Which hemisphere is the observer?

AI Thread Summary
The discussion centers on determining the hemisphere of an observer based on a celestial image without cardinal directions. Participants clarify that the left side of the image represents West, applicable to both hemispheres, but the observer's perspective indicates they are in the Southern Hemisphere. The plane of the ecliptic is discussed, highlighting how its orientation differs between hemispheres, with the Moon's position providing clues about the observer's location. Calculations regarding the Sun's position during the December solstice suggest a latitude of approximately 27° south. The conversation concludes with questions about the angles related to the Sun's position and the ecliptic, emphasizing the importance of understanding celestial mechanics in observations.
guv
Messages
122
Reaction score
22
Thread moved from the technical forums to the schoolwork forums
TL;DR Summary: Astro Olympiad Problem determining the latitude of an observer from a picture taken.

usaaao_2020_R2_V1_Q3.png

Well this question and answer are really confusing. There are no cardinal directions labelled on the picture. However because the Sun and the Moon should move on a circular path, the left side should be West. This is true for both north and south hemisphere dwellers. What's the reason this is taken by someone living in the southern hemisphere?
 
Physics news on Phys.org
guv said:
the left side should be West.
What do you mean 'left side'? You're looking towards the sunset. The West will be roughly in that direction.

A line drawn between the Moon and the Sun shows you the plane of the ecliptic. The orbit of the Moon is slightly inclined (~5 deg) to it, so it's not exact, but the question lets you ignore it anyway. If you extended that line to form a great circle across the sky, it would be the path the Sun, the Moon, and the planets take through the day (again, ignoring inclinations).
If you are on the Northern hemisphere, you'll see the plane of the ecliptic from 'above'. Conversely on the Southern hemisphere, you see it from 'below'.
Therefore, in the north, when looking to the west, the Moon will be angled to the south - which will be leftwards. In the south, its tilted to the north - to the right of you.
 
  • Like
Likes vanhees71 and guv
Based on the ecliptic argument, and personal experience down-under, the observer is clearly in the Southern Hemisphere.

The Sun appears 23.5° deg south at the December solstice, which is the southern summer solstice. Counting from overhead, the ecliptic is 50° degrees down, but 23.5° is due to season, which makes the latitude about 50° - 23° = 27° south.

Sidereal time zero is at high meridian, noon, on the March (vernal) equinox, at zero longitude, (London). The December solstice is 9 months after the March equinox, so the sidereal clock has lagged 24h * 9/12 = 18 hours, plus six hours since noon for the sunset observation, which makes 18 + 6 = 24 = 0 hours. The sidereal time is therefore about zero, but that ignores the fact that at 27° south in summer, the sun sets later than 6 PM, which puts the sidereal time into the early hours of the sidereal morning.

There must be at least one mistake in that reasoning, or maybe two that cancel out.
 
I looked at this in Stellarium before posting, unfortunately I was mislead by the pictures without extending ecliptic into an line. Now it's obvious.
 
For the 2nd question, looking at their solution, I wonder how they draw the conclusion ##\angle K' Sun = 90## and ##\angle K' S Sun = 180##. I came up with this picture. My K is the same as K' (ecliptic pole).The intersections between the Sun's path (slanted dotted path) and the horizon (H) are the sunrise (front) and sunset (back) points. How is ##\angle K Sun = 90##? Any line from K to a point on the ecliptic will be perpendicular to the ecliptic, but the Sunset point is not on the ecliptic, it's on the horizon. What's wrong with the picture I draw?
 

Attachments

  • Screenshot from 2023-03-12 11:23:20.png
    Screenshot from 2023-03-12 11:23:20.png
    40.6 KB · Views: 110
  • 20230312_111936.jpg
    20230312_111936.jpg
    22.4 KB · Views: 99
guv said:
but the Sunset point is not on the ecliptic, it's on the horizon.
The Sun is always on the ecliptic. At sunset, the horizon and the ecliptic intersect where the Sun is.
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top