Why does the ISS track appear to be sinusoidal?

Question

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.

Answer 1

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.

This answer is a rewrite using my original answer and the technical details from where_is_tftp's answer.

Answer 2

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:

Answer 3

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.

Answer 4

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!

Answer 5

Bear in mind that the orientation of an axis of rotation or of an orbit is essentially fixed in space. This means that there is a fixed relationship between the ISS orbit and the Earth's polar axis. Its orbital plane must be centered on Earth's center of mass; as it happens, the ISS orbit is inclined with respect to the Earth's equator (the ISS orbital parameters are by design), so the ISS ground track will appear to move up and down across the equator. Since the map is laid out such that the Earth's surface is unfolded into a flat sheet with North always up, the ground track will appear sinusoidal (it's a math/geometry thing).

For what it's worth: the track shifts from orbit to orbit because the Earth has moved (rotated) under it. The ISS orbital period is about 92 minutes, so when the ISS has completed one orbit, the Earth has rotated about 23 degrees under it.

The track doesn't undergo seasonal change because neither the Earth's axis nor the ISS orbit change with the seasons. What changes through the seasons is the Sun's aspect relative to the Earth due to Earth's axial tilt relative to its orbital path around the Sun. The key point is how I introduced my answer - the ISS orbit is fixed in its relationship to Earth's axis, which is all that matters for this question.

Answer 6

I'll repost my answer to Analytical expression for the ground track of the International Space Station to supplement these answers and show what the the ISS (or other near-circular) orbit ground track is.

Not only is it not a sine wave it's not even a closed curve since the Earth slowly rotates below the ISS.

(I) use a parametric equation.

If the Earth were not rotating, then we would have something like

\begin{align} x & = \cos \omega (t-t_0)\\ y & = \sin \omega (t-t_0) \ \cos i\\ z & = \sin \omega (t-t_0) \ \sin i\\ \end{align}

where the radius of the orbit is 1, $\omega$ is $2 \pi/T$ and $T$ is the orbital period, and $i$ is the inclination of the orbit.

Then we would have

\begin{align} lon & = \arctan2(y, x) + const\\ lat & = \arcsin(z)\\ \end{align}

If the Earth is rotating then

$$lon = \arctan2(y, x) - \omega_E (t-t_0) + const$$

where $\omega_E$ is $2 \pi/T_D$ and $T_D$ is a sidereal day (23h, 56m, 4s roughly).

Solving this for longitude as a function of latitude looks like some serious work and I am not sure there is an analytical solution.

Instead you can try the parametric equation approach where you first make a hidden table of times, and then solve for $lon(t)$ and $lat(t)$ and plot $lat$ vs $lon$

Here is a plot, I haven't adjusted $t_0$ or $const$ and just used rough values for $\omega$, $\omega_E$ and $i$ but it should be enough to get you stared.

$t_0$ and $const$ represent the known starting conditions of the spacecraft that you are trying to plot; $t_0$ is the time at which it crosses the equator going north, and $const$ is the longitude on the Earth below the spacecraft at that time.

Here is some further reading:

• 5 Orbit and Ground Track of a Satellite
• Describing Orbits
• Theory of satellite ground-track crossovers

Python script:

import numpy as np
import matplotlib.pyplot as plt

twopi = 2*np.pi
to_degs, to_rads = 180/np.pi, np.pi/180.

omega = twopi/(92*60)
omega_E = twopi/(23*3600 + 56*60 + 4)

time = 60 * np.arange(101.) # 100 minutes

t0 = 1000. # arbitrary, you can fit this later
inc = 51.
const = 1.0  # arbitrary, you can fit this later

x = np.cos(omega * (time-t0))
y = np.sin(omega * (time-t0)) * np.cos(to_rads*inc)
z = np.sin(omega * (time-t0)) * np.sin(to_rads*inc)

lon = np.arctan2(y, x) - omega_E * (time-t0) + const
lat = np.arcsin(z)

if True:
    plt.figure()
    plt.plot(to_degs*lon, to_degs*lat, '.k')
    plt.xlim(-180, 180)
    plt.ylim(-60, 60)
    #plt.gca().set_aspect('equal')
    plt.show()