Considering the answer of user Mark Adler here, to calculate how much ΔV is needed to raise orbit we can simply approximate it as the change in orbital velocity.

Starlink satellites seem to be deployed at around 270km.

So, from 270 km to reach 550 km we have:

$u_{270} = \sqrt{\frac{GMe}{a}} = 7.743 km/s$

$u_{550} = 7.58 km/s $

$ΔV = 163 m/s ~ 170 m/s $

Τo maintain orbit at 550 km we have to calculate atmospheric drag expected at this altitude and as per here and here we have:

$F_D = 0.5 ρ u^2 C_D A = 0.387 mN.$

Cd: ~1 (coeff)

A: assuming the solar panel is deployed and facing perpendicular to the wind direction when operating (to at least figure out the max area/max drag), $A_{panel} = 3.4 m * 10m $ + $A_{body} = 3.5m * 0.2m = 0.7$. To approximate density at this altitude I used this table.

Now that we have the force we need to counter to stay at the same altitude we can do:

$F = ma $ with initial mass 260 kg and after burning some fuel to reach there approx. 257 kg. using the Tsiolkovsky rocket equation with Isp = 1600 as per Musk's tweet.

That means that $a = 1.5 * 10^{-6}$ m/s**2 Let's say $a * 1 year = 47.45 m/s/year.$ is the acceleration we need.

The incident as described in my other post says that the atmospheric drag on those days increased by approx. 50% and the altitude of the satellites was 210 km.

Same source as above for density at 210 km during mean solar activity gives $2.2 * 10^{-10} kg/m3$. With that:

$Fd(mean solar) = 4.82 mN$ at 210 km

$Fd = 7.24 mN$ at 210 km (incident w/ 50% increase)

However, this outcome does not seem correct, because from what I read here, Figure 3 or Figure 4 where they compare specific impulse with thrust power at 1600 Isp we get at least 45 mN/kW.

Typical krypton ion engines with Isp = 1600 produce 100 mN which is way more than what they experienced during the incident. Approximating the power can give is something between 5-7 kW.

So, how come they couldn't overcome these conditions? Is it something wrong in the calculations?

  • 2
    $\begingroup$ Cool question! There's more sources for atmospheric density here but your value seems good if the Sun was in a good mood that day. You might check to see what mood the Sun was actually in during that time; the atmosphere can get quickly heated by transient solar events and density at this altitude can deviate remarkably from "mean values", see for example Which LEO satellite lost over 30 km of altitude in the geomagnetic storm of 13-14 March 1989? $\endgroup$
    – uhoh
    Jan 18, 2023 at 0:15
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    $\begingroup$ First: for objects in space calculate with a Cd = 2.2 (because of the low density, hypersonic flux, this a used value regardless of the actual shape of the object). Second: Its a year ago, so I do not know where I have read this, but the satellites have been shut down to safemode during the peak of the storm, so no propulsion $\endgroup$
    – CallMeTom
    Jan 18, 2023 at 8:42
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    $\begingroup$ 5-7kW for propulsion? 30 m² solar array per satellite plus power for communication? This seems to be a huge overestimate. $\endgroup$
    – asdfex
    Jan 18, 2023 at 11:52
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    $\begingroup$ Also note that LEO satellites are in Earth shadow almost 50% of time. $\endgroup$
    – asdfex
    Jan 18, 2023 at 11:53
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    $\begingroup$ @uhoh I have checked what the mood of the Sun was actually in during that time. In my other post, which I linked above, I explain the incident and link a couple of papers that show the increase in percentage. This is where I get this approximation for 50%. $\endgroup$ Jan 20, 2023 at 10:57

2 Answers 2


I see quite a couple of problems with this calculation, let me list them for further investigation:

  • The difference between air drag at 550 km and 210 km seems to be off. Air density changes by a factor of 55 (taking mean values for both of them, not taking the storm into account)

  • As CallMeTom notes, your drag coefficient is lower than the commonly assumed value for operations in a near-vacuum at orbital speeds.

  • You take the "increase by 50%" stated in some source as 50% higher than mean drag. According to the citation in the linked question it says "50% higher than on previous launches". This may refer to the highest density ever encountered. Excursions from the mean value often reach factors above 10.

  • Your calculation assumes that engines can fire continuously. This is likely not the case or would require huge additional batteries to cover the half of the orbit in darkness.

  • You assume that the total power of 5 - 7 kW is available to propulsion. Most of it will be dedicated to communications. The engine will likely be on the order of 1 kW or even less, or it would need to be oversized for normal operations

  • You assume that during the storm the engine was used to actively fight the drag. This was not the case as satellites were put into a mode with panels oriented edge on to minimize drag. In this configuration the ability to generate power for propulsion is greatly reduced.

  • $\begingroup$ Thank you for the remarks, I will make calculations again and come back. The first 2 points will be addressed. The 3rd I don't get it. I have linked 3 papers in my other post where they look at the storm and they indicate, for example on this one: agupubs.onlinelibrary.wiley.com/doi/10.1029/2022SW003152 Figure 4, a maximum of 40% increase. /// For remark 4: How could I use this in my calculations? /// Remark 5: Even with 1kW, based on the above calculations, the thrust would be sufficient to reach 550km. // $\endgroup$ Jan 20, 2023 at 11:09
  • $\begingroup$ Remark 6: I don't assume that the engine was actually used. I am simply trying to calculate a maximum thrust that can be produced by Starlinks. So, if the available thrust power was enough to overcome the storm, they would have used it. But it seems that is wasn't enough. So if I calculate the drag on that day, I can say that the upper limit for thrust is equal to that. $\endgroup$ Jan 20, 2023 at 11:11

There is another way to estimate the thrust of starlink satellites (if that is what you are trying to find).

Through some rearranging of the rocket equation (and dimensional analysis) we can arrive to the following approximating expression:

$$F = \frac{m_p}{t} \cdot {I_{sp}\cdot g_{0}}$$

In the above $F$ is the thrust (assumed constant), $m_p$ is the propellant mass needed for the transfer, $I_{sp}$ is the specific impulse, and $t$ is the total thrusting time.

As pointed out in the answer from @asdfex, the satellite will repeatedly enter in and out of eclipse during the transfer to the final orbit. During this time in eclipse we can assume that the engines are not active as this would result in an extreme oversizing of the satellite batteries. To get the total thrusting time we therefore need to make an assumption on the duty cycle (i.e. what fraction of the transfer time the engine is active). Let's assume a duty cycle of 50% as a first guess. This would cover time in eclipse plus some additional margin for outages. According to this link starlink satellites can take around two to three months to reach their final orbit after launch.

We can use these numbers (and those indicated in your original question) to get a ballpark estimate of the thrust output of starlink satellites. Plugging the numbers in we can guess the thrust of a starlink satellite as being somewhere between 11 mN and 17 mN.

  • $\begingroup$ Thank you, this is useful @Armadillo. These value, let's say 15mN is the thrust needed to be produced by the satellite over the time period t (so 2-3 months) to go from one orbit to another. But, this doesn't relate somehow to the max thrust the thrusters can produce. With this equation we can calculate that Starlinks need to produce 15 mN of Force to orbit raise, and some additional to stay there after they reached their operational altitude. $\endgroup$ Jan 22, 2023 at 12:11
  • $\begingroup$ For them to get away from the storm, and orbit raise to higher altitudes with lower densities, they would need to produce about 15mN of thrust, just to counteract the atmospheric drag and then additional force to achieve some DV to orbit raise. What I am trying to do is understand the max they could (or couldn't) produce but not over a 3 month period, over some days, just so they could "escape" the bad conditions found at 210km. $\endgroup$ Jan 22, 2023 at 12:15
  • $\begingroup$ It might not be a correct to assume that the satellites were lost just because they could not produce sufficient thrust. While the instantaneous thrust of the satellites in normal operating conditions might be enough, there may have been several reasons why the satellites could not be commanded to provide the required total impulse in sufficient time due to exceptional circumstances and operational constraints. It might be worth exploring what these mitigating conditions could be to get an answer. $\endgroup$
    – Armadillo
    Jan 22, 2023 at 13:40
  • $\begingroup$ Concerning the maximum thrust achievable: If starlink satellites could produce more than ~15 mN thrust during standard orbit raising, then this would probably ordinarily be exploited to reach their final orbit in less time. Typically satellite operators would like their satellites to reach their operational orbit as soon as possible to save on costs. $\endgroup$
    – Armadillo
    Jan 22, 2023 at 13:46

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