What is the maximum latitude from which a satellite in a geostationary orbit around Saturn would be visible to an observer on the planet?

Question

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.

Answer 1

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

From Kepler's 3rd 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, and a value for the rotation rate on the longitude page.

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°

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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:"

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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

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Verification:

Using Cosmographia to generate a view of Saturn's horizon at $\phi$_g = 60.1°, you can see just into the B-ring:

Answer 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 $p$.

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