Why does the ISS track appear to be sinusoidal?

Okay, time for a stupid question, but it keeps me up at night trying to figure it out.

The ISS 'route' appears on paper as such: I am imagining it has to do with a combination of its orbit and the tilt of the Earth, but if this is the case, on spring/fall equinox it should be a flat line, no?

Can someone explain why it's in a sine wave? I've been playing with my globe trying to see how the tilt of the Earth might affect the path.

• What is the sine? The y coordinate of a circle. Oct 7 '16 at 21:12
• FYI, the yellow line is called the 'ground track', but I'm not up for answering the question. en.wikipedia.org/wiki/Ground_track Oct 7 '16 at 21:29
• Why would the equinox make a difference? Oct 8 '16 at 13:55
• @uhoh Apologies - I understood a sinusoid to be a repetitive wave. Feel free to edit it as necessary for the appropriate language. Dec 11 '16 at 7:50
• @Mikey wow now you have me stumped! I'm not sure if there's a standard, accepted name or adjective for the ground track of even a simple, circular orbit with non-zero inclination. No need for apologies here. How about naming what it is about the shape that you are curious about, and what shape you might have expected? "Can someone explain why the ISS's orbit looks a bit like a sine wave but flatter on the top and bottom when projected onto a flattened map?" That's just an example. Maybe you just want to ask "Why does it oscillates up and down instead of being straight on a flat Earth map?"
– uhoh
Dec 11 '16 at 8:21

The trajectory of the ISS only appears to be sinusoidal when it is mapped onto the flat plane. In reality the orbit around Earth looks like that: The answer to why the ISS' orbit appears to be looking like a sine wave is harmonic motion and circular motion. The ISS is moving along the circumference of a circle (the orbit) around the earth: $\omega$ is the angular speed, $\Phi$ is the initial angle (phase), $r$ is the radius of the orbit and $P$ is the position of the ISS. The angle at time t is given by $\omega t + \Phi$.

If you observe the position $P$ at time $t$ you can express the $x$- and $y$-coordinate as

$\ P_{x}(t)=rcos(\omega t+\Phi)$ and similarly $\ P_{y}(t)=rsin(\omega t+\Phi)$

In your case the ISS moves in circular motion around Earth and it's position is simply $P$ at time $t$. If you observe this point, standing at the origion $O$ of the above circle and project that onto the $OY$ axis you get $\ y(t)=rsin(\omega t+\Phi)$. Then plotting the function you get something like this: Note that the border between day and night also appears to be curved due to the same reason. The line isn't closed because while the ISS revolves around Earth once, Earth itself has rotated by a bit.

When mapping the surface of a sphere onto the plane you can only keep one line straight. The equator of the earth appears as a straight line in both pictures. Also keep in mind that this is a simplification of different projection methods used and only explains why the orbit seems to look like a sine-wave. In reality more complex projection methods are used.

• Thanks. I was trying to simulate it with a marker on a globe from an orbit around the equator as the Earth begins to tilt every which way. I did not know it was such a skewed orbit (in relation to the equator). Oct 7 '16 at 21:58
• The day-night border also shows up as a wavy curve on the map and is known to be a circle/orbit around the planet. It might help to imagine that the ISS would fly exactly along that border.
– null
Oct 7 '16 at 23:21
• @Mikey There is a fun online simulator here en.homasim.com/orbitsimulation.php Enter the parameters and it draws the groundtrack for you. (Choice of parameters is somewhat limited but it's a good learning tool) Oct 8 '16 at 0:41
• @Mikey The reason the orbit is so highly inclined is so that the Russians can collaborate on the station. Because of how near the poles it reaches, the ISS passes over the Baikonur launch site every day. Oct 8 '16 at 4:40
• @Sykler Yeah that is a bit misleading in my answer as the two pictures actually show different orbits. Oct 8 '16 at 13:31

An orbit takes place on a single 2d plane, through the center of the planet. A sphere and a plane through its center intersect in a circle. See diagram: There is no actual sine wave movement going on, the ISS moves around the planet above the red line, in a circle. The apparent sine motion of the ground track is entirely due to the Mercator projection being used when the map is 'unfolded' from globe to flat surface. Hopefully this image allows you to more easily see the projection, note how the red line "curves downwards" at the front of the image, crosses the equator, and goes up around the back of the sphere. When the checkerboard surface is stretched out into a 2d grid, the red line gets stretched out to be a sine.

Note that the sinusoidal motion is also due to the orbit being almost circular. If the orbit was a highly eccentric ellipse, like this: you get a ground track like this: This is very useful for observing the ground in the northern hemisphere, since the satellite spends a lot of time in that hemisphere.

Going to even further extremes, the ground track of a geosynchronous satellite is a point. For something nearly in geosynchronous orbit, you can get figure-8 ground tracks.

• In this case, maybe :) a more practical use lies behind it though. The soviet union had a hard time launching to geosynchronous orbits since they had no launch sites close to the equator, they used three of these molyina orbits to keep at least one satellite above the horizon at all times. But yes, this is used as a spy satellite orbit too, for sure. Nov 7 '21 at 20:36

See pp. 14 & 15 in this FAA guide for an explanation, including these great diagrams: I also recommend sketching it out on a roll of paper:  • +1 for the toilet paper alone May 4 '21 at 12:22
• Brilliant with the toilet paper, that is very helpful! If you're comfortable sharing, what interesting contraption are you building with the button-clicky thing, the zip-tie and the flash drive? May 6 '21 at 20:52
• @Mikey, thanks! The stuff on my desk (toilet paper, fidget cube, and all) is just desk clutter. May 9 '21 at 5:56

The answer is harmonic motion and circular motion.

When you observe a point moving in circular motion over the circumference and you map this move into the diameter of length $x_m$ in $OX$ axis - you get a harmonic oscillation equation:

$$x(t) = x_m cos(\omega t + \phi)$$

If you map this into $OY$ axis you get the same motion, just seen differently as:

$$y(t) = x_m sin(\omega t + \phi)$$

In your case the ISS moves in circular motion and it's coordinate point is $P'$ in a time $t$. But the projection of this point (seen from observer located on the Earth in the middle $O$ of our circle) onto $OY$ axis is $$y(t) = x_m sin(\omega t + \phi)$$ and when observed continuously the point seem to be moving in harmonic motion.

You have just rediscovered what Galileo Galilei found in 1610, so you may well be on a good way to other discoveries - good luck!

• This should be the accepted answer and not mine as it is more detailed, maybe the OP should change that. Oct 8 '16 at 13:30
• @JannikPitt no it shouldn't. Yours answers the question more directly, with relevant direct illustrations. This answer does this a bit more abstractly (i.e. without showing the globe etc.). Oct 8 '16 at 14:25
• @Ruslan The answer starts nicely as general explanation of how circular motion can be seen as a harmonic motion and then applies this knowledge to the problem: "In your case the ISS moves in circular motion and it's coordinate point is P′ in a time t (...)" Oct 8 '16 at 15:11
• @where_is_tftp Yeah maybe those two answers should be merged into one, using both the illustrations and the little bit maths that is behind it. Should I do that? Oct 8 '16 at 15:29