An astronaut in a spacesuit travels around the Earth at the same altitude as the ISS. Let's say the astronaut leads the ISS by 1000 meters along the same orbit.

After one orbit, how much velocity has the astronaut lost or gained with respect to the ISS due to atmospheric drag? How much closer or farther will they be?

Assume a quiet sun.

Assume the astronaut is in any configuration you like; prone, broadside, or tumbling, curled up, etc.

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    $\begingroup$ The force due to drag will be proportional to the cross-sectional area of the object, while the acceleration will be inversely proportional to the mass. Essentially, this question boils down to a comparison of the area/mass ratio for an astronaut and the ISS. $\endgroup$ Feb 10, 2020 at 14:32
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    $\begingroup$ You could probably find some information related to SuitSat (en.wikipedia.org/wiki/SuitSat) $\endgroup$ Feb 10, 2020 at 16:12
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    $\begingroup$ Two words: ballistic coefficient. $\endgroup$ Feb 10, 2020 at 16:32
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    $\begingroup$ @Uwe or perhaps just the magnitude of drag is so small that working outside the ISS is possible. $\endgroup$
    – uhoh
    Feb 11, 2020 at 0:30
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    $\begingroup$ Oh! I got your point! :-) You can estimate BC also from a random number generator, but I doubt that it would be of any help. Ok for the simulation, I'll post some graphs. $\endgroup$
    – Cristiano
    Feb 13, 2020 at 16:34

3 Answers 3


Here’s a “real case” simulation, in the sense that I read the ISS initial state from a TLE+SGP4, then I propagate that TLE forward in time until I find a point 1 km away from the initial state (that point is the astronaut and he is exactly along the ISS orbit).

The ISS ballistic coefficient is obtained from a best fit of the ISS position during two reboosts; I chose the TLEs from 20002.55629799 to 20023.25950111 (they are smooth enough to obtain a "stable" BC estimate). To calculate the initial states for this simulation I used the TLE 20005.24658230.

After about 18 minutes, the program finds a ballistic coefficient of about $179\,kg/m^2$ (it’s not fixed, because the drag coefficient varies with the air composition), but keep in mind that it’s a fitting parameter (like the BSTAR used in the TLEs) and its value is good only for my simulation and I have no idea about its accuracy. If the simulation perfectly simulates the real world, the BC is exact, but since no simulation can be perfect, the BC is not exact.

The BC of the astronaut is simply calculated by putting mass= 90 kg, area= 1.8 x 0.8 = 1.44 m2 and cd= 1.2 (but it varies with the air composition and I use an additional parameter to set that variation). The result of that variation is an average BC= $46.3\,kg/m^2$.

Without doing any calculation, we already know that the astronaut will decay faster than the ISS (the smaller the BC, the faster the decay rate) and the distance between the two objects will become greater and greater.

Unfortunately I cannot post a 3D graph here, but the following 3 graphs should explain what happens.

The first one should answer your question:

distance /range rate

it shows the distance between the ISS and the astronaut and the range rate. I plotted two orbits to show that the range rate tends to increase. With this particular TLE, the distance after 93 minutes (about 1 orbit) is 1009.289 m.

The second graph shows the ISS radius vector and the difference between the magnitude of the astronaut radius vector and the ISS radius vector (the astronaut flies lower than the ISS):

radius vector

The last graph shows the ISS ECI speed and the difference between the magnitude of the astronaut and ISS ECI velocities (the astronaut flies faster than the ISS in an inertial reference frame):


As a final note, here’s the usual details about the simulation. It includes the Newtonian and the relativistic accelerations of all the planets, Sun and Moon.
The Earth's gravity field is modeled with the SGG-UGM-1 gravity model (computed using EGM2008 derived gravity anomaly and GOCE observation data) truncated to the degree and order 15.
For the calculation of the air density, I use the NRLMSISE-00 model along with an updated data file for the solar and geomagnetic indices: www.celestrak.com/spacedata/SW-All.txt.

EDIT: it seems interesting to me to add a graph of the drag:


EDIT #2: following the OrganicMarble's comment, I calculated the RIC components (radial, intrack, crosstrack) of the astronaut.
While the radial and intrack components don't add any useful information (given the small distance between the objects, the radial component is about the same as ||Rsat|| - ||Riss|| and the intrack component is about the same as the distance between the two objects), the crosstrack component could be interesting:

xtrack component

I've extended the propagation to 10 orbits to show that the astronaut oscillates left and right of the ISS orbit, although after 10 orbits the oscillation is still less than 1 meter (xtrack > 0 is towards the angular momentum direction).

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    $\begingroup$ This is great stuff! In a station-centered frame, what would be the astronaut's motion? $\endgroup$ Feb 13, 2020 at 19:55
  • $\begingroup$ @OrganicMarble I think that you can consider the first graph as ISS-centered; in that frame, the graph shows the distance and the range rate of the astronaut wrt the ISS. But probably I don't understand what you mean. $\endgroup$
    – Cristiano
    Feb 13, 2020 at 22:30
  • $\begingroup$ It would be interesting to see the motion of astronaut as seen by a camera on the ISS....does she drift more ahead, below, off the side, or what? $\endgroup$ Feb 13, 2020 at 23:15
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    $\begingroup$ Really nice! Thanks. $\endgroup$ Feb 14, 2020 at 13:24
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    $\begingroup$ @uhoh Tomorrow I'll post both the simulation w/o the atmosphere and the result of your procedure with the density obtained from the NRLMSISE-00 model. $\endgroup$
    – Cristiano
    Apr 15, 2020 at 17:46

[I add another answer to avoid too much mess.]

Here are the graphs that show the differences in distance and speed with and w/o the atmosphere:


Distance with the atmosphere: 1008.3116 m
Distance w/o the atmosphere: 999.7882 m


||Vast|| - ||Viss|| with the atmosphere: 2.2493 mm/s
||Vast|| - ||Viss|| w/o the atmosphere: 0.6093 mm/s

Your procedure

Since the air density and CD are not constant, I need to integrate FD / m against the time.
Even if your value for the astronaut cross section is more realistic, I kept my value (1.44 m2).

$\Delta v_{ISS} = 0.381916\,mm/s$
$\Delta v_{astronaut} = 1.47378\,mm/s$

$x_{ISS} = 1.06555\,m$
$x_{astronaut} = 4.11185\,m$


How much of a drag is it, orbiting the Earth in a space suit?... After one orbit, how much velocity has the astronaut lost or gained with respect to the ISS due to atmospheric drag?

tl;dr: It depends on the mood that the Sun is in, but you'd loose about 23 meters per orbit or 340 meters per day at 400 km, plus/minus a factor of 10.

After looking at several images of astronauts outside the ISS and guessing at dimensions, I've come to the conclusion that a suited astronaut with full backpack (life support system) is at most about 2.0 square meters in cross section viewed from the back or front with arms and legs outstretched. Viewed top-down it's probably half that.

The drag equation at high velocity is just

$$F_D = \frac{1}{2} \rho v^2 C_D A.$$

With $a=F_D/m$ and $\Delta v = a \Delta t$ this becomes:

$$\Delta_v = \frac{F_D}{m} \Delta t = \frac{\rho v^2 C_D A \Delta t}{2m}.$$

Assuming 100 kg for the astronaut and hardware, a 92 minute orbit and $C_D$ of 1 and an orbital velocity of 7670 m/s at 400 km, all we need is the density. However this is by far the largest uncertainty.

At the ISS altitude of 400 km the website http://www.braeunig.us/space/atmos.htm (see also this answer) gives 5E-13, 4E-12, and 5E-11 kg/m^3 for Low, Mean, and Extremely High solar activity.

If we choose the middle or Mean value we get $\Delta v$ of only 0.013 m/s per orbit.

That corresponds to a change in altitude of about 23 meters per orbit or 340 meters per day.

It will be about a factor of 10 faster or slower if the extremely high or low solar activity densities are assumed, about a factor of 2 slower if you can maintain a prone, head-first attitude.

How much closer or farther will they be (to the ISS)?

Along track relative to the ISS we can use $x = \frac{1}{2} a t^2$ which ends up as $\frac{1}{2} v_{final} \Delta t$ which is about a 36 meter change if the ISS didn't lose any velocity, or zero if the ISS loses velocity at the same rate as the astronaut. To that I will defer to the other answer which examines the ISS' trajectory in detail.


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