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This answer to Is there a name for the great circle where latitude and longitude are equal? says (in part):

More generally, a clélie is the name given to any spherical curve where the longitude $\varphi$ and colatitude $\theta$ have the relationship $\varphi=c\theta,\quad c>0$, and the curve of Viviani corresponds to the locus of a geosynchronous orbit, $c=1$.

Is it really true that a polar geosynchronous orbit (displayed in a synodic or rotating frame) is described by Viviani's curve? Can this be demonstrated mathematically?

File:Viviani-fenster-1.svg construction of Viviani's curve Source

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Brief answer: yes, it's possible. Here's a somewhat scrappy answer: this is pretty much a transcription of what I wrote down when working it out, so it's a bit messy: sorry.

First of all I'll use what I think is the mathematicians' version of spherical polar coordinates (apparently physicists use the two angle names swapped). So starting from a right-handed cartesian coordinate system $(x, y, z)$, the spherical polars are $(r, \theta, \phi)$, where

$$ \begin{align} x &= r \sin\phi \cos\theta\\ y &= r \sin\phi \sin\theta\\ z &= r \cos\phi \end{align} $$

(So here, $\theta$ is the angle in the $x$-$y$ plane, and $\phi$ is the angle to the $z$ axis).

So let's set up an inertial coordinate system whose origin is the centre of the Earth (so this system is not rotating with the Earth). Obviously we'll take the Earth to be perfectly spherical, its density will be a function only of $r$ and the orbit of the satellite will be perfectly circular.

The satellite will orbit about the $y$ axis, so the coordinates of the satellite, if we pick $t=0$ suitably, are

$$(r_s, 0, \omega_s t)$$

OK, so now pick an Earth-based coordinate system, which does rotate with the Earth, $(R, \Lambda, \Phi)$. Here:

$$ \begin{align} R &= r\\ \Lambda &= \theta - \omega_E t\quad\omega_E > 0\\ \Phi &= \phi \end{align} $$

Again I've ignored a boring constant which defines the planet's angle at $t = 0$.

So we can project the satellite's motion down onto the surface of the Earth, and the projected path, in the Earth-based coordinates, is then

$$(R_E, -\omega_E t, \omega_s t)$$

And this all makes sense: the satellite moves in the $-\lambda$ direction over time – westward – and it moves in the positive $\Phi$ direction.

This curve is periodic if $\omega_E/\omega_s$ is rational. The simplest case of this is $\omega_E = \omega_s$, in which case the curve looks like $(R_E, -\omega t, \omega t)$.

So then the trick is to project this last curve down onto the $X$-$Y$ plane (using capital letters for Earth-centred coordinates again). Using polar coordinates $(\rho, \Lambda)$ in that plane (I'm running out of variants of '$r$' here), then

$$ \begin{align} \rho &= R \sin\Phi\\ \Lambda &= \Lambda \end{align} $$

So the projection of the curve followed by the satellite on the surface of the Earth looks like $(R_E \sin\omega t, -\omega t)$. To see that this is a circle, change to cartesian coordinates $(X, Y)$:

$$ \begin{align} X &= \rho\cos\Lambda\\ Y &= \rho\sin\Lambda \end{align} $$

(I confess at this point to have lost track over whether these are the same $(X, Y)$ as above, but it does not matter).

So

$$(X, Y) = R_E(\sin\omega t \cos\omega t, -\sin\omega t \cos\omega t)$$

And now we can use some trig identities:

$$ \begin{align} \sin\theta\cos\theta &= \frac{\sin 2\theta}{2}\\ \sin\theta\sin\theta &= \frac{1}{2} - \frac{\cos 2\theta}{2} \end{align} $$

to get, finally,

$$(X, Y) = \frac{R_E}{2}(\sin 2\omega t, \cos 2\omega t -1)$$

Which is the equation of a circle in the $(X, Y)$ plane, radius $R_E/2$ and centre $(0, -R_E/2)$.

And that means that yes, it's Viviani's curve.

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  • $\begingroup$ This is really a wonderful answer! I've gotten almost all the way through; as far as "looks like $(R_E \sin\omega t, -\omega t)$. This is a circle radius $R_E$, centre $(R_E/2, 0)$." But for some reason I can't see how that is the circle, and I can't risk ingesting any further coffee. Is it possible to add a hint to help me with this last step? $\endgroup$ – uhoh Mar 29 at 2:57
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    $\begingroup$ @uhoh: I confess that I did the last step by getting a program to draw the thing, seeing it was, and convincing myself I understood why it's a circle. However I have the centre wrong, at least! I'll add some working. Sorry! $\endgroup$ – tfb Mar 29 at 10:24
  • $\begingroup$ oh no worries, it's a cool little problem isn't it! $\endgroup$ – uhoh Mar 29 at 10:40
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    $\begingroup$ @uhoh: yes, it is. It's also very classical geometry: you can imagine people in 15xx working this sort of thing out (OK, not the orbits then). And I like that there are now about three questions in two different SEs all of which touch on each other but are not the same problem. $\endgroup$ – tfb Mar 29 at 11:55
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    $\begingroup$ @uhoh: yes, or rather, since I projected down onto the XY plane and we know the curve is traced on the surface of the sphere that happens automagically. $\endgroup$ – tfb Mar 29 at 16:58
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Is it really true that a polar geosynchronous orbit (displayed in a synodic or rotating frame) is described by Viviani's curve? Can this be demonstrated mathematically?

Yes, and it's simple to show. At least if the satellite is in a perfectly circular orbit, Earth is perfectly spherical and there are no perturbations from any other source. Earth rotates at a constant speed and the satellite moves at a constant speed, hence:

  • The latitude of the satellite changes linearly in time, doing one full revolution each day.
  • The longitude of the satellite changes linearly in time (Earth rotates constantly), doing one full revolution each day.

E.g. if we start at (0,0):

  • after 12 hours Earth rotated by 180° in longitude and the satellite moved by 180° in latitude.

  • after 5.99 hours Earth rotated by 89.x° and the satellite moved by the same 89.x°.

  • after exactly 6 hours the satellite is over the pole and longitude is not well-defined, but this is only a single point at which the curve is not defined.

So, we can conclude that the satellite follows a Clelia curve with $|\phi| = |\theta|$ . I'll leave the proof that this Clelia curve is actually a Viviani's curve to the maths' textbooks.

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  • $\begingroup$ Thanks! I'll hold off for an answer that does the maths fro the Viviani's curve but this is certainly helpful! $\endgroup$ – uhoh Mar 26 at 9:29

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