Need some help...

Hi people!
Just started an astronomy class this semester and here are some of the
questions that we have to answer and I just wanted you to check out my
answers to what I knew and if any of you can help me out a bit, I'd
appreciate it a lot, but even for checking, thank you all so much. Wherever
you see only one answer to the question, that means that I am pretty sure
that's the correct answer out of the 4 possible answers that are given on
the page. Please comment on them if you see something fishy about my answers
and these are all the answers I could dig up from the book, because in my
lectures, we just look at the movies and stuff like that, which is pretty
fun.

1st Q: In modern astronomy, the constellations are...
Ans: 88 non overlapping sky regions, covering the whole sky.

2nd Q: Suppose you look at the photograph of many galaxies. Assuming that
all galaxies formed at about the same time, which galaxy in the picture is
the youngest?
Ans: The one that is farthest away from you.

3rd Q: One major difference between the Sun and the Moon in our sky is
that...
Ans: The Sun emits light while the Moon merely scatters and reflects it.

4th Q: What is the ecliptic?
Ans: The Sun's apparent path along the celestial sphere.

5th Q: Which of the following statements about the celestial equator is true
at all latitudes?
Ans: It's an extension of the Earth's equator onto the celestial sphere.

6th Q: An asteroid that is moving south in your local sky, from the North
Celestial Pole toward the Celestial Equator can be described as having
its...
A: Declination decrease with time B: Right ascension decrease with
time
C: Right ascension increase with time D: Declination increase with
time
( I don't know the answer or any of these terms to be honest )

7th Q: Over the course of one night, an observer at any given location on
the Earth sees the constellation gradually shift across the sky from east to
west. This is caused primarily by:
A: The inherent rotation of the Universe B: The rotation of the Earth
around its own axis
C: Precession of the spin axis of the Earth D: The orbital motion of
the Earth around the Sun
( Just don't know the answer here)

8th Q: Which of the following statements will be correct about the behavior
of your zenith over a period of one year if you were standing upon the
Earth's equator?
Ans: It will always be perpendicular to the Earth's spin axis
( Please check make sure this makes sense to you, because I could be wrong,
because as it seems, I am kind of crappy when it comes to the orientation
and so on)

9th Q: Over an interval of 6 months, the tilt of the Earth's spin axis with
respect to the background starts will change by...
Ans: 0* (* = degrees)

10th: If the Sun moved through a full range of declination of 47*
(from -23.5* to +23.5*) over the whole year, what would be the declination
of the Sun at the central lone spot in the figure 8 that it would make
across the sky?
A: +14.9* B: 0*, on the celestial equator C: -8.6* D) +8.6*
( I can draw you a little picture in paint or something if you don't
understand this question, because we have a little picture of it in the
book)

11th Q: The approximate date around March 21 represents which season for
people living in the New Zealand?
Ans: Beginning of autumn.

12th Q: Which of the following objects will never be seen form Earth's
crescent?
A: Mercury B: Mars C: Venus D: the Moon
(I am not sure what they mean by this question, but it sounds interesting
and if someone can explain what it means to me, once again great big thanks)

13th Q: If the Moon is setting at noon, the phase of the Moon must be...
A: full B: waning crescent C: half D: waxing crescent

14th Q: There is about a 5* angle between the orbit of the Moon and the...
And: Plane of the ecliptic, or the Earth's orbit

15th Q: The Earth's shadow at a distance of the Moon's orbit from the Earth
is...
A: considerably wider than the Moon B: Slightly less wide than the
size of the Moon
C: Almost exactly as wide as the Moon D: extremely small, leaving
only narrow shadow band of the Moon during eclipse

Now just two more word problems that I didn't solve and the 1st one it the
most important one.

1. Derive the equation from which you could calculate speeds of someone on
the surface of the Earth, depending on their latitude. The equation will
depend on person's latitude, (Q), the radius of the Earth, (Re), and the
period of rotation, (T).
( I tried it, but nothing.) (Hints: the speed of a person on a North pole is
0 km/hr and speed of a person on the Equator is 1650 km/hr)

2. Suppose we discover an Earth like planet in another solar system. It
revolves around a star like our Sun along a circular orbit of radius two
astronomical units and an axis tilt of 35*. Would you expect this plant to
experience seasons? If so, would you expect them to be more or less extreme
than the season on Earth ( meaning cold periods and hot periods would last a
lot longer than here on Earth)?

Once again thank you all who ever has some valuable input on any of these
questions and I just started the course, so I am sure I'll have some more
questions about astronomy later on.

Zarkovic wrote:

[snip qq. 1-5]

All the answers you gave for the first five seem fine, although
without seeing the other choices it's hard to be sure there were no
better choices. For the following I'll just provide some hints; if
they don't help please ask for clarification.


6th Q: An asteroid that is moving south in your local sky, from the North
Celestial Pole toward the Celestial Equator can be described as having
its...
A: Declination decrease with time B: Right ascension decrease with
time
C: Right ascension increase with time D: Declination increase with
time
( I don't know the answer or any of these terms to be honest )

RA and Dec. are respectively like longitude and latitude on earth,
but of course on the celestial sphere. RA is conventionally given in
hours rather than degrees, and is measured from the "first point of
Aries" (AKA the "vernal equinox"), the point where the ecliptic
crosses the equator from south to north. Declination is sometimes
given in degrees north or south of the celestial equator, but often
as a signed number, positive declinations being to the north of the
equator and negative ones to the south.

On earth a southbound ship in the northern hemisphere will have a
decreasing latitude, as it gets closer to the equator. Does the
analogy help?

7th Q: Over the course of one night, an observer at any given location on
the Earth sees the constellation gradually shift across the sky from east to
west. This is caused primarily by:
A: The inherent rotation of the Universe B: The rotation of the Earth
around its own axis
C: Precession of the spin axis of the Earth D: The orbital motion of
the Earth around the Sun
( Just don't know the answer here)

The main clue here is "over the course of one night": the rotational
motion involved must have a period on the order of a day. The periods
of the above cycles a A) ? -- not believed to occur, or at least
too slow to detect; B) 24 hours WRT the sun, or about 23h56m WRT the
fixed stars; C) about 26,000 years; D) 365.2422 days. Given these
data the answer should be pretty obvious.

8th Q: Which of the following statements will be correct about the behavior
of your zenith over a period of one year if you were standing upon the
Earth's equator?
Ans: It will always be perpendicular to the Earth's spin axis
( Please check make sure this makes sense to you, because I could be wrong,
because as it seems, I am kind of crappy when it comes to the orientation
and so on)

That sounds reasonable, but again I can't be sure that none of the
other choices was a better answer.

9th Q: Over an interval of 6 months, the tilt of the Earth's spin axis with
respect to the background starts will change by...
Ans: 0* (* = degrees)

Very nearly. Unless there were other choices giving very small
angles, it's almost certainly the best one.

10th: If the Sun moved through a full range of declination of 47*
(from -23.5* to +23.5*) over the whole year, what would be the declination
of the Sun at the central lone spot in the figure 8 that it would make
across the sky?
A: +14.9* B: 0*, on the celestial equator C: -8.6* D) +8.6*
( I can draw you a little picture in paint or something if you don't
understand this question, because we have a little picture of it in the
book)

See www.analemma.com for some very nicely presented illustrations,
graphs, and animations.

11th Q: The approximate date around March 21 represents which season for
people living in the New Zealand?
Ans: Beginning of autumn.

I prefer to regard the equinoxes and solstices as the midpoints,
rather than the beginnings, of their respective seasons -- but I'm in
the minority; your answer reflects the most commonly heard
descriptions of these dates.

12th Q: Which of the following objects will never be seen form Earth's
crescent?
A: Mercury B: Mars C: Venus D: the Moon
(I am not sure what they mean by this question, but it sounds interesting
and if someone can explain what it means to me, once again great big thanks)

The question seems a bit garbled, and should probably read "... never
be seen from Earth as a crescent." Try looking up "inferior" and
"superior" planets; if this doesn't provide any illumination -- so to
speak -- I'll try to explain further.

13th Q: If the Moon is setting at noon, the phase of the Moon must be...
A: full B: waning crescent C: half D: waxing crescent

Picture the situation: assuming we're in the northern hemisphere the
sun will be due south and the moon will be in the west, so the two
bodies must be near 90° apart. Does that help?

14th Q: There is about a 5* angle between the orbit of the Moon and the...
And: Plane of the ecliptic, or the Earth's orbit

Sounds fine.

15th Q: The Earth's shadow at a distance of the Moon's orbit from the Earth
is...
A: considerably wider than the Moon B: Slightly less wide than the
size of the Moon
C: Almost exactly as wide as the Moon D: extremely small, leaving
only narrow shadow band of the Moon during eclipse

See http://www.MrEclipse.com/Special/LEprimer.html.

Now just two more word problems that I didn't solve and the 1st one it the
most important one.

1. Derive the equation from which you could calculate speeds of someone on
the surface of the Earth, depending on their latitude. The equation will
depend on person's latitude, (Q), the radius of the Earth, (Re), and the
period of rotation, (T).
( I tried it, but nothing.) (Hints: the speed of a person on a North pole is
0 km/hr and speed of a person on the Equator is 1650 km/hr)

You need to do a little trigonometry; try drawing a diagram of the
situation as viewed from above the North Pole, looking down on the
earth. I've got to go to bed, so haven't the time right now ... if
you're still stuck tomorrow and no-one else has chimed in I'll see if
I can walk you through the calculation. What have you tried so far?

2. Suppose we discover an Earth like planet in another solar system. It
revolves around a star like our Sun along a circular orbit of radius two
astronomical units and an axis tilt of 35*. Would you expect this plant to
experience seasons? If so, would you expect them to be more or less extreme
than the season on Earth ( meaning cold periods and hot periods would last a
lot longer than here on Earth)?

Have you any ideas on this one? What causes the seasons on Earth? The
Analemma site I referenced above may provide some clues. Again, if
you haven't made any progress by tomorrow I'll try to provide some
more help then.

--
Odysseus

Zarkovic wrote:

[snip qq. 1-5]

All the answers you gave for the first five seem fine, although
without seeing the other choices it's hard to be sure there were no
better choices. For the following I'll just provide some hints; if
they don't help please ask for clarification.


6th Q: An asteroid that is moving south in your local sky, from the North
Celestial Pole toward the Celestial Equator can be described as having
its...
A: Declination decrease with time B: Right ascension decrease with
time
C: Right ascension increase with time D: Declination increase with
time
( I don't know the answer or any of these terms to be honest )

RA and Dec. are respectively like longitude and latitude on earth,
but of course on the celestial sphere. RA is conventionally given in
hours rather than degrees, and is measured from the "first point of
Aries" (AKA the "vernal equinox"), the point where the ecliptic
crosses the equator from south to north. Declination is sometimes
given in degrees north or south of the celestial equator, but often
as a signed number, positive declinations being to the north of the
equator and negative ones to the south.

On earth a southbound ship in the northern hemisphere will have a
decreasing latitude, as it gets closer to the equator. Does the
analogy help?

7th Q: Over the course of one night, an observer at any given location on
the Earth sees the constellation gradually shift across the sky from east to
west. This is caused primarily by:
A: The inherent rotation of the Universe B: The rotation of the Earth
around its own axis
C: Precession of the spin axis of the Earth D: The orbital motion of
the Earth around the Sun
( Just don't know the answer here)

The main clue here is "over the course of one night": the rotational
motion involved must have a period on the order of a day. The periods
of the above cycles a A) ? -- not believed to occur, or at least
too slow to detect; B) 24 hours WRT the sun, or about 23h56m WRT the
fixed stars; C) about 26,000 years; D) 365.2422 days. Given these
data the answer should be pretty obvious.

8th Q: Which of the following statements will be correct about the behavior
of your zenith over a period of one year if you were standing upon the
Earth's equator?
Ans: It will always be perpendicular to the Earth's spin axis
( Please check make sure this makes sense to you, because I could be wrong,
because as it seems, I am kind of crappy when it comes to the orientation
and so on)

That sounds reasonable, but again I can't be sure that none of the
other choices was a better answer.

9th Q: Over an interval of 6 months, the tilt of the Earth's spin axis with
respect to the background starts will change by...
Ans: 0* (* = degrees)

Very nearly. Unless there were other choices giving very small
angles, it's almost certainly the best one.

10th: If the Sun moved through a full range of declination of 47*
(from -23.5* to +23.5*) over the whole year, what would be the declination
of the Sun at the central lone spot in the figure 8 that it would make
across the sky?
A: +14.9* B: 0*, on the celestial equator C: -8.6* D) +8.6*
( I can draw you a little picture in paint or something if you don't
understand this question, because we have a little picture of it in the
book)

See www.analemma.com for some very nicely presented illustrations,
graphs, and animations.

11th Q: The approximate date around March 21 represents which season for
people living in the New Zealand?
Ans: Beginning of autumn.

I prefer to regard the equinoxes and solstices as the midpoints,
rather than the beginnings, of their respective seasons -- but I'm in
the minority; your answer reflects the most commonly heard
descriptions of these dates.

12th Q: Which of the following objects will never be seen form Earth's
crescent?
A: Mercury B: Mars C: Venus D: the Moon
(I am not sure what they mean by this question, but it sounds interesting
and if someone can explain what it means to me, once again great big thanks)

The question seems a bit garbled, and should probably read "... never
be seen from Earth as a crescent." Try looking up "inferior" and
"superior" planets; if this doesn't provide any illumination -- so to
speak -- I'll try to explain further.

13th Q: If the Moon is setting at noon, the phase of the Moon must be...
A: full B: waning crescent C: half D: waxing crescent

Picture the situation: assuming we're in the northern hemisphere the
sun will be due south and the moon will be in the west, so the two
bodies must be near 90° apart. Does that help?

14th Q: There is about a 5* angle between the orbit of the Moon and the...
And: Plane of the ecliptic, or the Earth's orbit

Sounds fine.

15th Q: The Earth's shadow at a distance of the Moon's orbit from the Earth
is...
A: considerably wider than the Moon B: Slightly less wide than the
size of the Moon
C: Almost exactly as wide as the Moon D: extremely small, leaving
only narrow shadow band of the Moon during eclipse

See http://www.MrEclipse.com/Special/LEprimer.html.

Now just two more word problems that I didn't solve and the 1st one it the
most important one.

1. Derive the equation from which you could calculate speeds of someone on
the surface of the Earth, depending on their latitude. The equation will
depend on person's latitude, (Q), the radius of the Earth, (Re), and the
period of rotation, (T).
( I tried it, but nothing.) (Hints: the speed of a person on a North pole is
0 km/hr and speed of a person on the Equator is 1650 km/hr)

You need to do a little trigonometry; try drawing a diagram of the
situation as viewed from above the North Pole, looking down on the
earth. I've got to go to bed, so haven't the time right now ... if
you're still stuck tomorrow and no-one else has chimed in I'll see if
I can walk you through the calculation. What have you tried so far?

2. Suppose we discover an Earth like planet in another solar system. It
revolves around a star like our Sun along a circular orbit of radius two
astronomical units and an axis tilt of 35*. Would you expect this plant to
experience seasons? If so, would you expect them to be more or less extreme
than the season on Earth ( meaning cold periods and hot periods would last a
lot longer than here on Earth)?

Have you any ideas on this one? What causes the seasons on Earth? The
Analemma site I referenced above may provide some clues. Again, if
you haven't made any progress by tomorrow I'll try to provide some
more help then.

--
Odysseus

6th Q: An asteroid that is moving south in your local sky, from the
North
Celestial Pole toward the Celestial Equator can be described as having
its...
A: Declination decrease with time (this is my answer according to
your explanations, is it right?)

7th Q: Over the course of one night, an observer at any given location
on
the Earth sees the constellation gradually shift across the sky from
east to
west. This is caused primarily by:
B: The rotation of the Earth around its own axis (again I used
your reasoning for this, so tell me if I messed up)

10th: If the Sun moved through a full range of declination of 47*
(from -23.5* to +23.5*) over the whole year, what would be the
declination
of the Sun at the central lone spot in the figure 8 that it would make
across the sky?
A: +14.9* B: 0*, on the celestial equator C: -8.6* D) +8.6*
I went to the site and I checked out all the movies and diagrams, but
still I couldn't quite get it. I am thinking it's either 0*, but something
tells me that's not true, or that it is either 8.6 plus or minus, but I'd
say 60% sure that it's 0*, so answer B



12th Q: Which of the following objects will never be seen form Earth's
crescent?
A: Mercury B: Mars C: Venus D: the Moon
I'd say Mars, because the Moon, Mercury and Venus are all in front of the
Earth, so that's how I sort of deducted that answer. What do you say about
that?

13th Q: If the Moon is setting at noon, the phase of the Moon must be...
A: full B: waning crescent C: half D: waxing crescent

According to the picutre in our text book, it says that Third Quater phase
sets at noon, but mabye answer B would be correct here.

15th Q: The Earth's shadow at a distance of the Moon's orbit from the
Earth
is...
A: considerably wider than the Moon (my guess from the picture
that's on this website below)

http://www.MrEclipse.com/Special/LEprimer.html

Now just two more word problems that I didn't solve and the 1st one it
the
most important one.

1. Derive the equation from which you could calculate speeds of someone
on
the surface of the Earth, depending on their latitude. The equation will
depend on person's latitude, (Q), the radius of the Earth, (Re), and the
period of rotation, (T).
( I tried it, but nothing.) (Hints: the speed of a person on a North
pole is
0 km/hr and speed of a person on the Equator is 1650 km/hr)

I tried the trig part of it. created a triangle ABC and angle BAC was north
pole, angle ABC was the intersection between the latitude and longitude
lines both at 0 degrees and angle ACB was my angle that led from the (0,0)
angle to the right of Earth's equator, and I tried all that, but I couldn't
come up with it. Please, please help on this one. MOST URGENT AND IMPORTANT
QUESTION

2. Suppose we discover an Earth like planet in another solar system. It
revolves around a star like our Sun along a circular orbit of radius two
astronomical units and an axis tilt of 35*. Would you expect this plant
to
experience seasons? If so, would you expect them to be more or less
extreme
than the season on Earth ( meaning cold periods and hot periods would
last a
lot longer than here on Earth)? I learned that seasons would be more
extreme in class yesterday. If you can fill me in on this with any other
ideas then sure, but I think i got this question in the bag.

Thanks a lot man, and hopefully I wont bore you to death and mabye you can
even help me sometime in future.

6th Q: An asteroid that is moving south in your local sky, from the
North
Celestial Pole toward the Celestial Equator can be described as having
its...
A: Declination decrease with time (this is my answer according to
your explanations, is it right?)

7th Q: Over the course of one night, an observer at any given location
on
the Earth sees the constellation gradually shift across the sky from
east to
west. This is caused primarily by:
B: The rotation of the Earth around its own axis (again I used
your reasoning for this, so tell me if I messed up)

10th: If the Sun moved through a full range of declination of 47*
(from -23.5* to +23.5*) over the whole year, what would be the
declination
of the Sun at the central lone spot in the figure 8 that it would make
across the sky?
A: +14.9* B: 0*, on the celestial equator C: -8.6* D) +8.6*
I went to the site and I checked out all the movies and diagrams, but
still I couldn't quite get it. I am thinking it's either 0*, but something
tells me that's not true, or that it is either 8.6 plus or minus, but I'd
say 60% sure that it's 0*, so answer B



12th Q: Which of the following objects will never be seen form Earth's
crescent?
A: Mercury B: Mars C: Venus D: the Moon
I'd say Mars, because the Moon, Mercury and Venus are all in front of the
Earth, so that's how I sort of deducted that answer. What do you say about
that?

13th Q: If the Moon is setting at noon, the phase of the Moon must be...
A: full B: waning crescent C: half D: waxing crescent

According to the picutre in our text book, it says that Third Quater phase
sets at noon, but mabye answer B would be correct here.

15th Q: The Earth's shadow at a distance of the Moon's orbit from the
Earth
is...
A: considerably wider than the Moon (my guess from the picture
that's on this website below)

http://www.MrEclipse.com/Special/LEprimer.html

Now just two more word problems that I didn't solve and the 1st one it
the
most important one.

1. Derive the equation from which you could calculate speeds of someone
on
the surface of the Earth, depending on their latitude. The equation will
depend on person's latitude, (Q), the radius of the Earth, (Re), and the
period of rotation, (T).
( I tried it, but nothing.) (Hints: the speed of a person on a North
pole is
0 km/hr and speed of a person on the Equator is 1650 km/hr)

I tried the trig part of it. created a triangle ABC and angle BAC was north
pole, angle ABC was the intersection between the latitude and longitude
lines both at 0 degrees and angle ACB was my angle that led from the (0,0)
angle to the right of Earth's equator, and I tried all that, but I couldn't
come up with it. Please, please help on this one. MOST URGENT AND IMPORTANT
QUESTION

2. Suppose we discover an Earth like planet in another solar system. It
revolves around a star like our Sun along a circular orbit of radius two
astronomical units and an axis tilt of 35*. Would you expect this plant
to
experience seasons? If so, would you expect them to be more or less
extreme
than the season on Earth ( meaning cold periods and hot periods would
last a
lot longer than here on Earth)? I learned that seasons would be more
extreme in class yesterday. If you can fill me in on this with any other
ideas then sure, but I think i got this question in the bag.

Thanks a lot man, and hopefully I wont bore you to death and mabye you can
even help me sometime in future.

Zarkovic wrote:

6th Q: An asteroid that is moving south in your local sky, from the
North
Celestial Pole toward the Celestial Equator can be described as having
its...
A: Declination decrease with time (this is my answer according to
your explanations, is it right?)

Yes.

7th Q: Over the course of one night, an observer at any given location
on
the Earth sees the constellation gradually shift across the sky from
east to
west. This is caused primarily by:
B: The rotation of the Earth around its own axis (again I used
your reasoning for this, so tell me if I messed up)

That's right.

10th: If the Sun moved through a full range of declination of 47*
(from -23.5* to +23.5*) over the whole year, what would be the
declination
of the Sun at the central lone spot in the figure 8 that it would make
across the sky?
A: +14.9* B: 0*, on the celestial equator C: -8.6* D) +8.6*
I went to the site and I checked out all the movies and diagrams, but
still I couldn't quite get it. I am thinking it's either 0*, but something
tells me that's not true, or that it is either 8.6 plus or minus, but I'd
say 60% sure that it's 0*, so answer B

The analemma is asymmetrical, because it's produced by two
independent sinusoidal cycles: one from the eccentricity of the
earth's orbit, and the other from the tilt of the earth's axis. The
'crossover' point is therefore some distance from the equator, as
seen in the second diagram in the "summation - effect" frame at
www.analemma.com. This shows that at 8.6°N latitude, the sun passes
directly overhead at mean solar noon -- the equation of time then
being near zero -- on two different dates every year, once in
mid-April and once near the end of August.


12th Q: Which of the following objects will never be seen form Earth's
crescent?
A: Mercury B: Mars C: Venus D: the Moon
I'd say Mars, because the Moon, Mercury and Venus are all in front of the
Earth, so that's how I sort of deducted that answer. What do you say about
that?

Yes: Mars doesn't exhibit a crescent phase as seen from earth. Venus
and Mercury appear as crescents near when they are closest to us, on
either side of their "inferior conjunctions" with the Sun, and of
course the moon appears as a crescent every month, just before and
after the New Moon.

13th Q: If the Moon is setting at noon, the phase of the Moon must be...
A: full B: waning crescent C: half D: waxing crescent

According to the picutre in our text book, it says that Third Quater phase
sets at noon, but mabye answer B would be correct here.

The terminology for the moon's phases is a little confusing; at
"first quarter" and "third quarter" the visible disk is *half*
illuminated -- so I think the best answer is C. A crescent moon sets
just before or after the sun does, and the full moon sets near sunrise.


15th Q: The Earth's shadow at a distance of the Moon's orbit from the
Earth
is...
A: considerably wider than the Moon (my guess from the picture
that's on this website below)

http://www.MrEclipse.com/Special/LEprimer.html

Yes. That's why lunar eclipses are more frequent than solar; the
shadow cast by the moon on the earth is comparatively very small.

Now just two more word problems that I didn't solve and the 1st one it
the
most important one.

1. Derive the equation from which you could calculate speeds of someone
on
the surface of the Earth, depending on their latitude. The equation will
depend on person's latitude, (Q), the radius of the Earth, (Re), and the
period of rotation, (T).
( I tried it, but nothing.) (Hints: the speed of a person on a North
pole is
0 km/hr and speed of a person on the Equator is 1650 km/hr)

I tried the trig part of it. created a triangle ABC and angle BAC was north
pole, angle ABC was the intersection between the latitude and longitude
lines both at 0 degrees and angle ACB was my angle that led from the (0,0)
angle to the right of Earth's equator, and I tried all that, but I couldn't
come up with it. Please, please help on this one. MOST URGENT AND IMPORTANT
QUESTION

Start by picturing the earth as viewed from above the north pole.
Over the course of a day every point on the earth's surface traces
out a circle. The size of the circle depends on the latitude: at the
pole (90° lat.) the radius is zero, so a person standing there is
stationary, while a person on the equator (which is of course a great
circle) travels the full circumference of the earth, for a speed of

s = 2 * pi * R_e / T = 2 * 3.14 * 6380 km / 24 h = 1670 km/h.

A person at any other latitude will be travelling at some speed
between these two extremes, depending on the radius r of the small
circle he traces during the day:

s = 2 * pi * r / 24 h.

So we need to find a way of determining r from the latitude to
produce a general formula. To keep things simple treat the earth as a
perfect sphere. We need another diagram, this time looking at a
cross-section of the earth from the plane of the equator: draw a
circle with a vertical diameter representing the earth's axis, from
the north pole through the centre (labelled C) to the south pole. A
horizontal diameter through C represents the equator; label its
right-hand end E. Now choose a point P on the circle and draw a
radius CP joining that point to the centre; the angle ECP is its
latitude, Q. The radius of the small circle traced by P in its
diurnal motion is a horizontal line from P to the axis, which it
meets at L, forming a right angle. Assuming my description wasn't too
hard to follow, you should now have a right triangle CLP, with
hypoteneuse CP = R_e, the angle at C equal to 90° - Q (the
"colatitude"), and the opposite leg LP = r. Since

r / R_e = sin(90° - Q),

r = R_e * cos(Q).

Plugging this into the previous formula gives

s = 2 * pi * R_e * cos(Q) / 24 h.

When Q = 0°, cos(Q) = 1, giving the expected result for the equator,
and likewise for the poles where cos(Q) = 0. For a latitude of 45°
you should get s = 1180 km/h. Note that Q = -45° (for 45°S) gives the
same result as well. It's hard explaining without being able to draw
the diagrams; I hope you were able to follow my descriptions.

2. Suppose we discover an Earth like planet in another solar system. It
revolves around a star like our Sun along a circular orbit of radius two
astronomical units and an axis tilt of 35*. Would you expect this plant
to
experience seasons? If so, would you expect them to be more or less
extreme
than the season on Earth ( meaning cold periods and hot periods would
last a
lot longer than here on Earth)? I learned that seasons would be more
extreme in class yesterday. If you can fill me in on this with any other
ideas then sure, but I think i got this question in the bag.

The circular orbit is a red herring; it seems to be a common
misconception that our seasons are caused by the eccentricity of the
earth's orbit. At any rate a planet with a larger axial inclination
than the earth's will experience more pronounced seasons. With a
wider, slower orbit the hypothetical planet's seasons will be not
only more extreme than the earth's, but longer as well.

--
Odysseus

Zarkovic wrote:

6th Q: An asteroid that is moving south in your local sky, from the
North
Celestial Pole toward the Celestial Equator can be described as having
its...
A: Declination decrease with time (this is my answer according to
your explanations, is it right?)

Yes.

7th Q: Over the course of one night, an observer at any given location
on
the Earth sees the constellation gradually shift across the sky from
east to
west. This is caused primarily by:
B: The rotation of the Earth around its own axis (again I used
your reasoning for this, so tell me if I messed up)

That's right.

10th: If the Sun moved through a full range of declination of 47*
(from -23.5* to +23.5*) over the whole year, what would be the
declination
of the Sun at the central lone spot in the figure 8 that it would make
across the sky?
A: +14.9* B: 0*, on the celestial equator C: -8.6* D) +8.6*
I went to the site and I checked out all the movies and diagrams, but
still I couldn't quite get it. I am thinking it's either 0*, but something
tells me that's not true, or that it is either 8.6 plus or minus, but I'd
say 60% sure that it's 0*, so answer B

The analemma is asymmetrical, because it's produced by two
independent sinusoidal cycles: one from the eccentricity of the
earth's orbit, and the other from the tilt of the earth's axis. The
'crossover' point is therefore some distance from the equator, as
seen in the second diagram in the "summation - effect" frame at
www.analemma.com. This shows that at 8.6°N latitude, the sun passes
directly overhead at mean solar noon -- the equation of time then
being near zero -- on two different dates every year, once in
mid-April and once near the end of August.


12th Q: Which of the following objects will never be seen form Earth's
crescent?
A: Mercury B: Mars C: Venus D: the Moon
I'd say Mars, because the Moon, Mercury and Venus are all in front of the
Earth, so that's how I sort of deducted that answer. What do you say about
that?

Yes: Mars doesn't exhibit a crescent phase as seen from earth. Venus
and Mercury appear as crescents near when they are closest to us, on
either side of their "inferior conjunctions" with the Sun, and of
course the moon appears as a crescent every month, just before and
after the New Moon.

13th Q: If the Moon is setting at noon, the phase of the Moon must be...
A: full B: waning crescent C: half D: waxing crescent

According to the picutre in our text book, it says that Third Quater phase
sets at noon, but mabye answer B would be correct here.

The terminology for the moon's phases is a little confusing; at
"first quarter" and "third quarter" the visible disk is *half*
illuminated -- so I think the best answer is C. A crescent moon sets
just before or after the sun does, and the full moon sets near sunrise.


15th Q: The Earth's shadow at a distance of the Moon's orbit from the
Earth
is...
A: considerably wider than the Moon (my guess from the picture
that's on this website below)

http://www.MrEclipse.com/Special/LEprimer.html

Yes. That's why lunar eclipses are more frequent than solar; the
shadow cast by the moon on the earth is comparatively very small.

Now just two more word problems that I didn't solve and the 1st one it
the
most important one.

1. Derive the equation from which you could calculate speeds of someone
on
the surface of the Earth, depending on their latitude. The equation will
depend on person's latitude, (Q), the radius of the Earth, (Re), and the
period of rotation, (T).
( I tried it, but nothing.) (Hints: the speed of a person on a North
pole is
0 km/hr and speed of a person on the Equator is 1650 km/hr)

I tried the trig part of it. created a triangle ABC and angle BAC was north
pole, angle ABC was the intersection between the latitude and longitude
lines both at 0 degrees and angle ACB was my angle that led from the (0,0)
angle to the right of Earth's equator, and I tried all that, but I couldn't
come up with it. Please, please help on this one. MOST URGENT AND IMPORTANT
QUESTION

Start by picturing the earth as viewed from above the north pole.
Over the course of a day every point on the earth's surface traces
out a circle. The size of the circle depends on the latitude: at the
pole (90° lat.) the radius is zero, so a person standing there is
stationary, while a person on the equator (which is of course a great
circle) travels the full circumference of the earth, for a speed of

s = 2 * pi * R_e / T = 2 * 3.14 * 6380 km / 24 h = 1670 km/h.

A person at any other latitude will be travelling at some speed
between these two extremes, depending on the radius r of the small
circle he traces during the day:

s = 2 * pi * r / 24 h.

So we need to find a way of determining r from the latitude to
produce a general formula. To keep things simple treat the earth as a
perfect sphere. We need another diagram, this time looking at a
cross-section of the earth from the plane of the equator: draw a
circle with a vertical diameter representing the earth's axis, from
the north pole through the centre (labelled C) to the south pole. A
horizontal diameter through C represents the equator; label its
right-hand end E. Now choose a point P on the circle and draw a
radius CP joining that point to the centre; the angle ECP is its
latitude, Q. The radius of the small circle traced by P in its
diurnal motion is a horizontal line from P to the axis, which it
meets at L, forming a right angle. Assuming my description wasn't too
hard to follow, you should now have a right triangle CLP, with
hypoteneuse CP = R_e, the angle at C equal to 90° - Q (the
"colatitude"), and the opposite leg LP = r. Since

r / R_e = sin(90° - Q),

r = R_e * cos(Q).

Plugging this into the previous formula gives

s = 2 * pi * R_e * cos(Q) / 24 h.

When Q = 0°, cos(Q) = 1, giving the expected result for the equator,
and likewise for the poles where cos(Q) = 0. For a latitude of 45°
you should get s = 1180 km/h. Note that Q = -45° (for 45°S) gives the
same result as well. It's hard explaining without being able to draw
the diagrams; I hope you were able to follow my descriptions.

2. Suppose we discover an Earth like planet in another solar system. It
revolves around a star like our Sun along a circular orbit of radius two
astronomical units and an axis tilt of 35*. Would you expect this plant
to
experience seasons? If so, would you expect them to be more or less
extreme
than the season on Earth ( meaning cold periods and hot periods would
last a
lot longer than here on Earth)? I learned that seasons would be more
extreme in class yesterday. If you can fill me in on this with any other
ideas then sure, but I think i got this question in the bag.

The circular orbit is a red herring; it seems to be a common
misconception that our seasons are caused by the eccentricity of the
earth's orbit. At any rate a planet with a larger axial inclination
than the earth's will experience more pronounced seasons. With a
wider, slower orbit the hypothetical planet's seasons will be not
only more extreme than the earth's, but longer as well.

--
Odysseus

Hey thanks a lot guy. Sorry to bug you the last couple of times. Let me know
if there is any way I can repay you for your help. Thanks again man.

Hey thanks a lot guy. Sorry to bug you the last couple of times. Let me know
if there is any way I can repay you for your help. Thanks again man.