Mintaka is a star almost on the celestial equator, right ascension, a = 5h 31m 57.3s, and declination, d = -00 o 18' 02" - if I observed it from one week to the next how far will it have traveled? Anyone?

Thanks!

Tetsuwan_Atom wrote:

Mintaka is a star almost on the celestial equator, right ascension, a = 5h 31m 57.3s, and declination, d = -00 o 18' 02" - if I observed it from one week to the next how far will it have traveled? Anyone? Thanks!

Since it is a stationary object unlike the moon and planets, asteroids and comets, It will have remained at the same Right Ascension (RA) and the same Declination (DEC).. Although It's altitude and location in the night sky for any giving time will be different from one night to the next.. Since our time is standard to the earths rotation in correlation with the sun and not the Deep Space Celestial objects. a Solar day is longer than a celestial day would be. So objects will rise in the sky a few minutes earlier each night..
  Currently the Altitude is about 19 or 20 degrees, in a week it will be about 24 to 25 degrees..With minimal change in declination..Only a matter of about 0.9 arc seconds difference..

Over extremely long periods of time do to wobble in the Earths axis, the RA and DEC will change slightly.. That is why ever so many years star charts are updated. Currently we are on Epoch 2000.

I see what you're saying but I'm still trying to recreate this scenario...

If I looked at Mintaka on 7/15 at 4:00am and then again at 4:30am it would appear to have moved ____ degrees.

Then if I checked again on 7/22 it would have appeared to move ____ degrees.

This is an astronomy101 class assignment but I forgot to wake up at 4am ):

Thank you!

Tetsuwan_Atom wrote:

I see what you're saying but I'm still trying to recreate this scenario... If I looked at Mintaka on 7/15 at 4:00am and then again at 4:30am it would appear to have moved ____ degrees. Then if I checked again on 7/22 it would have appeared to move ____ degrees. This is an astronomy101 class assignment but I forgot to wake up at 4am ): Thank you!

The night sky rotation is15 degees per hour. So from 4am to 4:30am it would have moved 7.5 degrees..

If you checked again in 7 days it would be at the same RA and Declination.. Only in a week it would be in the same visual location in the sky about a half hour earlier..  And due to the rotation of the earth and the orbit around the sun with the axis angle of the earth the altitude will change slightly.

at 7/15 at 4:00am the location of Mintaka is:

RA: 5hr 32m 15.934s,  Dec: -00deg 17' 32.021" 
Azimuth of: 80deg 21m 40s altitude of: -14 06m 47s

at seven days from that time and date 7/22 4:00am it would be

RA: 5hr 32m 16.134  Dec: -00deg 17' 31.024"
Azimuth: 84deg 26m 21s Altitude: -08 deg 29m 22s

The more noticeable change would be in the altitude and azimuth in the night sky. Other than that you are only talking about fractions of an arc second in RA or Declination. And that is all  due to the Earths rotation on it axis and time of rotation as I already discussed.

Tetsuwan_Atom wrote:

This is an astronomy101 class assignment but I forgot to wake up at 4am ):

Well, I was working on a response which would have made you work for the solution as opposed to having it handed to you.  Make sure you thank Tim when you are graded on this quiz.

tmefferd wrote:

Tetsuwan_Atom wrote: This is an astronomy101 class assignment but I forgot to wake up at 4am ): Well, I was working on a response which would have made you work for the solution as opposed to having it handed to you.  Make sure you thank Tim when you are graded on this quiz.

Yea, But I wouldn't use my figures for any test.  Mine is only an example. If he uses them it will be wrong... My figures are close for my location around that time. (Latitude, Longitude and elevation)..  His would be different for his location at the time specified.. Plus I think mine is more than he really needs.. I just did that to show that the differences in the RA and DEC are very minimal. The most obvious visual difference would be in the Altitude and Azimuth..

To figure everything out exactly you would need to take into account  your exact location of observation,, the time of earths rotation (1 earth day).  The angle of the earths axis to the object,  the length of a sidereal day,  period of time it takes for objects to move a specific distance through the sky etc etc..
All these factors can determine that down to the arc second if you want..  If you want to go even much further you can even figure the earths wobble into that..
 
I don't think he needs to be that accurate. I think what he wants is the visual location in the night sky. (Alt-Az) So I think all he needs to know is the speed or motion of an object moves through the sky. Then he will need to figure the difference from a solar day and a sidereal day. Which I did not give an answer to. He needs to research to find what the differences are.(Which isn't much). Once that is known then it is easy to figure out...

Take the difference between a (mean)solar earth day and a celestial (sidereal) day. Multiply that by seven. that will give the amount of time difference needed to calculate how many degrees of difference there will be for an objects location/ movement in the night sky..

Thank you thank you!

Yeah, I live in Arizona and I'm just trying to make an educated guess.  Since our calibrated measuring device is our fingers everyone should get slightly different answers.

Hmm...I hope my teacher isn't a member here.  o_O;;