<p>In the common case asteroids will rotate around a nearly fixed axis, like the Earth. That axis of rotation defines a north pole and south pole for that asteroid. The north pole is the one where, when a viewer is seeing the asteroid from above it, the asteroid is rotating counter-clockwise. Once you have north and south poles, you can define an equator and, thus, a system of latitude. I have no idea what is done for asteroids that are undergoing chaotic tumbling.</p>

<p>The "sub-radar latitude" is the asteroid's latitude that is directly facing the radar viewer. A good example from the literature that uses the term many times, and includes a short definition, is <A HREF="http://arxiv.org/abs/1101.3794">arXiv:1101.3794</A>. A short excerpt:</p>

<blockquote>
  <p>[...] implying significant change in the object’s sub-radar latitude (the angle between the asteroid-Earth line and the object’s equatorial plane).</p>
</blockquote>

CC BY-SA 3.0
2016-09-28T05:16:27.777Z
So basically the radar hits a point on the asteroid and the latitude at that point is the sub-radar latitude?
Robert Christian
CC BY-SA 3.0
2016-09-28T23:20:37.327Z
If the asteroid were a sphere it would be the first point hit by the radar wave. By the time any radar from Earth reaches an asteroid the beam will probably have spread out enough to hit the entire asteroid. If the asteroid were a sphere, though, it would be the part of the asteroid that reflects most strongly. Very few asteroids are spherical, though, but the definition of sub-radar latitude doesn't change since the latitude is defined by the spin axis and center of mass.
Sean Lake
CC BY-SA 3.0
2016-09-29T12:39:04.090Z
*"The "sub-radar latitude" is the asteroid's latitude that is directly facing the radar viewer"*.  Does that mean: consider the two vectors LOS line of sight from radar to asteroid, and AA asteroid axis of rotation.  Consider the angle ("Lean") between LOS and AA (ie, acos of dot product).  If "Lean" is 90 degrees, the "SRL" is 0° - correct?  If "Lean" is say 100 degrees, the "SRL" is -10° latitude ... correct?
Fattie
CC BY-SA 3.0
2016-09-30T21:33:38.843Z
@JoeBlow Yes, you're just talking about the relationship between polar and latitude-like angles.
Sean Lake
CC BY-SA 3.0
2016-09-28T04:58:18.003Z
2016-09-28T04:58:18.003Z
Sean Lake
<p>How to find sub-radar latitude in reference to radar telescopes hitting near earth asteroids?</p>

CC BY-SA 3.0
2016-09-28T03:03:33.363Z
More context of what part of astronomy this is related to would help, making an informed response more likely.
Sean Lake
CC BY-SA 3.0
2016-09-28T04:13:42.287Z
Radar telescope calculating density of asteroids.
Robert Christian
CC BY-SA 3.0
2016-09-28T14:57:00.127Z
Please edit the question and fill in relevant details so that people stumbling across it will have some idea as to what the question is really asking.
Donald.McLean
CC BY-SA 3.0
2016-09-30T13:28:55.770Z
It seems to me "sub radar latitude" is nothing more than the off-square tilt of the asteroid's axis.  (Obviously, per our line of sight - ie, obviously the radar beam is of course nothing more than our line of sight).  So, it's the "forwards-backwards tilt" per us.  ("Left-right" tilt, per us, would not affect this measure.)
Fattie
CC BY-SA 3.0
2016-10-02T15:58:14.587Z
A bit more detail about what you mean would help here.
ProfRob
<telescope><astrophysics>
Not clear on how to calculate sub-radar latitude
Robert Christian
# Not clear on how to calculate sub-radar latitude

(Asked by: username_0 on 2016-09-28)

How to find sub-radar latitude in reference to radar telescopes hitting near earth asteroids?

**Tags:** telescope, astrophysics
> **Comments:**
> **username_1**: More context of what part of astronomy this is related to would help, making an informed response more likely.  
> **username_0**: Radar telescope calculating density of asteroids.  
> **username_2**: Please edit the question and fill in relevant details so that people stumbling across it will have some idea as to what the question is really asking.  
> **username_3**: It seems to me "sub radar latitude" is nothing more than the off-square tilt of the asteroid's axis.  (Obviously, per our line of sight - ie, obviously the radar beam is of course nothing more than our line of sight).  So, it's the "forwards-backwards tilt" per us.  ("Left-right" tilt, per us, would not affect this measure.)  
> **username_4**: A bit more detail about what you mean would help here.  

---

## 1 answers

### Answer by username_1 on 2016-09-28. Score: 4 | Accepted answer

In the common case asteroids will rotate around a nearly fixed axis, like the Earth. That axis of rotation defines a north pole and south pole for that asteroid. The north pole is the one where, when a viewer is seeing the asteroid from above it, the asteroid is rotating counter-clockwise. Once you have north and south poles, you can define an equator and, thus, a system of latitude. I have no idea what is done for asteroids that are undergoing chaotic tumbling.

The "sub-radar latitude" is the asteroid's latitude that is directly facing the radar viewer. A good example from the literature that uses the term many times, and includes a short definition, is [arXiv:1101.3794](http://arxiv.org/abs/1101.3794). A short excerpt:

> [...] implying significant change in the object’s sub-radar latitude (the angle between the asteroid-Earth line and the object’s equatorial plane).

> **Comments:**
> **username_0**: So basically the radar hits a point on the asteroid and the latitude at that point is the sub-radar latitude?  
> **username_1**: If the asteroid were a sphere it would be the first point hit by the radar wave. By the time any radar from Earth reaches an asteroid the beam will probably have spread out enough to hit the entire asteroid. If the asteroid were a sphere, though, it would be the part of the asteroid that reflects most strongly. Very few asteroids are spherical, though, but the definition of sub-radar latitude doesn't change since the latitude is defined by the spin axis and center of mass.  
> **username_3**: *"The "sub-radar latitude" is the asteroid's latitude that is directly facing the radar viewer"*.  Does that mean: consider the two vectors LOS line of sight from radar to asteroid, and AA asteroid axis of rotation.  Consider the angle ("Lean") between LOS and AA (ie, acos of dot product).  If "Lean" is 90 degrees, the "SRL" is 0° - correct?  If "Lean" is say 100 degrees, the "SRL" is -10° latitude ... correct?  
> **username_1**: @JoeBlow Yes, you're just talking about the relationship between polar and latitude-like angles.  

---

