TLDR: Yes, but they aren’t going to. Perhaps 52-71m/s dV is practical.
My model of a rocket exhaust (in a vacuum) in as cone, that extends out behind the rocket while it fires, which applies pressure on the flat circle of the cone, as the gas molecules slam into whatever they impact.
By my measurement, this vacuum raptor engine bell https://twitter.com/SpaceX/status/1302038129990279168/photo/1 is about 14cm tall and 11cm wide – note that we care about the ratio here, not the absolute measurement. That gives us a radius to height for this engine’s exhaust plume at about 14:5.5, or 2.5:1. According to the wikipedia entry, Starship will have ~12MN of thrust in a vacuum.
So – at some distance away the engine bell there will be a point where the plume would apply 1N per square meter, or one pascal. There’s effectively no chance that this could cause any risk to the starship in question – it’s safe. Using area =πr2, this gives us 12,000,000 = πr2, or r = 1954m. How far away is that? Using the cone from earlier, that’s about 4886m away.
What does a pressure of 1Pa give us? Best case, we’re trying to push something like a centaur upper stage – large surface area, but low mass – travelling in an identical orbit, but a little distance behind the Starship. According to wikipedia, a Centaur III is about 3.05m diameter, 12.68m long and has a dry mass of 2247KG.
Assuming optimal positioning without rotation, that’s 3.05x12.68=38.7m2. So 38.7N exerted on a 2247KG object gives us, um, 0.017ms-2 – that’s...not much. If sustained for a whole minute – which is a brave assumption, given the starship would be accelerating away – would give us 1m/s of deltaV. Per minute.
Now I’ve demonstrated methodology, here’s a table I calculated, which ends when the target object is subject to about one atmospheric pressue - which given it was constructed on Earth should be be fine - well, at least not catastophic. Atmospheric pressure assumed to be 100,000Pa, not 101325Pa.
Pressure (Pa) |
Force (N) |
Area (m2) |
Radius (m) |
Distance |
Target area |
Target Force |
Target acceleration |
1 |
12000000 |
12000000 |
1954.41004761168 |
4886.0251190292 |
38.7 |
38.7 |
0.0172229639519359 |
10 |
12000000 |
1200000 |
618.038723237103 |
1545.09680809276 |
38.7 |
387 |
0.172229639519359 |
100 |
12000000 |
120000 |
195.441004761168 |
488.60251190292 |
38.7 |
3870 |
1.72229639519359 |
1000 |
12000000 |
12000 |
61.8038723237103 |
154.509680809276 |
38.7 |
38700 |
17.2229639519359 |
10000 |
12000000 |
1200 |
19.5441004761168 |
48.860251190292 |
38.7 |
387000 |
172.229639519359 |
100000 |
12000000 |
120 |
6.18038723237103 |
15.4509680809276 |
38.7 |
3870000 |
1722.29639519359 |
Huh. So as long as SpaceX was willing to let the target get within 150-500m of the Starship, a meaningful amount of DeltaV could be imparted to our perfect target. Outside of 500m the thrust is negligible,
Next question – how long would a Starship be close enough to do this. After all, it’s burning to goto Mars. Thrust of 12MN, mass of around 1320T. 12,000,000/1320000 = 9.1ms-2. Assuming constant mass for the first few seconds, and that once the starship is 500m away the thrust is negligible, starting at 150m - you have meaningful thrust for the first 350m of the Starship’s burn. Ignoring orbital mechanics, as this will be over in seconds – the formula s = ut + 0.5at2 gives a total time of… 8.8 seconds until it’s out of meaningful range.
I don’t trust my maths to integrate this entire mess to give a better estimate, but back of the envelope excel work gives me an estimate of range 52-71m/s, based on a series of calculations based on 1s intervals as they move apart.
52-71m/s isn’t enough to de-orbit from many orbits – but it’s enough to significantly decreased the orbital life of an object at 400km height, as per previous answers.
All you need is persuade SpaceX to let their rocket get within 150m of orbital spacejunk about as big as it is…