The recently launched Starlink satellites are designed to use their HAL Ion thrusters to seek their own independent orbits, evade collisions, and de-orbit at the end of their lives. That's a lot of maneuvering.

Does anyone know, or can make an educated guess what the Delta-V is for one of these?

Here is a rough estimate.

tl;dr:

raising 445 to 550 km        58 m/s
keeping it there             20 m/s
bringing it down            112 m/s

Total                       190 m/s


using about two kilograms of krypton, which is about 5 liters at 100 atmospheres.

## Raising them up to 550 km

I looked at the current TLEs and plotted eccentricity versus altitude and they seem to all be in nearly circular orbits at about 445 kilometers.

I used (from here)

$$T \approx \frac{24 \times 3600}{\text{revs per day}}$$ $$a \approx \left( \frac{T^2 GM_E}{4 \pi^2} \right)^{1/3}$$

to estimate the semimajor axis, then subtracted 6378137 meters to obtain an altitude (from here).

Orbital velocity is given from the vis-viva equation

$$v=\sqrt{GM_E \left( \frac{2}{r} - \frac{1}{a} \right)}$$

which reduces to

$$v=\sqrt{\frac{GM_E}{a}}$$

for a circular orbit. At 445 km altitude the velocity will be about 7643 m/s. At it's final altitude of 550 km the orbital velocity will be about 7585 m/s. That's 58 m/s slower, and as it turns out it happens to require about 58 m/s delta-v thrust forward, to raise (and slow down) the orbit from 445 km to 550 km. (currently looking for @MarkAdler's answer that first points this out)

update: Found them! 1, 2

## Keeping them up at 550 km

Let's estimate how much delta-v is necessary to keep a Starlink satellite at 550 km for say one year starting with an estimate of drag force

$$F_D = \frac{1}{2} \rho v^2 C_D A.$$

Let's use a coefficient of drag $$C_D$$ of 1 and use a cross-sectional area of 3.5 x 0.2 meters. Interpolating http://www.braeunig.us/space/atmos.htm at 550 km gives atmospheric densities of 2.3E-14, 3.4E-13 and 1.0E-11 kg/m^3 for Low, Mean, and Extremely High solar activity. Using the mean value, we get about 7E-06 Newtons.

With a mass of about 227 kg, that's an acceleration of 3E-07 m/s^2. Over five years, that's only about 5 m/s! However, let's say 10% of the time is high solar activity (30x higher density) and call this part of the budget 20 m/s.

## Bringing them back down again (all the way!)

At E would like to bring them well below the ISS's orbit and most other satellites in LEO, let's be aggressive and say we need to go from 550 km to 350 km to ensure rapid decay. The velocity there is 7697 m/s, so that's a delta-v of 112 m/s.

raising 445 to 550 km        58 m/s
keeping it there             20 m/s
bringing it down            112 m/s

Total                       190 m/s


Choosing an arbitrary Isp of 2000 seconds (exhaust velocity of 20,000 m/s) it means that at least 1% of the satellite's mass would have to be krypton to do this, or at least 2.3 kg.

At 3.8 g/liter at standard atmosphere, a bottle at 100 atmospheres for example would have to be say five liters, and also fairly heavy to safely hold the pressure.

The krypton system is a significant component of the whole spacecraft in terms of both volume and mass!

• According to lizard-tail.com/isana/lab/orbital_decay 350km isn't going to give you prompt decay -- try 180km or so. – Russell Borogove May 30 '19 at 1:15
• @RussellBorogove I'm sure that is correct but I'm not sure they need prompt decay so badly that they'd add the extra krypton mass and associated bottle mass. My thinking is that they would be comfortable getting down to 350 km, then tilting them broadside "into the wind" to give a factor of 5 or 10 increase in drag, and letting nature take its course. Presumably a non-directional low gain antenna would still allow command and control. – uhoh May 30 '19 at 1:19
• This answer assumes instantaneous delta-v. Starlink uses EP which will not have the same dalta-v as an instantaneous approximation especially this close to a large body. This answer is more of a lower bound, but I would expect the delta-v for the orbit raising and lowering to be a factor of 3-10 larger. – Knudsen Number Jun 3 '19 at 21:33
• @KnudsenNumber OK I found the Mark Adler links that I allude to in the answer; 1, 2. In the limit of low thrust the calculation is correct. My guess is that for these satellites it's going to be very close to correct; within a few percent which is certainly fine for what I label in my first sentence as a "rough estimate". They'll take weeks, using hundreds of orbits to get there. – uhoh Jun 3 '19 at 22:17
• You actually won't need that much fuel to deorbit. A deorbit that will happen anytime in a year's time is usually okay. – PearsonArtPhoto Jun 4 '19 at 1:37