<p>Your question is valid and is a common operation. When computing when an object will be visible, many observers want to be able to draw a globe or map that is marked with the position from which an object like Pluto is directly overhead. Such a position on the globe at a given instant (and you do say “at the current time”) is, all other things being equal, the best-situated place from which to observe the object.</p>

<p>Unfortunately, the “libastro” C library that PyEphem wraps seems to only provide one instance of this concept: Earth satellites have <code>.sublat</code> and <code>.sublong</code> attributes, because it is so common to want to draw the path of a satellite on a globe or map.</p>

<p>But libastro does not generalize the concept. From what I can see, there is no way to generate the point on the Earth that is directly below any other Solar System object. And since I have not been in the habit of trying to extend libastro, it is likely that PyEphem will not gain this ability.</p>

<p>However, I have been developing a replacement for PyEphem that is written in pure Python and that I will be free to extend, called Skyfield. I will hopefully have this concept working there soon, and when I do so I plan to make it work for all objects, and not limit it to Earth satellites!</p>

<p>(In the meantime, as pointed out in the comments: you could try asking PyEphem for the geocentric RA and dec and, if you could adjust RA by the Earth's current hour angle, then you could turn those into an approximate latitude and longitude.)</p>

CC BY-SA 3.0
2016-01-22T19:14:27.517Z
Thanks! Do you mean "iterating until the *altitude* becomes very close to 90°"? Do I need to check the azimuth if I'm just interested in the point on Earth and not where to look up?
Hugo
CC BY-SA 3.0
2016-01-23T19:37:43.817Z
Unfortunately not, because the planet will have a different RA from a different location on Earth, because they are looking at it from a slightly different angle against the backdrop of stars.
Brandon Rhodes
CC BY-SA 3.0
2016-01-27T00:02:28.670Z
Oh! The *geocentric* RA and dec. Yes. Those coordinates would, if the RA were adjusted by the current Earth rotation angle, given an approximate position on the Earth if one approximates it as a sphere.
Brandon Rhodes
CC BY-SA 3.0
2016-01-27T09:06:03.543Z
This is the code I came up with based on this answer: https://github.com/brandon-rhodes/pyephem/issues/95#issuecomment-174019314 and for now gives `lat=-20.94, lon=25.38`. @barrycarter's code gives very similar results: `lat=-20.9443019471, lon=25.3799291351 `.
Hugo
CC BY-SA 3.0
2016-01-27T14:04:56.627Z
Great! To help people find this, try editing the question, add an "Update:" line and then a quick paragraph pointing to these solutions. Thanks!
Brandon Rhodes
CC BY-SA 3.0
2016-01-22T15:25:44.830Z
2016-01-27T00:03:39.380Z
2016-01-27T00:03:39.380Z
Brandon Rhodes
<p>The orbital speed of Earth around the Sun is 30,000 m/s, the highest mountain is about 9,000 m and the rotational speed of the surface around the polar axis is about 450 m/s. So this makes me conclude that topography such as height over the Sea has to be of little importance.</p>

<p>The polar axis around which the Earth turns is tilted by 23 degrees relative to the ecliptic plane of the planets' average. Pluto's orbit is 17 degrees tilted so every region on Earth between +40 and -40 degrees latitude should, sequentially in time, be the one spot on Earth's surface which is closest to Pluto. Spread out over looong times. That's about between New York and Sydney. For the least inclined planets the nearest point on Earth would be somewhere between Houston and Rio de Janeiro, i.e. every place on Earth within 24 degrees north or south of the equator (except very local depressions).</p>

CC BY-SA 3.0
2016-01-22T19:16:04.370Z
Thank you, this is useful for sanity checking my calculations.
Hugo
CC BY-SA 3.0
2016-01-22T16:21:44.393Z
2016-01-22T16:27:38.703Z
2016-01-22T16:27:38.703Z
LocalFluff
<p>I'd like to be able to find the latitude and longitude on Earth (assuming a sphere is fine) which is closest to Pluto at the current time (or where Pluto is directly overhead). </p>

<p>Is this possible with PyEphem?</p>

CC BY-SA 3.0
2016-01-18T21:15:06.757Z
As the Earth rotates this question does not mean much: the point that is closest changes all the time.
adrianmcmenamin
CC BY-SA 3.0
2016-01-19T04:56:40.380Z
The point moves at 100s of km /h, depending on latitude. I can't help wondering why you want to know this point though...
Michael
CC BY-SA 3.0
2016-04-06T15:46:59.353Z
@adrianmcmenamin The user is looking for a method of finding this point, not a single answer for a given time.
called2voyage
<earth><pluto>
Closest point on Earth to a planet?
Hugo
# Closest point on Earth to a planet?

(Asked by: username_0 on 2016-01-18)

I'd like to be able to find the latitude and longitude on Earth (assuming a sphere is fine) which is closest to Pluto at the current time (or where Pluto is directly overhead).

Is this possible with PyEphem?

**Tags:** earth, pluto
> **Comments:**
> **username_1**: As the Earth rotates this question does not mean much: the point that is closest changes all the time.  
> **username_2**: The point moves at 100s of km /h, depending on latitude. I can't help wondering why you want to know this point though...  
> **username_3**: @adrianmcmenamin The user is looking for a method of finding this point, not a single answer for a given time.  

---

## 2 answers

### Answer by username_4 on 2016-01-22. Score: 4

Your question is valid and is a common operation. When computing when an object will be visible, many observers want to be able to draw a globe or map that is marked with the position from which an object like Pluto is directly overhead. Such a position on the globe at a given instant (and you do say “at the current time”) is, all other things being equal, the best-situated place from which to observe the object.

Unfortunately, the “libastro” C library that PyEphem wraps seems to only provide one instance of this concept: Earth satellites have `.sublat` and `.sublong` attributes, because it is so common to want to draw the path of a satellite on a globe or map.

But libastro does not generalize the concept. From what I can see, there is no way to generate the point on the Earth that is directly below any other Solar System object. And since I have not been in the habit of trying to extend libastro, it is likely that PyEphem will not gain this ability.

However, I have been developing a replacement for PyEphem that is written in pure Python and that I will be free to extend, called Skyfield. I will hopefully have this concept working there soon, and when I do so I plan to make it work for all objects, and not limit it to Earth satellites!

(In the meantime, as pointed out in the comments: you could try asking PyEphem for the geocentric RA and dec and, if you could adjust RA by the Earth's current hour angle, then you could turn those into an approximate latitude and longitude.)

> **Comments:**
> **username_0**: Thanks! Do you mean "iterating until the *altitude* becomes very close to 90°"? Do I need to check the azimuth if I'm just interested in the point on Earth and not where to look up?  
> **username_4**: Unfortunately not, because the planet will have a different RA from a different location on Earth, because they are looking at it from a slightly different angle against the backdrop of stars.  
> **username_4**: Oh! The *geocentric* RA and dec. Yes. Those coordinates would, if the RA were adjusted by the current Earth rotation angle, given an approximate position on the Earth if one approximates it as a sphere.  
> **username_0**: This is the code I came up with based on this answer: https://github.com/brandon-rhodes/pyephem/issues/95#issuecomment-174019314 and for now gives `lat=-20.94, lon=25.38`. @barrycarter's code gives very similar results: `lat=-20.9443019471, lon=25.3799291351 `.  
> **username_4**: Great! To help people find this, try editing the question, add an "Update:" line and then a quick paragraph pointing to these solutions. Thanks!  

---

### Answer by username_5 on 2016-01-22. Score: 3

The orbital speed of Earth around the Sun is 30,000 m/s, the highest mountain is about 9,000 m and the rotational speed of the surface around the polar axis is about 450 m/s. So this makes me conclude that topography such as height over the Sea has to be of little importance.

The polar axis around which the Earth turns is tilted by 23 degrees relative to the ecliptic plane of the planets' average. Pluto's orbit is 17 degrees tilted so every region on Earth between +40 and -40 degrees latitude should, sequentially in time, be the one spot on Earth's surface which is closest to Pluto. Spread out over looong times. That's about between New York and Sydney. For the least inclined planets the nearest point on Earth would be somewhere between Houston and Rio de Janeiro, i.e. every place on Earth within 24 degrees north or south of the equator (except very local depressions).

> **Comments:**
> **username_0**: Thank you, this is useful for sanity checking my calculations.  

---

