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
• That's basically what ISS VASIMR, sadly canceled, was supposed to do. – TildalWave Nov 9 '15 at 16:46
• Does this answer your question? How hard does atmospheric drag push on the ISS? Is it more than one pound? – uhoh Mar 14 at 23:05
• @uhoh: That's a duplicate of this one, not the other way around. – Vikki Mar 15 at 0:32
• This question and answer is more concise than the new one and therefore the better one. The new question contains a lot of unnecessary fluff making it hard to read. – asdfex Mar 15 at 9:54

First of all, let's figure out what the drag actually is. For that, Heavens-above has a nice chart.

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 $$e_k=m \cdot v^2/2$$. The energy at 406km is 29400301 J/kg, and at 406.1 is 29399868 J/kg. Thus, 433 J/kg of energy is lost per day. The force is applied over all that time period to make that energy lost. $$F=m\cdot A$$ provides us that 433 J/kg / (Distance traveled in a day) = m * a. Thus, the constant acceleration that would keep it at the same orbit is about $$6.56\cdot 10^{-7}~\rm m/s^2$$, or $$0.656 ~\rm \mu m/s^2$$.

Given the station mass of 419455kg, the decelerating force would be 0.275 newtons.

• Possible reasons for changing drag rate include: differing density of atmosphere ISS is passing through at different altitudes; differing latitudes meaning ISS meets different upper-atmosphere conditions (different densities, temperatures, effects of winds beneath, effects due to Earth (and therefore its atmosphere) not being spherical); and of course changes to ISS configuration and orientation will change how streamlined it is, not to mention that different payloads will change the ratio of its momentum to its drag profile, so it'll be more or less affected by the same magnitude of force. – anaximander Nov 9 '15 at 16:41
• You didn't include the biggest one of all, which is the solar cycle. The solar cycle results in the higher atmosphere being more active at it's peak, and less active in the solar minimums. – PearsonArtPhoto Nov 9 '15 at 16:44

Here's an update to the answer by @PearsonArtPhoto with data from other periods. In that answer, the drop rate during the last solar cycle (#24) is used. In the current (2020) minimum of solar activity, the drop rate is substantially lower:

That's a mere 200 meter per month, 7 m/day and a decelerating force of 0.017 N.

Compared to that, at the peak of the second-to-last solar cycle (#23) in 2002 the activity was twice as high, and the ISS suffered a lot more:

That's a staggering drop of 12 km/month or 400 m/day, corresponding to a force of more than 1 N.

This is a change by a factor of 50 - although it can't be fully blamed to solar activity alone: The orbital height varied by 50km over the years, and also the operation mode of ISS varied. In 2002 "night glider" mode wasn't in use yet - it reduces the drag during night by optimizing the position of the solar arrays.

• Can you point to a reference that shows night glider is currently used? I like your answer otherwise, but the only source I have to that is very old. You could well be right, I just don't know. – Organic Marble Mar 15 at 11:53
• No, I don't have a recent source. It was introduced in 2003. I removed the 'nowadays' so the statement is right in either case. – asdfex Mar 15 at 12:11
• Perhaps @Tristan can weigh in with some current knowledge. – Organic Marble Mar 15 at 12:14
• Better yet, I'll just ask space.stackexchange.com/q/50773/6944 – Organic Marble Mar 15 at 12:34