Location of Solar System object

AI Thread Summary
To find the latitude and longitude of the point on Earth's surface directly under the Sun, the declination of the Sun corresponds to the latitude. The longitude can be determined by calculating the local noon time based on UTC, which aligns with the Sun being due south. Right Ascension (RA) is not strictly necessary for this calculation, as the local sidereal time can help establish the longitude. The discussion emphasizes the relationship between celestial coordinates and terrestrial positioning, particularly how the Sun's movement affects these calculations. Understanding these concepts is crucial for accurately determining the subpoint of solar system objects.
jbaratta
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Hi, I am a non physicist/astronomer and having a little difficulty navigating definitions. Apologies if the question has already been covered elsewhere.

If I calculate the RA/ Dec of say, the sun, on a certain date and time, how do I find the lat/long of the point on the Earth's surface directly under the sun. That is, the point on the Earth's surface on the line joining the centres of the Earth and sun?

Thanks
 
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jbaratta said:
Hi, I am a non physicist/astronomer and having a little difficulty navigating definitions. Apologies if the question has already been covered elsewhere.

If I calculate the RA/ Dec of say, the sun, on a certain date and time, how do I find the lat/long of the point on the Earth's surface directly under the sun. That is, the point on the Earth's surface on the line joining the centres of the Earth and sun?

Thanks

Aren't they the same? Declination of the Pole Star is +90 degrees. For it to be at zenith, your Earth location would have to be at +90 degrees. Likewise Declination of the celestial equator is 0. You'd have to be at Earth's equator for anything on the celestial equator to be at zenith.

Right Ascension is trickier...

Don't quote me on this.
 
DaveC426913 said:
Aren't they the same? Declination of the Pole Star is +90 degrees. For it to be at zenith, your Earth location would have to be at +90 degrees. Likewise Declination of the celestial equator is 0. You'd have to be at Earth's equator for anything on the celestial equator to be at zenith.

Right Ascension is trickier...

Don't quote me on this.

Thanks for the response.

I also think that declination and latitude are the same. RA is different and I think it involves somehow the Hour Angle
 
Not getting many responses to this question so I'll stick my neck out!

Would appreciate someone telling me if I'm on the right track.

The question is: given the RA and Dec of an object, what is the latitude and longitude of the point on the Earth's surface that lies on the line joining the centres of the object and earth?

I think that the latitude of this point is equal to the Declination. Is this correct?

I think that the longitude is found by setting the equation for Local Sidereal Time equal to the RA and solving for longitude. In other words, the Hour Angle must be equal to zero. Is this correct?
 
jbaratta said:
Not getting many responses to this question so I'll stick my neck out!

Would appreciate someone telling me if I'm on the right track.

The question is: given the RA and Dec of an object, what is the latitude and longitude of the point on the Earth's surface that lies on the line joining the centres of the object and earth?

I think that the latitude of this point is equal to the Declination. Is this correct?

I think that the longitude is found by setting the equation for Local Sidereal Time equal to the RA and solving for longitude. In other words, the Hour Angle must be equal to zero. Is this correct?
Ah well, if this is a homework assigment, you might want to post it in the homework fora.

Hang on a sec...
 
This is not homework. It's more of a hobby/ area of self study!
 
jbaratta said:
This is not homework. It's more of a hobby/ area of self study!

Even if it's for self study, it should be in the Homework Help forums. I'll move it there now.

This thread helps to explain our policy on schoolwork:

https://www.physicsforums.com/showthread.php?t=373889

.
 
Apologies for posting a question in the wrong area!
 
jbaratta said:
Apologies for posting a question in the wrong area!

No worries. Now back to the problem. Are you still having problems with it?
 
  • #10
Yes, I would appreciate any and all assistance!

Are my statements above on the right track?
 
  • #12
berkeman said:
I'm not able to answer your questions directly, but the wikipedia article seems pretty helpful:

http://en.wikipedia.org/wiki/Equatorial_coordinate_system

.

Thanks. I have read that article along with many others and I cannot see any information that answers my specific question.
 
  • #13
You don't really need the RA. If you have the time in UTC you have the longitude: The Sun is always due south at local noon, so you just calculate the longitude at which it is local noon based on the UTC.
 
  • #14
russ_watters said:
You don't really need the RA. If you have the time in UTC you have the longitude: The Sun is always due south at local noon, so you just calculate the longitude at which it is local noon based on the UTC.

Is this the same longitude that would make the Hour Angle equal to zero?
 
  • #15
Let's just talk about stars for starters, they don't have variable declination. Let's also go back to the medieval idea that they are fixed in a sphere around the earth. Now that sphere rotates around the Earth (not true but close enough). We have an observational longitude at Greenwhich. We draw a corresponding star map with an relative longitude at the Vernal Equinox and on this chart we measure the distance as Right Ascension (I prefer degrees to hours).

Now the subpoint of any star can be predicted from the observer's longitude and the time UCT (once GMT, but the USAF has called it Z for decades anyway).

As for the sun and the moon, their declination varies regularly; and the planets (wanderers) are even worse; but if you can predict where they are in the star map you can predict where their subpoint would be.

Perhaps it is necessary to add that West longitude needs to be converted to a 360 degree system: 170 W long is 190. In order to conform to the stellar map. Then you have to covert your calculations back to map coordinates.
 
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