How hard does atmospheric drag push on the ISS? Is it more than one pound?

Question

A comment under this question has me thinking; with it's huge main structure and giant solar panels the ISS presents a very large cross-section to Earth's rarified atmosphere at 400 km altitude. So much so that I have heard that every time the ISS goes into Earth's shadow the solar panels rotate to present a minimum drag surface, and some of them may do this even in the sunny parts of the orbit if full power is not needed to charge the batteries and run the systems currently drawing power.

How are the orientations of the ISS' eight independent solar arrays optimized?

Drag is bad because the ISS would soon reenter if not for regular shipments of rocket fuel from Earth to power engines that are burned to regularly re-raise the orbit, which sits at around 400 km these days.

Question: Roughly how hard does atmospheric drag push on the ISS? Is it more than one pound? The ISS is about 400,000 kg and so I'm guessing that the drag force must be quite substantial to make a dent in its velocity, even over months.

From here click for larger.

@asdfex When we close as duplicates we want to guide future readers to the best answer, not simply the oldest answer. So I've voted to close that as duplicate of this because I find this answer a lot more readable, informative and mathematically complete than the answer there. Interestingly the data there shows a 10x faster rate of descent as it's smack-dab in the middle of the late-2014 sunspot maximum associated with maximal atmospheric heating, so in this case perhaps merging the two will be the best way to serve future readers. I'll flag it and post something in meta about this. — uhoh, Mar 14 '21 at 23:14
I'd like this question more of it didn't mix SI units with pounds of force. — user2705196, Mar 15 '21 at 15:20
@user2705196 so I should change 400,000 kg to 880,000 lbs? Oh do you mean it should ask "Is it more than 4.5 Newtons?" To me it doesn't matter how the title ends; it could be "Is it more than a Henway?" It's arbitrary. What's important is that actual calculations (in questions or answers) don't mix units. I haven't calculated anything here, so units don't really matter. — uhoh, Mar 15 '21 at 15:28
My comment was mostly tongue-in-cheek! But yes, if you changed km to miles, kg to pounds (or stones?), and kept the pounds-force I would have found it less jarring. Of course, ideally as scientifically minded community, I'd stick to SI (and derived) units and thus simply ask for drag in Newton. No need to actually change the question though. I was mostly joking and should have added a ;-) to clarify! — user2705196, Mar 15 '21 at 15:58
Does this answer your question? How much of a drag is it, orbiting the Earth in a space suit? — Cristiano, Mar 20 '21 at 21:17
@Cristiano can you mention where on that page you think that the force on the ISS is listed? What is the value of the force you are seeing there? As far as I can tell, no, I don't see the answer to my question there. — uhoh, Mar 21 '21 at 01:08
@Cristiano Oh! 20 to 50 mN Yes I see now! Reading sideways I saw it reversed somehow. Yes it does provide an answer to this question, thanks! — uhoh, Mar 22 '21 at 14:47

Litho · Accepted Answer · 2021-03-14T12:35:19.550

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Here is a rough estimate. The ISS's height drops at the rate around 10 meters per day. The energy of a body of mass $m$ in a circular orbit of radius $r$ is $E=-\frac{\mu m}{2r}$, so $$ Fv=\frac{dE}{dt}= \frac{\mu m}{2r^2}\frac{dr}{dt}. $$ Since $v=\sqrt{\frac{\mu}{r}}$, $$ F=\frac{m v}{2r}\frac{dr}{dt}. $$ Substituting $m=4\cdot 10^5$ kg, $v=7800$ m/s, $r=6.8\cdot 10^6$ m, $\frac{dr}{dt} = 10/86400$ m/s, we get that $F$ is about $1/40$ of a Newton, or about $1/200$ of a pound-force.

Edit: 10 meters per day is the current rate, taken from the image (source and credit: Heavens-Above). It may be unusually low, see the comments.

edited Mar 14 '21 at 12:35

answered Mar 14 '21 at 10:20

Litho

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So given a quiet or normal Sun, about 25 mN. That's less than what's on most all-electric communications satellites that are put in GTO and circularize themselves then station-keep with Xenon thrusters. All-electric propulsion and multi-satellite launch system for communication satellites Converting at this point in time might be more work and risk than it's worth, but perhaps electric station keeping it will be built in to future space stations – uhoh Mar 14 '21 at 10:38
I got one order of magnitude more using the station keeping dv of 2 m/s/month and two orders of magnitude more using the specific orbital energy equation... to be checked. – asdfex Mar 14 '21 at 10:52
With $GM=3.986 \times 10^{14} \ \text{m}^3/\text{s}^2$ and $a= 6.778\times 10^{6} \ \text{m}$ I get a velocity of $7668.6314 \ \text{m/s}$. At 10 meters lower I get a velocity that is $5.68 \ \times 10^{-3} \ \text{m/s}$ higher. If that happens in 86400 seconds then the acceleration is $6.57 \times 10^{-8} \ \text{m/s}^2$. With $m=4 \times 10^5 \ \text{kg}$ that's a force of $0.0263 \ \text{N}$ or about 1/38 of a Newton, in excellent agreement with "about 1/40 of a Newton". – uhoh Mar 14 '21 at 11:04
@asdfex I might have used a wrong value for the height drop rate. I took it from https://www.heavens-above.com/IssHeight.aspx , which shows the rate of about 300 m per month. However, the answer https://space.stackexchange.com/questions/9087/how-often-does-iss-require-re-boosting-to-higher-orbit (which even refers to the same site!) shows an order of magnitude higher rate. I don't know which one is correct. Or does it depend on the solar cycle? – Litho Mar 14 '21 at 12:15
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@uhoh To be honest, that the increase rate of the orbital speed is the same as the deceleration due to the braking force is somewhat of a coincidence. If the gravitational potential was proportional not to $r^{-1}$, but to some other power, they would be different. Though still, they would differ only by a constant factor. – Litho Mar 14 '21 at 12:20
@Litho wow I didn't know that, very cool! – uhoh Mar 14 '21 at 12:25
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@uhoh Just in case: it applies only when the orbit stays almost circular. – Litho Mar 14 '21 at 12:28
@Litho so I've just asked What are the limitations to using this “trick” to quantitatively relate altitude changes with applied forces that cause them? – uhoh Mar 14 '21 at 15:08
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Litho, I used the same 10m/ day or 300m/ month. The solar cycle has a huge influence. But maybe a factor of 5 at most. – asdfex Mar 14 '21 at 15:09
Interesting that the solar cycle would show up here. Drag isn't the only force acting on the vehicle: solar radiation would be another, but the direction of solar radiation would oscillate back and forth as the ISS revolves around the earth... acting to accelerate it when directed along the velocity vector, then to decelerate it when directed opposite the velocity vector. The effect would cancel and the oscillations would show up with a period of the ISS (90 min?) – Mar 14 '21 at 16:08
Maybe the effect of the solar cycle is to change air density at great altitudes, which would directly impact drag (the dominant force)? Drag always acts opposite the velocity vector, so its effect wouldn't cancel out over the period of a revolution, and any periodicity in its magnitude would likely come from changes in air density, it seems? Does air density change with the intensity of solar radiation? – Mar 14 '21 at 16:13
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Air density changes a lot with radiation. Even day/night side is clearly distinguishable – asdfex Mar 14 '21 at 20:04
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So, I guess the reason they turn the solar panels at "night" is because shipping rocket fuel from the surface is so dang expensive, not because the drag is substantial. – jpaugh Mar 14 '21 at 23:08
@asdfex not just a factor of 5. The 2015 answer shows a sustained rate of over a factor of 10 faster than this, and there's even an example of over 400 meters in one day. See also How steady is the atmospheric drag force experienced by the ISS? – uhoh Mar 15 '21 at 00:15
@asdfex wow I see your new answer and have mentioned it here That's quite an amazing range of drags! – uhoh Mar 18 '21 at 10:48

jpaugh · Answer 2 · 2021-03-15T00:44:13.030

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While Litho's answer addresses the first step in the question, of how much drag the ISS experiences, it doesn't directly address why that amount of drag might be worth worrying about, or spending money developing a technique to adjust the solar panel's angle to account for.

In fact, even if the amount of drag saved by rotating the solar panels were insignificant on a daily basis, designing that feature could still have made sense while developing the IIS, from a project-management perspective, because the ISS is a massive, long-term endeavor.

For one thing, shipping fuel to space is very expensive! Circa 2010, it cost about $210 million per year to keep the ISS in orbit. That's a huge incentive to invest in "minor efficiencies." It follows that NASA would be willing to spend a little less than \$1 million per year to save even half a percent of that fuel.¹ And, considering the longevity of the ISS, they would have been willing to spend significantly more up-front.

_{¹ They might be willing to spend more than the dollar amount of fuel saved; the "launch weight" budget of any mission is far more constrained than the "launch cost" budget.}

I'm not sure how long NASA originally planned to keep this thing in space; but this NASA-RSA agreement from 1998 uses phrases like permanent human presence; mature operations; and, provide [...] maintenance plans five years in advance.

Additionally, because the ISS already needed a system to rotate the solar panels to track the sun, adding additional features to that big, unavoidable requirement would have been comparatively cheap.

Up to a point, adding more sophisticated logic to the controller would not have increased development costs substantially, since controller had to be built and coded, anyway. Perhaps it required upgrading the hardware to get better range-of-motion that was strictly necessary to track the sun.

That might affect the launch weight, which is normally a no-no; but if it saves weight on future launches, it might still be worth it. "One extra feature" would have added a small additional cost to the overall build budget.

Verification that the feature works and that it doesn't conflict with other design requirements could easily been more expensive than the feature itself; you have to make sure everything else still works when you go tweaking the power availability. Yet, the total cost from a man-hours perspective would have been amortized across hundreds of other features that were already intrinsic to the the design, and already had to be cross-checked against each other for mutual compatibility.

Furthermore, the entire point of the ISS was to be a long-running, upgradable system. They could have released a more sophisticated (but less critical) update to the panel control system well after the 1998 launch date --- at a time when schedules, budgets and oversight constraints had all relaxed somewhat. I'm sure they do over-the-air updates for some software-only upgrades today, although I wouldn't guess about how it worked in 1998.

edited Mar 15 '21 at 00:44

answered Mar 15 '21 at 00:24

jpaugh

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Does not even attempt to answer the question that was asked "Roughly how hard does atmospheric drag push on the ISS?" – Organic Marble Mar 15 '21 at 00:26
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@OrganicMarble There's already a great answer covering the most explicit part of the question; however, I was speaking more to the motivation of the question than the phrasing. – jpaugh Mar 15 '21 at 00:27
Stack Exchange sites work on a strict question and answer system. They aren't for discussions; the answer button is reserved for posting answers only. – Organic Marble Mar 15 '21 at 00:33
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@OrganicMarble Then, please read the whole question: "Roughly how hard does atmospheric drag push on the ISS? Is it more than one pound? The ISS is about 400,000 kg and so I'm guessing that the drag force must be quite substantial to make a dent in its velocity, even over months." From Litho's other answer, that drag force turns out not to be very substantial. What it doesn't address, is why bother adding this feature when the impact on drag is so small? – jpaugh Mar 15 '21 at 00:35
@uhoh I'm find with a down-vote if you disagree that I've contributed a reasonable answer. But, please, don't put words in my mouth. – jpaugh Mar 15 '21 at 00:38
@jpaugh I see that you are contributing something very important here and I appreciate it! You may want to copy/paste this or parts of it as an answer to How are the orientations of the ISS' eight independent solar arrays optimized? But I am having a hard time believing that the impact on drag is "so small". Can you cite some numbers or authoritative sources? Prose can't establish a ratio quantitatively. – uhoh Mar 15 '21 at 00:39
Also consider that drag is a lot worse near solar maximum – uhoh Mar 15 '21 at 00:39
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@jpaugh "Too long to be a comment:" is a standard phrase used by Stack Exchange answer authors to help readers accept a post that might not necessarily directly answer the question in some people's opinion but yet provides helpful and insightful additional information for the benefit of future readers. It's a standard thing. – uhoh Mar 15 '21 at 00:41
@uhoh I base that on Litho's answer, which states that the entire drag force on the IIS is "1/40 of a Newton, or about 1/200 of a pound-force.". That doesn't sound like much in the way of drag; although it is a lot in terms of cost and fuel. – jpaugh Mar 15 '21 at 00:41
@jpaugh "a lot" and "a little" have no absolute meaning here. Small force over a long time (drag down) is the same magnitude and opposite sign as large force over a short time (reboost burn). – uhoh Mar 15 '21 at 00:44
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@uhoh That is true. I do believe I addressed the "a long time" bit better than the first answer; although my answer does not rely as much on math as it does the some-what softer area of project management. It's important to remember that space is not just a physics equation. It's a place that people go. Everything comes into play. – jpaugh Mar 15 '21 at 00:45
@jpaugh if you scroll through these results you will see just how often and how freely we use "too long for a comment" at the beginning of an answer. I don't mean that you should keep it here, I just want to show that it's a commonly used expression here. – uhoh Mar 15 '21 at 01:17

How hard does atmospheric drag push on the ISS? Is it more than one pound?

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