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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top