Local Sidereal Time, RA of a star

  • Thread starter Thread starter bunchedupwalrus
  • Start date Start date
  • Tags Tags
    Local Star Time
AI Thread Summary
To determine when a star with a right ascension of 16h00m00s will be at its highest point on February 29, 2016, local sidereal time (LST) must equal the star's right ascension. The calculation involves understanding that LST is offset from solar time, with a rough estimate of a 4-minute difference per day. The Vernal Equinox marks 0h LST, but the sun is at its highest point at 12h solar time, resulting in an additional 12-hour offset. Therefore, the star will reach its highest point around 4:04 AM on March 20, 2016, not 2 PM as initially estimated. Accurate calculations and sketches can aid in visualizing the celestial mechanics involved.
bunchedupwalrus
Messages
9
Reaction score
0

Homework Statement


*A star on the celestial equator has right ascension of 16h00m00s. At what time of day will this star be at it's highest point, on Febuary 29th 2016. *

Homework Equations


HA = LST - RA

The Attempt at a Solution



When LST=RA, it'll be at it's highest point. How do I determine LST? There are many different explanations online. As a rough estimate, solar is off from sidereal by 4 minutes. 0 hour is on March 20th. Subtracting 4 minutes per day for 20 days gives me a difference of about 80 minutes.

So it'd be 16h - (80minutes), roughly 2pm?
 
Physics news on Phys.org
bunchedupwalrus said:
When LST=RA, it'll be at it's highest point.
Correct.
As a rough estimate, solar is off from sidereal by 4 minutes.
Correct.
0 hour is on March 20th.
Not correct. The sun is at 0h RA at the Vernal Equinox, roughly March 20. So when the sun is at its highest point on Mar 20, this is 0h LST. But the sun is at its highest point at 12h solar time, not 0h solar time. So there is an additional 12h offset.
 
I'd try drawing a sketch of the situation.

Also, consider this: on March 20, the star will be highest in the sky at 4 am. A day earlier, it would have risen 4 minutes later, so it would be highest around 4:04 am, right? Your answer of 2 pm can't be right, even ignoring the am/pm difference.
 
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