# Apparent Motion

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Hi all, given the R.A. and Dec apparent motion I’ve been trying to work out how Sky Safari arrived at the Total apparent motion. I originally thought I could use Pythagoras and then realised seconds of arc are not the same as seconds, so allowed for that, but I’m still failing. I guess it’s more complicated than I imagined but I’m now curious.

Can anyone enlighten me.

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That's a really difficult computation. You have a moving Earth and a moving comet.

The orbital data for both are available from various sites (JPL, minor planet centre, etc). The program will download that data and calculate the epheremides of the comet (in this case).

There are pre-written routines to do this calculation, I use Pyephem and Skyfield. Both are Python modules but use C to do the maths.

You could try it the hard way https://adsabs.harvard.edu/full/1982JRASC..76..157T

When the App says total AP. They mean adding in precession and nutation etc, which in this case are microscopic in relation

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10 minutes ago, AstroKeith said:

That's a really difficult computation. You have a moving Earth and a moving comet.

The orbital data for both are available from various sites (JPL, minor planet centre, etc). The program will download that data and calculate the epheremides of the comet (in this case).

There are pre-written routines to do this calculation, I use Pyephem and Skyfield. Both are Python modules but use C to do the maths.

You could try it the hard way https://adsabs.harvard.edu/full/1982JRASC..76..157T

When the App says total AP. They mean adding in precession and nutation etc, which in this case are microscopic in relation

Many thanks Keith,

I have the answer, I was just wondering how they arrived at it, but it sounds too difficult from your reply.

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One thing you need to appreciate is that the RA is measured close to the Celestial Pole (at a declination of almost 80 degrees) so you have

Round figures :

7909 seconds of RA per day = about 2h 11.8m of RA (32.95 degrees)

But this is at 80 degrees N, so the distance across the sky is reduced by Cos (80 degrees) = ~5.72 degrees

This is 20600" along the Small Circle at 80 degrees Latitude

Along with the 5325" along the Great Circle of Longitude, you can apply Pythagoras (it's a small triangle) to get the 21,500" ish daily motion

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.

Edited by Scooot
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1 hour ago, Gfamily said:

One thing you need to appreciate is that the RA is measured close to the Celestial Pole (at a declination of almost 80 degrees) so you have

Round figures :

7909 seconds of RA per day = about 2h 11.8m of RA (32.95 degrees)

But this is at 80 degrees N, so the distance across the sky is reduced by Cos (80 degrees) = ~5.72 degrees

This is 20600" along the Small Circle at 80 degrees Latitude

Along with the 5325" along the Great Circle of Longitude, you can apply Pythagoras (it's a small triangle) to get the 21,500" ish daily motion

Ah that makes sense, thanks very much.

I started looking at it because I wanted to assess the exposure I could take as I knew the comet was travelling very fast and one thing led to another

Over the course of a minute the RA didn’t change much at the time I was about to image but the Dec changed by about 8.6”. I just used Sky Safari to fast forward the time by a minute to work it out. As my kit images at 2.06 arc second per pixel i guessed I’d be ok at 20 second exposures without the comet trailing. Probably could have got away with longer but that’s what I decided to do in the end and the subs seem to be ok. Don’t know if this is the simplest way to work it out, or even if it’s right but I’ve got a bunch of subs to process

Anyway, thanks again.

Edited by Scooot
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On 30/01/2023 at 01:14, Scooot said:

Hi all, given the R.A. and Dec apparent motion I’ve been trying to work out how Sky Safari arrived at the Total apparent motion. I originally thought I could use Pythagoras and then realised seconds of arc are not the same as seconds, so allowed for that, but I’m still failing. I guess it’s more complicated than I imagined but I’m now curious.

Can anyone enlighten me.

In general, try

sqrt( (15*v_RA)^2 * [1 + COS(2*Dec) ]/2 +v_Dec^2 ).

I haven't had time to check my result.

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5 minutes ago, George Jones said:

In general, try

sqrt( (15*v_RA)^2 * [1 + COS(2*Dec) ]/2 +v_Dec^2 ).

I haven't had time to check my result.

Great, I’ll give that a go thanks. What’s v in the formula?

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21 minutes ago, Scooot said:

Great, I’ll give that a go thanks. What’s v in the formula?

Silly me. What I wrote is equivalent to

edit

sqrt(  [15*v_RA *  cos(Dec) ]^2 +v_Dec^2 )

It seems that @Gfamily's result is exact.

RA speed is v_RA = -7909.5669

Dec speed is v_Dec = 5325.382

Dec = 79 + 52/60 + 58.5/3600 =79.8829167

Edited by George Jones
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1 hour ago, Scooot said:

Great, I’ll give that a go thanks. What’s v in the formula?

It's just an indicator that it's the Velocity in each direction

• v_RA  is velocity in RA (seconds per day)
• v_Dec is velocity in Declination (" per day)
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46 minutes ago, Gfamily said:

It's just an indicator that it's the Velocity in each direction

• v_RA  is velocity in RA (seconds per day)
• v_Dec is velocity in Declination (" per day)

Oh I see thanks.

I just worked through your formula using Pythagoras and arrived at 21,277 so very close to Sky Safari. Adjusting for the 80° North makes sense now so thanks for explaining.

Edited by Scooot
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When I substitute the exact numbers given in the original post into the formula, I get 21510.57.

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