# What is the reason Starlink satellites took 4 days to re-enter during the accident on February 2022?

I am reviewing the incident that knocked out 39 Starlink satellites earlier this year. As I explain in this thread, there was a modest magnetic storm on the 3rd of February, which increased atmospheric drag and the satellites that were orbiting at 210 km altitude at the time, couldn't raise to their final orbit at 340 km.

After 4 days (on Feb. 7) the satellites entered the atmosphere and the event was even captured in Puerto Rico: video here.

My questions are:

• Why, if the thrust wasn't enough to raise, the satellites took so long to de-orbit? Did they slowly lose altitude? How does this work?
• What is a formula I could use to calculate how much more thrust the satellite would need with an increase of 50% in atmospheric drag?
• I believe there was some issue, where the Starlinks have been unable to keep a good attitude because the drag was too much so they tumbled. I need to look it up further before writing an answer tho, it is also possible I'm completely wrong. Commented Dec 25, 2022 at 19:35
• "the satellites took so long to de-orbit": what is your basis for saying four days is a notably long time for them to do so? Commented Dec 25, 2022 at 23:35
• Note that the satellite will be "lost" when it would take too much fuel to recover--the satellite can very well still be fully operational at that point, just unable to carry out it's mission. Commented Dec 26, 2022 at 4:01
• – uhoh
Commented Jan 8, 2023 at 13:38
• Potentially interesting, April 9, 2023 Inverse: NASA Reveals What Made an Entire Starlink Satellite Fleet Go Down
– uhoh
Commented Apr 10, 2023 at 8:33

Forces add. Drag forces simply subtract from the thrust force. In order to remain in orbit, you need a thrust of magnitude equal to the drag in the opposite direction. If you are already in orbit and drag increases by 50%, your thrust should immediately increase by 50% too if you wish to remain in that exact orbit. You can calculate the drag force on an object from the drag equation: $$F_D = \frac{1}{2}\rho C_D A_D v^2$$ Where $$\rho$$ is the local atmospheric density, $$v$$ is the satellite velocity, $$A_D$$ is the surface area contributing to drag (cross sectional area), $$C_D$$ is a constant called drag coefficient that factors for the shape of the body.