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Wouldn't it be useful to catch few meter sized near-Earth asteroids whenever opportunity knocks and "lock them up" at the L4 Langrangian point of our Moon ?

Although at the L4 and L5 Lagrangian points stable equilibrium exists, stability at the Earth-Moon points is greatly complicated by solar gravitational influence.
Nevertheless so called Kordylewski clouds exist at both the L4 and L5 points.

If gradually more and more asteroids could be collected at the L4 point, and supposing they would be placed close to each other, at what total mass they would aggregate naturally because of mutual gravitation ?

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    $\begingroup$ They can only really aggregate if they can lose kinetic energy somehow by friction and heating during many many collisions. Gravity alone will alter their orbits, but as they start to affect each other they'll speed up relative to their center, rather than slow down. They'll need to "cool" somehow before they can slow down and aggregate. $\endgroup$
    – uhoh
    Commented Dec 9, 2019 at 17:26
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    $\begingroup$ @uhoh When you place them really close to each other they don't have to collide. I I could imagine that the Sun's gravity could tear the asteroids apart a bit. $\endgroup$
    – Cornelis
    Commented Dec 9, 2019 at 17:29
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    $\begingroup$ Relevance to space exploration? $\endgroup$
    – Organic Marble
    Commented Dec 9, 2019 at 17:30
  • $\begingroup$ @OrganicMarble A new, lighter moon at a stable point near Earth could be an important "stepping stone" for space exploration ! $\endgroup$
    – Cornelis
    Commented Dec 9, 2019 at 17:35
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    $\begingroup$ @Cornelisinspace oh I see, this is about "manual" collection and careful placement, not natural aggregation. Got it! You just need enough mass to overcome tidal and other multi-body forces and solar pressure for example. $\endgroup$
    – uhoh
    Commented Dec 9, 2019 at 17:37

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Not touching upon how "useful" this could be as a stepping stone, as that's difficult to answer, but the more narrow question "Will a pile of rocks stick together at EML4, or drift apart due to tidal forces?" is quite answerable.

An object near EML4 is acted upon by several forces, mostly the Earth's gravity balancing inertia in the rotating frame of reference, with a slight pull from the Moon completing the stability. In addition there are some complications due to the Sun's perturbations.

As a first order approximation, the tidal gradient can't be larger than the Earth's gravitational field. And it turns out that's enough to give a straight-forward answer.

The Earth's gravitational gradient at that distance is $1.4 \cdot 10^{-11 } s^{-2}$

For a reference asteroid, let's use an object like Deimos. Deimos has a surface gravity of $3\cdot 10^{-3} m/s^2$. With a radius of 6km, the tidal gradient of Earth would result in an acceleration of $8.4 \cdot 10^{-8} m/s^2$, far less than the surface acceleration provided by the object.

Surface gravity of objects scales with the radius, provided they have the same shape and density. Similarly, the tidal acceleration also scales with distance, so the question simply boils down to density. Deimos density (1.5 kg/l) already provides about 4 orders of magnitude of safety, so:

Objects, of any reasonable density, placed right next to each other near ELM4, will stick together purely by gravitational forces, provided they are large enough to not be blown apart by radiation pressure

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  • $\begingroup$ Could you give the expression and the calculation for the tidal acceleration ? $\endgroup$
    – Cornelis
    Commented Dec 10, 2019 at 9:18
  • $\begingroup$ @Cornelisinspace it's the field gradient, the first derivative of Earth's gravitational field. It will locally be close to linear. $\endgroup$
    – SE - stop firing the good guys
    Commented Dec 10, 2019 at 12:04
  • $\begingroup$ But what would the minimum mass be for the collection of asteroids to stay together ? With a payload of 100 tonnes for each rocket the catched asteroid would be 4 x 4 x 4 m$³$ , so for a 100 x 100 x 100 m$³$ stepping stone already 15.000 rockets would be needed ! $\endgroup$
    – Cornelis
    Commented Dec 10, 2019 at 14:53
  • $\begingroup$ @Cornelisinspace It's independent of mass and just depends on density, as far as gravitational forces are concerned. $\endgroup$
    – SE - stop firing the good guys
    Commented Dec 10, 2019 at 16:41
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    $\begingroup$ @Cornelisinspace the max payload you can lift to orbit is not directly correlated to the mass of the asteroid you can move to your lagrange point. There is no such thing as minimum mass to stay together as long as your object are dense enough. C type asteroids with all those volatiles, would do just fine. All that water... $\endgroup$
    – Khay
    Commented Dec 11, 2019 at 16:56

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