Astronomy homework, right ascension

AI Thread Summary
The discussion clarifies the relationship between a star's right ascension and its visibility in the night sky. It notes that while the star is 15 minutes and 39.3 seconds behind the sun, it will set shortly after the sun, making it invisible during dark hours at the specified latitude. The right ascension of a star remains constant, while the sun's position changes throughout the year. By calculating the local sidereal time (LST) at midnight, it is determined that visible right ascensions range from 04h to 08h. This understanding helps in accurately answering astronomy homework questions related to star visibility.
petha1
Messages
5
Reaction score
1
Homework Statement
Will a star with right ascension of 14h,15m,39,3s be visible on October 21 from a latitude of 56 degrees 53 minutes, 0 seconds? Assume that night is between [22.00,02.00]
Relevant Equations
Right ascension measures how many hours ahead or behind the sun a star is at the vernal equinox, decreses by 2 hours/month.
March 21 - October 21 = 7 months. So the star is only 15m 39,3 seconds behind the sun. This means that the star won't be up when the sun is down.

Answer: No, the star will set only 15 minutes after the sun has set. Therefore it won't be on the sky at the specified latitude during dark hours.

Is this correct reasoning?
 
Physics news on Phys.org
petha1 said:
Relevant Equations:: Right ascension measures how many hours ahead or behind the sun a star is at the vernal equinox, decreses by 2 hours/month.
Your answer is correct, but this statement is misleading. The right ascension of a star does not change. It is the position of the sun which changes through the year.
 
Thanks for the help. I think I figured out the correct way to answer this question.
October 21 is 9 months since the vernal equinox. That means that the LST at noon is 18, leading to
the LST at midnight beeing 18+12 = 30 = 6 (mod 24) so the visible RAs are in the range [6-2,6+2] = [4,8]so the visible right ascensions are between 04h and 08h.
 
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