What if some future Falcon Heavy's launch put a regular boulder in orbit around Earth (as a secondary payload)? Well, not arbitrary one that lies by the edge of the road, rather something resembling potentially useful asteroids, with water ice, carbon and metal inclusions and so on. Tesla roadster is no doubts cool advertisement, but a boulder in the orbit can be a test yard for developing microgravity anchoring, drilling techniques, tugging, maybe even resource processing, and all that for a small price, so even students with cubesats technically could participate. If there was a boulder, then probably Philae lander components could have been tested before departing to the rendezvous with a comet, and we'd also knew better how to make use of NEO resources. So, the question is: do you think it's a worthy thing to do?

  • $\begingroup$ Comets and asteroids tend to be very loose aggregations, Launching a pile of rubble and subjecting it to launch vibrations and acceleration will probably compact the material and make it less representative. $\endgroup$
    – Hobbes
    Jan 31, 2018 at 10:39
  • $\begingroup$ I'm confused what your question is. $\endgroup$
    – gerrit
    Jan 31, 2018 at 11:55
  • $\begingroup$ A little bolder weighing some hundred kg is something very different to 67P/Churyumov–Gerasimenko weighing about 10 trillion kg. You could not test lander components on such a tiny piece of dust if they should land on a comet with a mass of more than a 100 billion times bigger. $\endgroup$
    – Uwe
    Jan 31, 2018 at 14:29
  • $\begingroup$ @Uwe, is it that much different? Is 0.0001g something that you can't emulate by turning on a small ion drive? $\endgroup$
    – ZuOverture
    Jan 31, 2018 at 17:13
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    $\begingroup$ The whole point of this boulder being small enough to fit on a FH is exactly this: if someone can use it to design the anchoring system that works in almost zero gravity, then it will probably work at 0.0001g as well. So big boulders would only make the bills soar without reason. $\endgroup$
    – ZuOverture
    Jan 31, 2018 at 17:26

2 Answers 2


edit: Actually, now that I re-read your question and think about this happening in Earth orbit, there's another problem. The dynamics in Earth orbit are very different than in deep space, far from any local sources of gravity. The rock and the spacecraft are both in orbit around the Earth, and so their relationship will constantly be changing as they move in their own Earth orbits. If one is above, below, or to the side of the other, then they are not going to stay like that for more than a few minutes. Even if one is behind the other, once the spacecraft starts accelerating toward the rock, it will start drifting up, away from Earth. There are no real parallel Earth orbits, despite the discussion in the question Parallel orbits around the Earth - effectively?

If you get all the way out to a heliocentric orbit, then you can do your maneuvers over a period of hours without having to think much about the Sun.

It's always good to have a reminder just how weak gravity really is... so:

Gravity is really, really, weak!

$$G \approx 6.6741 \times 10^{-11} \ \ \text{m}^{3} \text{kg}^{-1} \text{s}^{-2}$$

For a body with spherical symmetry you can calculate the gravitational force as if all the mass were at the center, instead of integrating. So for a body of radius $R$ and density $\rho$, the gravitational acceleration (the thing that's 9.8 m/s^2 on Earth) would be:

$$M = \frac{4}{3} \pi R^3 \rho $$

$$a = GM \frac{1}{R^2} = \frac{4}{3}G \pi R \rho $$

Plug in an example "rock" density of 5 g/cm^3 (5,000 kg/m^3) and the 4.5 meter radius of a BFR (BFRocklifter) and you get 6.29E-06 m/s^2. That is a bit less than one millionth of Earth gravity, but more importantly it is only about 1% as strong as a characteristic gravitational attraction to the crazy-shaped object 67P that Philae landed on.

That is, if the BFRocklifter could even lift it.

And, you'd have to bring up two of them in order to simulate the crazily-shaped gravity field associated with 67P. See the image below.

Also, as @Uwe points out in comments, the escape velocity is going to be much lower as well. In fact it also scales linearly with radius (for a sphere of fixed density) so it will be 100 times lower than that near 67P as well.

$$v_E = \sqrt{\frac{2GM}{R}} = 2 R \sqrt{\frac{2}{3} G \pi \rho}$$

At this point, if it's a low mass cubesat with large solar panels, the photon pressure from the Sun is starting to become non-negligible.

enter image description here

  • $\begingroup$ I have to thank you! I should have done the calculation by myself before writing the comment. $\endgroup$
    – Uwe
    Jan 31, 2018 at 15:00
  • $\begingroup$ If I did it right, the escape velocity for the sphere with 9 m diameter is 7.52 mm/s, that is 133 times less than the 1 m/s of 67P. $\endgroup$
    – Uwe
    Jan 31, 2018 at 16:04
  • $\begingroup$ @Uwe yep, it scales linearly with radius as well for fixed density. See (several) edits. $\endgroup$
    – uhoh
    Jan 31, 2018 at 16:19
  • $\begingroup$ This answer mostly adresses maneuvers around small mass object. If that was the problem, it could have been solved by testing a spacecraft near ISS, because it's bigger than a boulder and has more complex shape than 67P. But that's a minor problem, it was already shown that spacecrafts can work near comets and asteroids (Hayabusa, Stardust), and smaller NEOs will only require a bit better propulsion precision. Meanwhile landing attempts, and more important, sampling in low gravity still is a significant challenge. $\endgroup$
    – ZuOverture
    Jan 31, 2018 at 18:29
  • $\begingroup$ @ZuOverture The ISS has a more complex shape indeed, but its mass is very, very small when compared with the huge mass of 67P. The mass of the ISS is about 455 t or 455,000 kg. But 67p has a mass of about 10,000,000,000,000 kg. The ISS is bigger than a boulder of 9 m diameter, but it is very, very small when compared with 67P. But the mass of the spherical bolder with 9 m diameter is 1,908,500 kg, more than 4 times the mass of the ISS. Only the length and width of the ISS is bigger than that of the bolder, but not the mass. $\endgroup$
    – Uwe
    Jan 31, 2018 at 21:23

There's probably no real reason for this:

  • to test landers, robots, that work on hard surfaces - you just test them on Earth!

  • as @Hobbes explains, actual comets/asteroids are in fact "loose bunches" of stuff. So your "Boulder Experiment #1 !" really wouldn't help.

Note that the one and only reason the Falcon guys had for adding a Tesla car, was publicity. It's exactly like running TV ads or simply painting a sign on the side - it was simply a "publicity stunt". (And a good one!)

It's worth noting too that boulders are incredibly heavier than Tesla cars, or any other robots / spacecraft.

A boulder the same size as a Tesla would weigh 10? 20? X more than a Tesla. A 1-tonne piece of stone is actually quite small. (Ask any stonemason!)

I would say that for some very specific use, your surprising idea mighty actually be implemented! (If we had to test some very specific robot, which, for some reason simply couldn't be tested in gravity.) I'd say it's unlikely to be useful as a sort of "general facility". And again, bear in mind that stone is extremely dense.

(Mind you, some folks argue that the ISS achieved - nothing - other than just showing it could be done, so - who knows!)

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    $\begingroup$ A cube made of granite with a weight of 1000 kg has an edge length of 0.71 m. Density 2800 kg/m³. Much smaller than a Tesla. $\endgroup$
    – Uwe
    Jan 31, 2018 at 23:36

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