
Geometrically, but NOT "physically", the
Sun orbits eastward around the
Earth once per year
on the celestial sphere.
The path of the
Sun
on the celestial sphere
is called the ecliptic.
The motion on the
ecliptic is superimposed on the daily
westward rotation of
the celestial sphere.
The two motions are, in fact, offset by an angle 23.4°.
The relatively short
orbital period of
the Sun means the
Sun moves fairly rapidly on the
celestial sphere
unlike the fixed stars.
The plane
of the Sun's
orbit
(or the Earth's
orbit from the
heliocentric perspective) is called the
ecliptic plane.
The line perpendicular to the
ecliptic plane
is called the
ecliptic axis.
The ecliptic,
ecliptic plane,
the ecliptic axis,
the celestial axis,
and
the Earth's axial tilt
are illustrated in the 2 figures below
(local link /
general link: ecliptic_plane.html;
local link /
general link: season_001_ecliptic.html):
1) from
heliocentric perspective,
2) from the geocentric
or celestial sphere perspective.
Complete the following sentences:
Sub Tasks:
Actually, the
Earth's axial tilt
is NOT constant in time.
There is a slow axial precession
which is a change in direction of the
Earth's axis
without a change in tilt angle.
We consider the
axial precession
below in section Axial Precession.
There is also a slow change in the size of the angle of tilt.
See the figure below
(local link /
general link: axial_tilt.html).
The time variation of the
Earth's axial tilt
is one of the Milankovich cycles
which have a profound effect on the
Earth's climate.
Sub Tasks:
Answer:
It's convenient to have a unit that gives the
Sun's
angular velocity as
approximately one unit per day. This permits easy approximate calculations.
Choosing 360 units does this: this is a first reason for their choice.
Why NOT 365.25 units which would give more nearly one unit per day?
365.25 units is an awkward number of units to deal with.
But 360 has 24 divisors
(see Wikipedia: 360 (number))
which often makes computations, particularly divisions, easy.
So there is a second reason for the choice: lots of divisors.
So for 2 reasons, the degree is a
natural unit
for solar astronomy
and its ancient adjunct astrology.
But we do NOT know for sure why the
ancient Babylonian astronomers
chose 360 units for the circle---they didn't
tell us why, they just did it.
For a more verbose explication of the reasoning of the
ancient Babylonian astronomers, see
babylonian_360_degrees.html.
Sub Tasks:
Answer:
It's easy to find the Sun
zodiac constellation.
Check the Wikipedia: Zodiac: Constellations (Table of Dates: scroll down ∼ 5%).
for the constellation boundaries
defined by IAU.
For the constellation opposite the
Sun just scroll ahead/back about 6 months.
(taken as 183 days in common years).
For Aug10--Sep15,
the Sun
constellation is
Leo ♌
(ordinal dates 222--258 containing
the Sun)
and the
midnight one is
Capricornus ♑
(ordinal dates 19--46 containing
the Sun),
Aquarius ♒
(ordinal dates 47--70 containing
the Sun),
or
Pisces ♓
(ordinal dates 71--108 containing
the Sun).
For Jan19--Feb15,
the Sun
constellation is
Capricornus ♑
(ordinal dates 19--46 containing
the Sun)
and the
midnight one is
Cancer ♋
(ordinal dates 202--221 containing
the Sun)
or
Leo ♌
(ordinal dates 222--258 containing
the Sun).
Note I've taken 6 months as 183 days in the calculations for the above results
and used common years
only and NOT leap years.
I didn't want to consider too many finicky special cases just to obtain results that must be
a bit approximate anyway.
What zodiac sign
contained the Sun when you
(or your group leader if appropriate) were born (i.e., what is
your zodiac sign)
and what
zodiac constellation
contained the Sun when you
(or your group leader if appropriate) were born?
See
Wikipedia: Zodiac: Twelve signs
(Table of Dates: Scroll down ∼ 5%)
(in the column for
tropical zodiac)
and
Wikipedia: Zodiac: Constellations
(Table of Dates: scroll down ∼ 5%)
(in the column for
IAU boundaries).
Answer:
Taurus ♉
and Taurus ♉.
But any answer is right.
How the night sky changes
as the Sun moves along the
ecliptic is explicated in
the figure below
(local link /
general link: zodiac_ecliptic.html)
this task
and applet figure above
(local link /
general link: naap_zodiac.html)
this task
(if it is working which it probably is NOT).
Sub Tasks:
Answer: Earlier---or as yours truly
likes to mnemonick it, earlier and earlier.
The actual amount earlier is about 4 minutes since the
sidereal day
is about 4 minutes shorter than the solar day
which is the nearly equal to
standard civil day.
Answer:
You will have to imagine the diagram, but it's much the same as the figure below
(local link /
general link: zodiac_ecliptic.html).
As the Sun orbits the
Earth eastward in the
geocentric perspective, the night sky
orbits the Earth eastward too opposite
the Sun.
So the night sky
moves pari passu
with the Sun an
angular velocity of about 1 degree/day.
Calculating
accurate and precise
times for
sunrise and
sunset is difficult.
One needs to use spherical trigonometry,
accurate knowledge of the Earth's orbit
and Earth's rotation,
conversions to standard time
from solar time,
and probably a host of minor effects.
The precision is NOT useless
since it allows accurate comparisons with more accurate
ways of determining sunrise times.
So there.
The approximate sunrise formula
for the Northern Hemisphere is
If the calculated t_sunrise ≤ 0 h, the Sun is
always above the horizon.
If the calculated t_sunrise ≥ 12 h, the Sun is
always below the horizon.
These events should happen at or north of the
Arctic Circle (i.e., at
latitude ≥ 66.6°), but the
formula being not-so-accurate gives the events at
latitude ≥ 75.4°).
For most latitudes, the formula
is accurate to within an hour.
However, for latitude ≅ 70, the
formula is only accurate to within about 2 hours.
For the Southern Hemisphere, "6 -" goes to "6 +"
and for sunset, t_sunset=24-t_sunrise.
Sub Tasks:
You will have convert today's date in to month count from the beginning of the
year with the
decimal fraction included.
For example, if it is June 21, t_month ≅ 5.67.
In your answer, you will have to convert time from hours to hours and minutes.
Answer:
For the example date June 21 and
the Las Vegas, Nevada
latitude 36° north,
yours truly gets sunrise at
4.868 am or 4:52 am.
Answer:
The USNO
sunrise time for some year
(probably 2014) for the example is
4:24 am. So the formula is off by 28 minutes which is NOT so bad given that it is a
crude approximation.
Horizontal coordinates
(AKA local coordinates) are local in time and space to an observer.
He/she is at the origin at the instant in time when
a specific set of
horizontal coordinates
apply.
The horizontal coordinates
are explicated in the figure below
(local link /
general link: horizontal_coordinates.html).
Sub Tasks:
Answer:
The horizontal coordinates
are local to the observer in time and space.
To expand a bit, they change continuously and rapidly with time and location on
Earth.
So they can't used in a catalog.
Answer;
They are good for the observer at their time and space location.
To expand a bit, they are good for locating
astronomical objects
in the sky
at the observer's immediate location and time.
Often one just points in the
azimuth direction
and then points up to the
altitude
or says it out loud.
In this section, we consider global and local views of the
celestial sphere
in order to get a unified understanding.
The figure below
(local link /
general link: celestial_sphere_004_day.html)
and
applet below that
(local link /
general link: naap_rotating_sky_explorer.html)
(if it is working which it probably is NOT)
illustrate the difference between the global and local
view of the
celestial sphere for an observer
at one point on the Earth.
Sub Tasks:
Northern Hemisphere,
which is longer day or
night on this day? ________________________
Answer: Day.
In the Northern Hemisphere,
which is longer day or
night on this day? ________________________
Answer: Night.
HINT: What
latitude circle
is always cut in half by the
Earth's
terminator.
Answer:
Following up the hint, the
location is the equator.
There are two ways of understanding this result.
First, the horizon plane at the
equator is aligned with the
the celestial axis.
So all the astronomical objects
have their daily small circle paths
on the sky cut in half.
So they all spend half their time above and half their time below the
horizon: i.e., 12 hours above
and 12 hours below.
The Sun, which is a
astronomical object, will
do the same.
So on the equator
day and
night are always 12 hours long?
Second, just consider that symmetry dictates that the
equator must always
intersect the terminator
on both sides of the Earth.
Both equator and
terminator are
great circles, so their intersections
cuts them both in half.
So rotation time from terminator
to terminator
on the equator
must be always be half a day: i.e., 12 hours.
Sub Tasks:
Diagram:
Answer: The proof is by inspection.
of the diagram.
The proof and the proofs of the general
formulae are given
in the figure below
(local link / general link: declination_altitude.html).
HINTS:
Answer:
It has to be within an angle
equal to the altitude
of the NCP.
That angle is all the
angular space available between the
NCP and the
northern horizon.
If the astronomical object
is farther from the NCP
than that angle,
the daily rotation of the
the celestial sphere
will cause it to dip below the northern
horizon and it won't be
circumpolar object.
Clearly, the declination
must be ≥ 90° - L, where L is the latitude.
For Las Vegas, Nevada, the
angle is ∼ 36°.
Answer:
For Las Vegas, Nevada:
Camelopardalis (CAM),
Cassiopeia (CAS),
Cepheus (CEP),
Draco (DRA),
Ursa Major (UMA),
Ursa Minor (UMI).
The general formulae relating
altitude AN/S
(upper/lower case for altitude
from due north/due south)
along the meridian,
declination (δ),
and latitude (L)
(with southern latitudes counted as
negative numbers)
are:
For proof of these formulae, see the figure above
(local link /
general link: declination_altitude.html).
Sub Tasks:
Answer:
AN(NCP) = AN(δ=90°) = 90° + (L-90°) = L
and AS(SCP) = AS(δ=-90°) = 90° - (L+90°) = -L
Answer:
Obviously A(zenith) = 90° always from either
due north or due south.
From the formula, one gets AN/S = 90°+(±)N/S(L-δ) = 90° since δ = L for zenith.
Answer:
Since on an equinox day,
the Sun is at δ = 0°,
we find L = 0° and you are on the
equator.
The intensity or
flux of a beam of
light is energy flow per unit time per
area perpendicular to the beam's direction.
A completely absorbing surface
perpendicular to a beam absorbs
power per unit area equal to the beam's intensity.
But as you tilt the absorbing surface away from the beam, the
power per unit area decreases and reaches zero when the
surface is parallel to the beam.
Sub Tasks:
The intensity of sunlight
at the mean orbital radius
is called the
solar constant.
It has an average value of 1367.6 watts per square meter (W/m**2)
(see NASA Earth fact sheet)
and it varies by only about
0.1 % NOT counting short-term variations due to
sunspots.
The solar constant
is a fundamental number for the
biosphere.
If you take the Earth's
cross-sectional area πR**2
and divide it by its surface area 4πR**2, you get the fraction 1/4.
This quantity is average insolation
at the top of the Earth's atmosphere.
About half of this power gets reflected or absorbed in transit to the ground,
leaving about 170 W/m**2 as the
ground-level average insolation.
This is what plants
and solar power both have
to live with.
Currently, the best readily available
commercial solar cells have efficiencies
ranging up to about 20 %
(see Wikipedia:
Solar cell efficiency: Comparison of Energy Conversion Efficiencies).
So they could harvest in a time-averaged world-averaged sense up to about 34 W/m**2.
This is a very low power density per unit area.
There is lots of solar energy
to harvest, but it will take extensive
solar power plants.
What about plants?
Their maximum efficiency for human-utilizable power is about 1 W/m**2.
This is why
biofuels can never be
the main source of commercial power.
There isn't enough land surface area that can be spared from other
uses (e.g., running ecosystems, supplying food)
to supply more than a fraction of the
world energy consumption.
The Earth's axial tilt
is approximately constant in with respect to the
fixed stars over time scales of order a human lifetime.
This means the average insolation
on
the Northern Hemisphere
and Southern Hemisphere
varies during the course of the year.
But this is irrelevant to the seasons.
Sub Tasks:
The Earth's axis
has a fixed direction in space, but NOT relative to the
Sun because of the
Earth's orbital motion.
The continuosly varying orientation of the
Earth's axis
continuously and inversely varies the amount of
heat energy
absorbed by the
Northern Hemisphere
and Southern Hemisphere.
The amount absorbed increases with tilting toward the
Sun because of more direct beaming of
light on the
Earth's surface.
The more heat energy absorbed
the higher temperature.
The upshot is that then a hemisphere
tilts toward/away from the Sun, it has
summer/winter.
In between the extremes, the
hemisphere has
spring and
fall.
So the
Earth's axial tilt
is the cause of the seasons.
It's really hard to give a concise, cogent answer.
Students should get full marks if they say anything reasonable.
Does the eccentricity
of the Earth's orbit
have any effect on the Earth's climate?
Explain.
Answer:
Well, it must have some effect, but evidently it's small because it doesn't override the effect of the
Earth's axial tilt effect
since perihelion is the coldest
time in the Northern Hemisphere
and aphelion the hottest.
So the Earth's axial tilt is
the very much the main cause of the seasons.
Answer:
In the Northern-Hemisphere summer,
Earth's axis
leans toward the Sun in the
north, and so the
North Pole is always on the
day side of the Earth.
In the Northern-Hemisphere winter, the
situation is reversed with the North Pole always
on the night side of the Earth.
The South Pole has the
mirror image behavior.
Sub Tasks:
Answer:
vernal equinox
summer solstice,
fall equinox,
winter solstice.
Note the terms
solstice and equinox
have two meanings: one is the event and the other
is the place on the
celestial sphere
where the event occurs.
Fill in the data in the table below following the instructions below.
Two-digit precision suffices.
The
Earth's axis
exhibits an
axial precession.
Precession
is the sweeping out of a cone or a
double cone
by a rotating body's rotation axis.
See the figure below
(local link /
general link: double_cone.html)
of a
double cone.
Now the Earth has a
rotational axis
and the gravity of the
Sun,
Moon,
and, to a much lesser degree, planets
cause the axis to have
precession: the
Earth's axial precession
In older jargon, the
Earth's axial precession
was called the
precession of the equinoxes
since it causes the
equinoxes
to slid westward along the
celestial equator.
The kinematics of the
Earth's axial precession
is explained in the figure below
(local link /
general link: axial_precession_akinematics.html).
The short answer is
gravitational perturbations.
The longer answer is given in the two figures below.
The
solar year and the
sidereal year,
and their difference caused
to the axial precession
are explicated in the figure below
(local link /
general link: axial_precession_year.html).
Sub Tasks:
What is the angular velocity in degrees per day of the
Earth? R_e = _________________________
Answer: R_e = 0.985609113115 degrees/day.
Using the exact period result
25,771.5 Julian years, what is the
angular velocity in degrees per day of
vernal equinox
westward along the
ecliptic? R_v = _________________________
Answer: R_v = 3.82448*10**(-5) degrees/day.
The Sun
moves eastward
and
the vernal equinox
moves westward.
Let's take eastward as positive
and westward as negative.
So the angular velocity of the
vernal equinox is minus the value you calculated above
(i.e., R_v = - |R_v|).
The time t for the Sun to lap the
vernal equinox satisfies
the equation
Answer: We have
Answer:
The solar year
or the tropical year.
This is the year from vernal equinox
to vernal equinox. It is the year
of the Earth's
seasons.
Explain why the equatorial coordinates
for astronomical objects beyond the
Solar System
have to be updated every few years. Give the main reason and a second reason.
Answer:
The main reason is because of the
axial precession.
The equatorial coordinate system
is tied to the Earth's
equator.
Thus, the
axial precession
will continuously, slowly, and systematically shift
equatorial coordinates of all
astronomical objects.
The shift is small over a human lifetime, but precision
astrometry requires
frequent updating of
equatorial coordinates.
In modern times, computers
update the
equatorial coordinates whenever
that is needed.
Reference catalog values traditionally get updated every 50 years. The last such update is referred
to as the J2000 epoch.
The second reason is that nearby extra-solar-system
astronomical objects
do have measurable proper motion
in space relative to the
to the Sun.
So precise astrometry requires occasional updates
of their equatorial coordinates.
Of course, astronomical objects
in the Solar System need large, continuous
updating for their
equatorial coordinates because
of their large proper motions in
angle.
Goodnight all.
Keywords:
altitude,
axial precession,
azimuth,
celestial axis,
celestial coordinate systems,
celestial equator,
celestial globe,
celestial meridian
(AKA the meridian),
celestial sphere,
declination (Dec),
ecliptic,
equatorial coordinates,
equinox,
horizon,
horizontal coordinates
(AKA local coordinates),
nadir,
north celestial pole (NCP),
orrery,
right ascension (RA),
season,
solstice,
south celestial pole (SCP),
TheSky
(TheSky6,
TheSkyX,
List of Tricks for TheSky,
TheSky Orientation),
transit,
zenith,
zodiac
(see Wikipedia: Zodiac: Constellations
(Table of Dates: scroll down ∼ 5%)).
Task Master:
EOF
End of Task
End of Task
The value varies a bit with reference.
Yours truly has given the
apparent V magnitude
from Wikipedia: β Hydri (β HYD).
The difference between
accurate and precise
is explicated in the figure below
(local link /
general link: accuracy_precision.html).
However,
a simple approximate formula of low accuracy for times can be derived using simplying assumptions---the main ones
being using ordinary planar trigonometry
instead of spherical trigonometry
and just leaving the result in solar time.
It's low accuracy, but
rather precise since formally
the formula has 3 signficiant figures.
The formula has no free parameters.
t_sunrise = 6 - (24/360)*23.4*sin[360*(t_month-2.59)/12]/tan(90-L)
= 6 - 1.56*sin[360*(t_month-2.59)/12]/tan(90-L) ,
where t_sunrise is in hours,
6 is 6 am is the ideal equinox sunrise time,
t_month is the time in the year in months (e.g., Dec31 midnight is 0,
Jan31 midnight is 1.000, etc.),
2.59 is the approximate time of the vernal equinox,
and
L is latitude.
AN(NCP) = L ,
where L is latitude
(counted as negative if
south latitude)
and the AN(NCP) is counted as negative if the
NCP is
below the horizon.
AN/S = (±)N/S(L - δ) + 90°
δ = L +(±)N/S(90° - AN/S)
L = δ +(±)N/S( AN/S - 90°)
the Sun
transits the meridian
at zenith.
What is your latitude
and where are you?
Multiply 1/4 by the solar constant
and you get the time-averaged power per unit area entering the top of the
Earth's atmosphere.
Actually the
Earth's axial tilt
direction does vary slowly as we will consider below in the
section Axial Precession.
The figures below explicate the
variation in the average insolation.
End of Task
AN/S = 90°+(±)N/S(L-δ) ,
where the upper/lower case is for
altitude from
due north/due south,
L is the local latitude
(counting southern latitudes as negative),
and δ is the declination (Dec)
of the astronomical object.
For the last column, you will need the lower case
(i.e., AS = 90°-N/S(L-δ))
since the
Sun at an
altitude measured from
due south.
_____________________________________________________________________
Table of Solstice and Equinox Data
_____________________________________________________________________
Sun Position Fiducial Date RA Dec Altitude of the Sun
(h) (degrees) in Las Vegas at
Solar Noon
(degrees)
_____________________________________________________________________
vernal equinox Mar21 0 0 54
summer solstice Jun21 6 23.4 77
fall equinox Sep21 12 0 54
winter solstice Dec21 18 -23.4 31
_____________________________________________________________________
It's actually very difficult to explain why
precession happens
even though it is a pretty common phenomenon.
It happens to
toy tops
as the figure below illustrates.
To explain the cause of precession,
it suffices to say for the moment that
if gravity
tries to topple (i.e., torque)
a rotating object, a precession of the
object's rotational axis can happen.
360° = (R_e - R_v)t
What is t in days?
Since the answer requires
double-precision math,
we will just give the result:
t = 360°/(R_e - R_v) = 360°/(0.985609113115 + 0.0000382448) = 365.2421904276 days.
The accepted
solar year = 365.2421897 days (J2000).
So the two values agree to 8 digit places. The calculation wasn't so bad.
This section is only
for remote instruction.
Post mortem comments that may often apply specifically to
Lab 2: The Sky: