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Calculating the motion of two dense spheres (in a frictionless vacuum) due to gravitational interaction is a classic physics problem that has come up numerous times.

Has any real-world space mission ever performed a demonstration of small body gravity? Set up an effectively weightless environment in high vacuum, nudge a pair of spheres very gently, and film their slow co-orbit with a time lapse camera.

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    $\begingroup$ Search terms: Cavendish, Eötvös. $\endgroup$ – rob Mar 23 at 1:13
  • $\begingroup$ @rob thanks, I have clarified my question. $\endgroup$ – Foo Bar Mar 23 at 13:09
  • $\begingroup$ What Rob meant was that yes it has been done, and to high accuracy. Not exactly as you are thinking. A dumbbell is suspended from a fine wire that is easy to twist. Two weights are put near the ends of the dumbbell. The force of gravity is weak, but enough to twist the wire. $\endgroup$ – mmesser314 Mar 23 at 14:11
  • $\begingroup$ Related. $\endgroup$ – rob Mar 24 at 4:19
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    $\begingroup$ Here's a simple Cavendish-like experiment, using lead rather than osmium, so it's much cheaper. ;) Bending Spacetime in the Basement. $\endgroup$ – PM 2Ring Mar 24 at 4:44
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If you're talking about an "effectively weightless environment", you're probably talking about a satellite in orbit. However, two objects in orbit aren't truly in a "weightless" frame; they will experience tidal acceleration due to the fact that they are slightly different distances from the Earth. It turns out that these effects would swamp the orbital motion you're looking for unless you got far enough away from the Earth.

Let's say that we have two spheres of density $\rho$, each of radius $r$, a distance $2r$ from each other. (Technically they'd have to separated by a distance $2r + \epsilon$, so that they weren't actually touching; but we'll ignore this.) The mass of each sphere is $\frac{4 \pi}{3} \rho r^3$, so the acceleration of each due to the other is $$ a_\text{sphere} = \frac{G (4 \pi/3) \rho r^3 }{ (2r)^2} = \frac{\pi}{3} G \rho r. $$

But now imagine that you're at the central point between the two spheres (crammed into that $\epsilon$-wide gap). You're orbiting along with the spheres. Since the spheres' centers of mass are not at the same location as you, you'll see them accelerate relative to you; this is the tidal acceleration I mentioned above. The exact magnitude & direction of this tidal acceleration differ depending on the positions of you, the masses, and the Earth; but in general, its order of magnitude is $$ a_\text{tidal} = \frac{G M_E r}{d^3} $$ where $d$ is the distance between you and the center of the Earth.

To cleanly observe the orbit the way you want to, you need this relative acceleration due to the Earth to be much smaller than the effects the spheres have on each other. In other words, their accelerations must satisfy $$ a_\text{sphere} > a_\text{tidal}. $$ After some algebra, this implies that $$ d > \sqrt[3]{M_E/\rho} $$ If $\rho$ is the density of osmium, this implies that $d \gg $ 6430 km or so. But the radius of the Earth is about 6370 km. In other words, to do this experiment, the distance between you and the center of the Earth must be much greater than the radius of the Earth.

This means that to do this experiment, a simple satellite in low Earth orbit wouldn't suffice; you'd need to launch into a relatively high orbit instead, since you want $a_\text{tidal}$ to be quite small compared to $a_\text{sphere}$. A geosynchronous orbit would be better; at that distance from the Earth, the effects of the tidal forces are only about 0.4% of the effects of the spheres on each other. But given the expense of launching to geosynchronous orbit, I suspect that scientists have decided that they have better things to do with their money.

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