<p>Here is a diagram with the quantities we need to solve this:</p>
<p><a href="https://i.sstatic.net/oZomm.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/oZomm.png" alt="enter image description here" /></a></p>
<p>From <a href="https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law" rel="nofollow noreferrer">Kepler's 3rd Law</a>, the &quot;kronostationary&quot; satellite---marked by a green dot---is located at <em>a</em> = 1.86 <em>R</em><sub>eq</sub>, where <em>R</em><sub>eq</sub> is the equatorial radius and <em>R</em><sub>pol</sub> is the polar radius (Python calculations dumped at the bottom of the post). The NASA PDS Atmospheres Encyclopedia gives values for the radius of Saturn and useful formulas on its <a href="https://pds-atmospheres.nmsu.edu/education_and_outreach/encyclopedia/latitude.htm" rel="nofollow noreferrer">latitude page</a>, and a value for the rotation rate on the <a href="https://pds-atmospheres.nmsu.edu/education_and_outreach/encyclopedia/longitude.htm" rel="nofollow noreferrer">longitude page</a>.</p>
<p>At the planetographic latitude (<span class="math-container">$\phi$</span><sub>g</sub>) of the location we want, the line of sight (green) forms a right angle with the local vertical (blue). This sets up a pair of similar triangles as shown. From the big one, we have:</p>
<p><span class="math-container">$$
 \tan{\phi_{\textrm{g}}} = \frac{a - R_{\textrm{cyl}}}{z}
$$</span></p>
<p>We can use the equation for an ellipse to express <em>z</em> in terms of <em>R</em><sub>cyl</sub>:</p>
<p><span class="math-container">$$
 \frac{R_{\textrm{cyl}}^2}{R_{\textrm{eq}}^2} + \frac{z^2}{R_{\textrm{pol}}^2} = 1
$$</span>
<span class="math-container">$$
 z = \frac{R_{\textrm{pol}}}{R_{\textrm{eq}}} \sqrt{R_{\textrm{eq}}^2 - R_{\textrm{cyl}}^2}
$$</span></p>
<p>We can then use the pink-red-red triangle to relate <span class="math-container">$\phi$</span><sub>c</sub> and <em>R</em><sub>cyl</sub>, eliminating <em>z</em> using the equation above:</p>
<p><span class="math-container">$$
 \tan{\phi_{\textrm{c}}} = \frac{z}{R_{\textrm{cyl}}} = \frac{R_{\textrm{pol}}}{R_{\textrm{eq}} R_{\textrm{cyl}}} \sqrt{R_{\textrm{eq}}^2 - R_{\textrm{cyl}}^2}
$$</span></p>
<p>The PDS Atmospheres formulas relate <span class="math-container">$\phi$</span><sub>g</sub> and <span class="math-container">$\phi$</span><sub>c</sub>:</p>
<p><span class="math-container">$$
 \tan{\phi_{\textrm{c}}} = \left(\frac{R_{\textrm{pol}}}{R_{\textrm{eq}}}\right)^2 \tan{\phi_{\textrm{g}}}
$$</span></p>
<p>Combining the first and last equations, and eliminating <em>z</em>, gives a surprisingly simple equation:</p>
<p><span class="math-container">$$
 R_{\textrm{cyl}} = \frac{R_{\textrm{eq}}^2}{a} = 32362\ \textrm{km}
$$</span></p>
<p>Plugging this value of <em>R</em><sub>cyl</sub> into the equations above gives:</p>
<ul>
<li><strong>Planetocentric latitude 54.8°</strong></li>
<li><strong>Planetographic latitude 60.1°</strong></li>
</ul>
<hr />
<p>Actual visibility and survivability of this satellite are not great. The kronostationary radius would put this satellite within the B ring, which is &quot;<a href="https://en.wikipedia.org/wiki/Rings_of_Saturn#B_Ring" rel="nofollow noreferrer">the largest, brightest, and most massive of the rings</a>:&quot;</p>
<p><a href="https://photojournal.jpl.nasa.gov/catalog/PIA11668" rel="nofollow noreferrer"><img src="https://i.sstatic.net/cEQiU.jpg" alt="enter image description here" /></a></p>
<hr />
<p><strong>Calculations:</strong></p>
<pre><code>import numpy as np

R_eq = 60268. # km
R_pol = 54364. # km
GMsat = 37931206.159 # km^3/s^2
P_sat = 10.656 * 60.**2. # s

# Kepler's 3rd law: a is the kronocentric distance (circular orbit)
a = (P_sat**2 * GMsat / 4 / (np.pi)**2)**(1/3.)
print(&quot;Kronocentric distance,&quot;)
print(&quot;   a (R_sat): &quot;, a / R_eq)
print(&quot;   a (km): &quot;, a)
print(&quot;   altitude (km): &quot;, a - R_eq)
print(&quot;  &quot;)

# solve for R_cyl
R_cyl = R_eq**2. / a
print(&quot;R_cyl: &quot;, R_cyl)
print(&quot;  &quot;)

# get latitude
lat_centric = np.arctan(R_pol / R_eq / R_cyl * (R_eq**2. - R_cyl**2.)**0.5)
lat_graphic = np.arctan(R_eq**2. / R_pol**2. * np.tan(lat_centric))
print(&quot;lat_centric: &quot;, lat_centric * 180./np.pi)
print(&quot;lat_graphic: &quot;, lat_graphic * 180./np.pi)
print(&quot;  &quot;)

Kronocentric distance,
   a (R_sat):  1.862330120959813
   a (km):  112238.91173000602
   altitude (km):  51970.91173000602
  
R_cyl:  32361.60942773073
  
lat_centric:  54.79195278661741
lat_graphic:  60.13756294902166
</code></pre>
<hr />
<p><strong>Verification:</strong></p>
<p>Using <a href="https://naif.jpl.nasa.gov/naif/cosmographia_components.html" rel="nofollow noreferrer">Cosmographia</a> to generate a view of Saturn's horizon at <span class="math-container">$\phi$</span><sub>g</sub> = 60.1°, you can see just into the B-ring:</p>
<p><a href="https://i.sstatic.net/PkH7b.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/PkH7b.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2022-01-05T06:11:14.670Z
Thank you, i can now die in peace without this bugging me :)
user177107
CC BY-SA 4.0
2022-01-04T21:41:36.620Z
2022-01-04T21:58:33.887Z
2022-01-04T21:58:33.887Z
giardia
<p>This is an interesting puzzle. The key is to work in the Cartesian plane.</p>
<p>Consider an oblate planet with an equatorial radius <span class="math-container">$a$</span>, and a polar radius <span class="math-container">$b$</span>. Because, as you point out, planets are slightly squashed toward the poles, <span class="math-container">$a &gt; b$</span> for all planets. If we look at the planet in profile, we can describe the shape of the planet as an ellipse with semi-major axis <span class="math-container">$a$</span>, and semi-minor axis <span class="math-container">$b$</span>.</p>
<p>If we place the centre of the planet at the origin of a plane, we can describe the surface of the planet that intersects the plane as a curve traced out by the set of points <span class="math-container">$p = (a\cos{\theta}, b\sin{\theta})$</span>, where <span class="math-container">$\theta$</span> is the angle subtended by the equator and a line from the equator to <span class="math-container">$p$</span>.</p>
<p><a href="https://i.sstatic.net/gVu22.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/gVu22.png" alt="Cartoon two-dimensional view of the problem" /></a>
If there is a satellite in a stationary orbit around the planet, it will orbit at an altitude <span class="math-container">$R$</span> above the equator. Its Cartesian coordinates are <span class="math-container">$(R, 0)$</span>.</p>
<p>For a given point <span class="math-container">$p$</span> on the surface of the planet, the slope of a line drawn between <span class="math-container">$p$</span> and the satellite is</p>
<p><span class="math-container">$$
 m = \frac{y_2 - y_1}{x_2 - x_1}= \frac{b\sin{\theta}}{a\cos{\theta} - R}
$$</span></p>
<p>With this slope in hand, all we need to do is choose the angle <span class="math-container">$\theta$</span> so that slope <span class="math-container">$m$</span> is perpendicular to a vector from the center of the planet to the point <span class="math-container">$p$</span>. The slope of this second vector, <span class="math-container">$m_2$</span>, is simply <span class="math-container">$b\sin{\theta} / a\cos{\theta}$</span>, and the two lines are perpendicular if the product of their slopes is -1</p>
<p><span class="math-container">$$
\frac{b\sin{\theta}}{a\cos{\theta} - R} \cdot \frac{b\sin{\theta}}{ a\cos{\theta}} = -1
$$</span></p>
<p>or
<span class="math-container">$$
b^2\sin^2{\theta} = -1(a^2\cos^2{\theta} - aR\cos{\theta} )
$$</span></p>
<p>Recognising that <span class="math-container">$\sin^2{\theta} = 1 - \cos^2{\theta}$</span>, and skipping a few lines of algebra, we get</p>
<p><span class="math-container">$$
(b^2 -a^2)\cos^2{\theta} + aR\cos{\theta} - b^2 = 0
$$</span></p>
<p>which we can solve quadratically to get <span class="math-container">$\theta$</span></p>
<p>This angle is the highest latitude where the satellite can be seen. Observers at different longitudes than the satellite will need to be closer to the equator</p>

CC BY-SA 4.0
2022-01-03T20:20:43.427Z
This is a nice start to an answer, but can you follow through - instead of writing "which we can solve quadratically... this angle is the highest latitude," can you explain and fully solve the problem so users who are less experienced with mathematics can understand?
WarpPrime
CC BY-SA 4.0
2022-01-04T01:53:19.013Z
I don't believe this answer is correct.  For an ellipse, one can't assume that a line tangent to a point on the ellipse is perpendicular to the segment from that point to the center of the ellipse.
Connor Garcia
CC BY-SA 4.0
2022-01-04T08:06:37.410Z
If I solved correctly, then the maximum latitude from which a geostationary satellite at 112,240 km would be visible is 62.75°, which seems wrong
user177107
CC BY-SA 4.0
2022-01-04T14:33:24.740Z
@connorgarcia, Oh, you are correct. I was assuming the tangent is perpendicular to the radial vector, but now that you point it out I realize that only applies to a circle. If you replace the slope $m_2$ with the correct slope to the tangent to the ellipse, the argument should hold. I'll figure out how to do that and update the answer.
Fergal
CC BY-SA 4.0
2022-01-04T17:09:19.290Z
@Fergal Yes, I think this change will make the answer correct.  Also your layout of this problem is quite nice.  Hint: If $P=(x_1,y_1)$, then the equation of the tangent line at $P$ will be $xx_1/a^2+yy_1/b^2=1$.
Connor Garcia
CC BY-SA 4.0
2022-01-03T19:56:45.773Z
2022-01-03T19:56:45.773Z
Fergal
<p>This question is part astronomy and part mathematics. I'm aware it involves “basic” trigonometry, but my brain is short-circuiting.</p>
<p>From my calculations, the distance from the center of Saturn where a satellite orbiting over the equator would have an orbital period equal to Saturn's rotational period is <span class="math-container">$112,240$</span> km.</p>
<p><em>If</em> Saturn were a perfect sphere, simple trignometry would suffice to find out at which latitude the planet's curvature gets in the way and stops the satellite from being visible. (If I'm not mistaken, 58.74° North or South is the maximum latitude from which said satellite would be visible if Saturn where spherical. This was calculated using Saturn's mean volumetric radius of 58,232 km as given by the NASA fact sheet.)</p>
<p>Unfortunately, Saturn is very much not spherical and is noticeably wider at the equator than at the poles. Since Saturn is an oblate spheroid, the maximum latitude from which a geostationary satellite  over the equator would be visible is <em>less</em> than 58.74°. But by how much, exactly?</p>
<p>I don't know, and that's why I'm asking. I have a very limited understanding of trignometry and it completely breaks apart when discussing an oblate spheroid instead of a sphere. This question probably fits better in the mathematics substack, but since its inception is completely based in astronomy I figured this is a good place as any to ask.</p>

CC BY-SA 4.0
2022-01-03T08:41:29.950Z
Different but very related question in Math SE: [Calculating surface of ellipsoid visible from any point outside of it](https://math.stackexchange.com/q/4339560/284619) There may be some helpful oblate Earth visibility from orbit answers in [Space SE](https://space.stackexchange.com/) as well.
uhoh
CC BY-SA 4.0
2022-01-03T11:16:21.167Z
@uhoh Thank you for pointing out that thread. I'd be lying if I said I understood anything, but it's a step in the right direction.
user177107
CC BY-SA 4.0
2022-01-03T11:20:05.080Z
ya those are there just as much to be helpful to those daring to try to answer and to future readers as they are for the OP (original post author, i.e. you). Hopefully they may help someone to arrive at a useful answer.
uhoh
CC BY-SA 4.0
2022-01-03T17:31:12.713Z
The tool for answering this is "planetographic latitude." This coordinate gives the direction of the surface normal (zenith direction), so the horizon angle would just be 90° from that. Wikipedia probably gives good planetographic/planetocentric definitions, and some good equations are here: https://pds-atmospheres.nmsu.edu/education_and_outreach/encyclopedia/latitude.htm
giardia
CC BY-SA 4.0
2022-01-03T18:42:14.780Z
Just re-do the calculation assuming Saturn is a sphere, but use the polar radius rather than the equatorial radius.
Greg Miller
CC BY-SA 4.0
2022-01-04T18:27:08.377Z
The maximum planetocentric latitude where the satellite would be visible (for oblate spheroidal Saturn) is 54.8°. I will write up a solution soon.
giardia
<orbit><orbital-mechanics><coordinate><saturn>
What is the maximum latitude from which a satellite in a geostationary orbit around Saturn would be visible to an observer on the planet?
user177107
# What is the maximum latitude from which a satellite in a geostationary orbit around Saturn would be visible to an observer on the planet?

(Asked by: username_0 on 2022-01-03)

This question is part astronomy and part mathematics. I'm aware it involves “basic” trigonometry, but my brain is short-circuiting.

From my calculations, the distance from the center of Saturn where a satellite orbiting over the equator would have an orbital period equal to Saturn's rotational period is $112,240$ km.

*If* Saturn were a perfect sphere, simple trignometry would suffice to find out at which latitude the planet's curvature gets in the way and stops the satellite from being visible. (If I'm not mistaken, 58.74° North or South is the maximum latitude from which said satellite would be visible if Saturn where spherical. This was calculated using Saturn's mean volumetric radius of 58,232 km as given by the NASA fact sheet.)

Unfortunately, Saturn is very much not spherical and is noticeably wider at the equator than at the poles. Since Saturn is an oblate spheroid, the maximum latitude from which a geostationary satellite over the equator would be visible is *less* than 58.74°. But by how much, exactly?

I don't know, and that's why I'm asking. I have a very limited understanding of trignometry and it completely breaks apart when discussing an oblate spheroid instead of a sphere. This question probably fits better in the mathematics substack, but since its inception is completely based in astronomy I figured this is a good place as any to ask.

**Tags:** orbit, orbital-mechanics, coordinate, saturn

> **Comments:**
> **username_1**: Different but very related question in Math SE: [Calculating surface of ellipsoid visible from any point outside of it](https://math.stackexchange.com/q/4339560/284619) There may be some helpful oblate Earth visibility from orbit answers in [Space SE](https://space.stackexchange.com/) as well.  
> **username_0**: @username_1 Thank you for pointing out that thread. I'd be lying if I said I understood anything, but it's a step in the right direction.  
> **username_1**: ya those are there just as much to be helpful to those daring to try to answer and to future readers as they are for the OP (original post author, i.e. you). Hopefully they may help someone to arrive at a useful answer.  
> **username_2**: The tool for answering this is "planetographic latitude." This coordinate gives the direction of the surface normal (zenith direction), so the horizon angle would just be 90° from that. Wikipedia probably gives good planetographic/planetocentric definitions, and some good equations are here: https://pds-atmospheres.nmsu.edu/education_and_outreach/encyclopedia/latitude.htm  
> **username_3**: Just re-do the calculation assuming Saturn is a sphere, but use the polar radius rather than the equatorial radius.  
> **username_2**: The maximum planetocentric latitude where the satellite would be visible (for oblate spheroidal Saturn) is 54.8°. I will write up a solution soon.  

---

## 2 answers

### Answer by username_2 on 2022-01-04. Score: 3 | Accepted answer

Here is a diagram with the quantities we need to solve this:

[![enter image description here](https://i.sstatic.net/oZomm.png)](https://i.sstatic.net/oZomm.png)

From [Kepler's 3rd Law](https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law), the "kronostationary" satellite---marked by a green dot---is located at *a* = 1.86 *R*eq, where *R*eq is the equatorial radius and *R*pol is the polar radius (Python calculations dumped at the bottom of the post). The NASA PDS Atmospheres Encyclopedia gives values for the radius of Saturn and useful formulas on its [latitude page](https://pds-atmospheres.nmsu.edu/education_and_outreach/encyclopedia/latitude.htm), and a value for the rotation rate on the [longitude page](https://pds-atmospheres.nmsu.edu/education_and_outreach/encyclopedia/longitude.htm).

At the planetographic latitude ($\phi$g) of the location we want, the line of sight (green) forms a right angle with the local vertical (blue). This sets up a pair of similar triangles as shown. From the big one, we have:

$$
\tan{\phi\_{\textrm{g}}} = \frac{a - R\_{\textrm{cyl}}}{z}
$$

We can use the equation for an ellipse to express *z* in terms of *R*cyl:

$$
\frac{R\_{\textrm{cyl}}^2}{R\_{\textrm{eq}}^2} + \frac{z^2}{R\_{\textrm{pol}}^2} = 1
$$
$$
z = \frac{R\_{\textrm{pol}}}{R\_{\textrm{eq}}} \sqrt{R\_{\textrm{eq}}^2 - R\_{\textrm{cyl}}^2}
$$

We can then use the pink-red-red triangle to relate $\phi$c and *R*cyl, eliminating *z* using the equation above:

$$
\tan{\phi\_{\textrm{c}}} = \frac{z}{R\_{\textrm{cyl}}} = \frac{R\_{\textrm{pol}}}{R\_{\textrm{eq}} R\_{\textrm{cyl}}} \sqrt{R\_{\textrm{eq}}^2 - R\_{\textrm{cyl}}^2}
$$

The PDS Atmospheres formulas relate $\phi$g and $\phi$c:

$$
\tan{\phi\_{\textrm{c}}} = \left(\frac{R\_{\textrm{pol}}}{R\_{\textrm{eq}}}\right)^2 \tan{\phi\_{\textrm{g}}}
$$

Combining the first and last equations, and eliminating *z*, gives a surprisingly simple equation:

$$
R\_{\textrm{cyl}} = \frac{R\_{\textrm{eq}}^2}{a} = 32362\ \textrm{km}
$$

Plugging this value of *R*cyl into the equations above gives:

* **Planetocentric latitude 54.8°**
* **Planetographic latitude 60.1°**

---

Actual visibility and survivability of this satellite are not great. The kronostationary radius would put this satellite within the B ring, which is "[the largest, brightest, and most massive of the rings](https://en.wikipedia.org/wiki/Rings_of_Saturn#B_Ring):"

[![enter image description here](https://i.sstatic.net/cEQiU.jpg)](https://photojournal.jpl.nasa.gov/catalog/PIA11668)

---

**Calculations:**

```
import numpy as np

R_eq = 60268. # km
R_pol = 54364. # km
GMsat = 37931206.159 # km^3/s^2
P_sat = 10.656 * 60.**2. # s

# Kepler's 3rd law: a is the kronocentric distance (circular orbit)
a = (P_sat**2 * GMsat / 4 / (np.pi)**2)**(1/3.)
print("Kronocentric distance,")
print("   a (R_sat): ", a / R_eq)
print("   a (km): ", a)
print("   altitude (km): ", a - R_eq)
print("  ")

# solve for R_cyl
R_cyl = R_eq**2. / a
print("R_cyl: ", R_cyl)
print("  ")

# get latitude
lat_centric = np.arctan(R_pol / R_eq / R_cyl * (R_eq**2. - R_cyl**2.)**0.5)
lat_graphic = np.arctan(R_eq**2. / R_pol**2. * np.tan(lat_centric))
print("lat_centric: ", lat_centric * 180./np.pi)
print("lat_graphic: ", lat_graphic * 180./np.pi)
print("  ")

Kronocentric distance,
   a (R_sat):  1.862330120959813
   a (km):  112238.91173000602
   altitude (km):  51970.91173000602
  
R_cyl:  32361.60942773073
  
lat_centric:  54.79195278661741
lat_graphic:  60.13756294902166

```

---

**Verification:**

Using [Cosmographia](https://naif.jpl.nasa.gov/naif/cosmographia_components.html) to generate a view of Saturn's horizon at $\phi$g = 60.1°, you can see just into the B-ring:

[![enter image description here](https://i.sstatic.net/PkH7b.png)](https://i.sstatic.net/PkH7b.png)

> **Comments:**
> **username_0**: Thank you, i can now die in peace without this bugging me :)  

---

### Answer by username_4 on 2022-01-03. Score: 2

This is an interesting puzzle. The key is to work in the Cartesian plane.

Consider an oblate planet with an equatorial radius $a$, and a polar radius $b$. Because, as you point out, planets are slightly squashed toward the poles, $a > b$ for all planets. If we look at the planet in profile, we can describe the shape of the planet as an ellipse with semi-major axis $a$, and semi-minor axis $b$.

If we place the centre of the planet at the origin of a plane, we can describe the surface of the planet that intersects the plane as a curve traced out by the set of points $p = (a\cos{\theta}, b\sin{\theta})$, where $\theta$ is the angle subtended by the equator and a line from the equator to $p$.

[![Cartoon two-dimensional view of the problem](https://i.sstatic.net/gVu22.png)](https://i.sstatic.net/gVu22.png)
If there is a satellite in a stationary orbit around the planet, it will orbit at an altitude $R$ above the equator. Its Cartesian coordinates are $(R, 0)$.

For a given point $p$ on the surface of the planet, the slope of a line drawn between $p$ and the satellite is

$$
m = \frac{y\_2 - y\_1}{x\_2 - x\_1}= \frac{b\sin{\theta}}{a\cos{\theta} - R}
$$

With this slope in hand, all we need to do is choose the angle $\theta$ so that slope $m$ is perpendicular to a vector from the center of the planet to the point $p$. The slope of this second vector, $m\_2$, is simply $b\sin{\theta} / a\cos{\theta}$, and the two lines are perpendicular if the product of their slopes is -1

$$
\frac{b\sin{\theta}}{a\cos{\theta} - R} \cdot \frac{b\sin{\theta}}{ a\cos{\theta}} = -1
$$

or
$$
b^2\sin^2{\theta} = -1(a^2\cos^2{\theta} - aR\cos{\theta} )
$$

Recognising that $\sin^2{\theta} = 1 - \cos^2{\theta}$, and skipping a few lines of algebra, we get

$$
(b^2 -a^2)\cos^2{\theta} + aR\cos{\theta} - b^2 = 0
$$

which we can solve quadratically to get $\theta$

This angle is the highest latitude where the satellite can be seen. Observers at different longitudes than the satellite will need to be closer to the equator

> **Comments:**
> **username_5**: This is a nice start to an answer, but can you follow through - instead of writing "which we can solve quadratically... this angle is the highest latitude," can you explain and fully solve the problem so users who are less experienced with mathematics can understand?  
> **username_6**: I don't believe this answer is correct.  For an ellipse, one can't assume that a line tangent to a point on the ellipse is perpendicular to the segment from that point to the center of the ellipse.  
> **username_0**: If I solved correctly, then the maximum latitude from which a geostationary satellite at 112,240 km would be visible is 62.75°, which seems wrong  
> **username_4**: @username_6, Oh, you are correct. I was assuming the tangent is perpendicular to the radial vector, but now that you point it out I realize that only applies to a circle. If you replace the slope $m_2$ with the correct slope to the tangent to the ellipse, the argument should hold. I'll figure out how to do that and update the answer.  
> **username_6**: @username_4 Yes, I think this change will make the answer correct.  Also your layout of this problem is quite nice.  Hint: If $P=(x_1,y_1)$, then the equation of the tangent line at $P$ will be $xx_1/a^2+yy_1/b^2=1$.  

---

