What is the velocity of the ISS relative to the Earth's surface?

Question

This was asked once before, but I believe there was no clear answer given and accepted.

The ISS has an orbital period of 92.9 minutes, or 5574 seconds. If it were in a completely equatorial orbit, this would give it a surface speed of 7190 m/s (40,075,000 / 5574). However, the ISS also has an orbital inclination of 51.64 degrees, meaning as it moves the surface of the earth also moves at a speed of roughly 464 m/s.

I've tackled this problem several times but my high school trigonometry fails me. How would one go about calculating this?

EDIT: Since asking this question I've realized the surface speed depends on the ISS's latitude, and varies quite a bit over an orbit. A more clear question to ask would be - how would one derive a function for surface velocity dependent on latitude?

Answer 1

JPL Horizons has trajectory data for the International Space Station, SPKID = -125544

Revised: Nov 23, 2022
Trajectory is TLE-based. Predicts run for 4 weeks into future, but are of low accuracy for times more than a few days past the revision date above.

Horizons can give position and velocity vectors of the ISS relative to the centre of the Earth in the ICRF frame. Unfortunately, it doesn't do the ECEF frame, so I improvised an approximate solution using the geometry of the WGS84 ellipsoid. The Earth's polar axis points in the Z direction of the ICRF frame.

First we find where the position vector intersects the ellipsoid surface. We need the distance from that intersection point to the Earth's axis to determine the rotation speed at that latitude. The rotation velocity vector at that point is perpendicular to the position vector and to the Earth's axis. Once we have that velocity vector we can subtract it from the velocity vector of the ISS. Hopefully, I haven't made any blunders with my geometry or algebra. ;)

The top graph below shows a typical result, over ~2 orbits, with a 1 minute time step. It gives a mean speed of ~7376 m/s. The second graph shows the raw ISS speed relative to the centre of the Earth. The third graph shows the Earth rotation speed. That is, the speed in the top graph is derived from the vector difference of the 2nd graph minus the 3rd graph.

Here's the ISS geocentric declination over the same timespan.

Note that times in the speed plots are in TDB, the declination plot uses UTC.

Here's my Sage / Python speed plotting script.

It can also be used on other Earth satellites. Horizons has several other satellites in its database, eg the Hubble has id# -48. Type mb into the target field for the full list of major bodies (spacecraft are towards the end). You can also type body names into the target field. Horizons will respond with a list of matching bodies if the name is ambiguous.

Here's the declination script. Any major body can be the observing center, and any body can be the target. Please see the Horizons docs for details on specifying bodies and times.

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Here's a little interactive 3D ellipsoid script I used to verify my maths.

Answer 2

The orbital velocity of the ISS is approximately 7.6 km/s, and the Earth's rotational velocity as a function of latitude is $465 \cos i$, where $i$ is the latitude.

Since the ISS is inclined 51.6° to the equator, its velocity vector relative to the ground below (ignoring rotation) is $\langle4720.723,5956.07\rangle \text{ m s}^{-1}$. The maximum latitudes that the ISS can reach are $\pm 51.6^\circ$, so the function relating position to surface velocity would be $v_s =\langle4720.723 - 465\cos i,5956.07\rangle \text{ m s}^{-1}$, whose magnitude will vary between the interval $\boxed{[7320.24,7424.04] \text{ m s}^{-1}}$.

Please let me know if you have any questions, as this may contain errors or be entirely incorrect.

Answer 3

Since none of the other answers actually showed how to compute the ground speed, the below Python script uses Skyfield to compute the ISS's velocity relative to the position on Earth it is directly above. You can likely find a library to do this is almost any programming language.

from skyfield.api import load, wgs84
import math
satellites = load.tle_file('http://celestrak.com/NORAD/elements/stations.txt')
print('Loaded', len(satellites), 'satellites')
iis = {sat.name: sat for sat in satellites}['ISS (ZARYA)']
t=load.timescale().now()
geocentric = iis.at(t)
lat, lon = wgs84.latlon_of(geocentric)
iisll=wgs84.latlon(lat.degrees,lon.degrees)
difference = iis - iisll
topocentric = difference.at(t)
v=topocentric.velocity.km_per_s
vkms=math.sqrt(v[0]v[0]+v[1]v[1]+v[2]v[2])
vms=vkms0.621371
print(str(vms) + " miles per second")

Answer 4

I remember reading something by Randall Munroe / XKCD about this. By wonderful coincidence, in the time it takes to listen to the song "500 Miles" by The Proclaimers, the ISS travels approximately 500 miles relative to the Earth's surface...or it might actually travel 500 miles plus 500 more, I'm afraid I've forgotten the detail, and as I'm posting this from work I can't access the XKCD website to check :-)