# What is the ISS drag?

ISS constantly loses altitude to air drag and other forces (tidal, electromagnetic). While finding that rate in the sources isn't that hard, with orbital mechanics of altitude loss actually increasing the linear speed despite growing drag, finding the decelerating force isn't nearly as easy.

Let's say we can develop a thruster of very high ISp that would keep the ISS at constant altitude without need for re-boosts from the delivery vehicles - what thrust would it need to have to let it maintain altitude?

• Do you want them to be constantly firing ? – Antzi Nov 9 '15 at 14:46
• @Antzi: Yes, non-stop. Say, it appears the EmDrive works as well as advertized. Or more realistically that's a very efficient ion engine. – SF. Nov 9 '15 at 14:48
• You'll need some mean of throttling to account for variation of drag induced by the environmental factors, and still requirer some stronger thrusters for evasive maneuvers. A variation of altitude can also be helpful for weaker launchers to reach the ISS (Or heavier cargos). – Antzi Nov 9 '15 at 14:52
• @Antzi: ISS has perfectly functional RCS for these, and if the thruster had a little surplus thrust it could be switched off for a time... nevertheless, treat this more as a thought experiment to explain what kind of force I mean instead of actual technical application. (ISS inside is microgravity; I want to know the strength of the drag component of that microgravity) – SF. Nov 9 '15 at 14:56
• That's basically what ISS VASIMR, sadly canceled, was supposed to do. – TildalWave Nov 9 '15 at 16:46

Of some note is the fact that the atmospheric drag rate changes over time, most notably with the solar cycle, but it can change for a variety of reasons, especially for a body as dynamic as the ISS. With the current altitude ranging around 400. The time that it took to go from 414 to 406 km was about 2.5 months, or, say, 75 days. That means that the drag on the spacecraft is about 106 m/day. Orbital energy can be calculated by $e_k=m * v^2/2$. The energy at 406km is 29400301 J, and at 406.1 is 29399868 J*mass. Thus, 433 J*mass of energy is lost per day. The force is applied over all that time period to make that energy lost. $F=m*A$ provides us that 433 J*mass / (Distance traveled in a day) = m*a. Thus, the constant acceleration that would keep it at the same orbit is about $6.56e-7 m/s^2$, or $0.656 \mu m/s^2$.