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The math, I read, is

$F = G\frac{m_{1}m_{2}}{r^{2}}$

What I'm unclear about is how does it work in practice.

Say there are two identical pebbles massing 1 kilogram each outside Sol's gravity well close enough to attract each other.

What happens then? And, How long does each 'stage' take?

Do the pebbles start to orbit around each other and eventually coalesce?

Do they, without coalescing, begin to attract other particles around them?

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Gravity's just a theory. ;) –  AdamRedwine Oct 16 '12 at 18:18
4  
"How does gravity work?" "Fine, thanks!" –  Keith Thompson Oct 16 '12 at 19:27
1  
The explanation of Feynman starts around minute 7 (youtube.com/watch?v=1SrHzSGn-I8) ;-) –  Fabian Oct 16 '12 at 23:19
    
Why the downvote? –  Everyone Oct 19 '12 at 19:59

1 Answer 1

up vote 6 down vote accepted

Assuming you've placed your two pebbles far from any other masses the only force they will feel is their mutual gravitational attraction. Suppose you place them one metre apart, then the force they feel is given by your equation, and since in your example $m_1$, $m_2$ and $r$ are all equal to one the force each pebble will feel is just $G$ or $6.673 \times 10^{-11}$ Newtons. Each pebble will therefore start accelerating towards the other pebble at $6.673 \times 10^{-11}$ ms$^{-2}$.

Lets say you've made your pebbles from granite, which has a density of about 2700 kg/m$^3$, so the radius of the (spherical) pebbles is 0.045 m. When the pebbles collide, after a bit of bouncing around they'll settle down and remain in contact at a spacing of 0.09m. At this spacing the force between them will be about $8.2 \times 10^{-9}$ Newtons.

You could make the pebbles orbit each other. The general equations for two bodies orbiting each other are a bit complex if you're not up to speed with calculus, but if you're happy with a circular orbit it's easy to calculate the orbital velocity. For an object moving in a circular orbit the acceleration towards the centre is simply $v^2/r$. In the example above we calculated the acceleration to be $6.673 \times 10^{-11}$ ms$^{-2}$, and the radius of the orbit is half the spacing so $r = 0.5m$. That means the orbital velocity is given by:

$$ 6.673 \times 10^{-11} = \frac{v^2}{0.5} $$

so:

$$ v = 5.78 \times 10^{-6} \space \text{m/sec} $$

So if you placed the pebbles a metre apart and set each one moving at 5.78 microns per second they would follow circular orbits about their centre of mass.

If you want to know more about how two bodies orbit each other have a look at the Wikipedia article on the two body problem.

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Just to add: The pebbles won't coalesce because the force pulling them together is very small - much smaller than the material strength. Once you get more than few 10km size lumps of rock their own gravity is strong enough to squash them into a sphere –  Martin Beckett Oct 16 '12 at 17:59
    
A quick follow-up for your leisure time - Once the two pebbles are at a separation of 0.09m, do any other pebbles nearby experience two separate gravity 'field', or a single one? I would think the latter –  Everyone Oct 16 '12 at 18:30
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@Everyone Why would you expect anything like that? There is nothing magic about the "1" in 1 meter. The length of the meter is entirely arbitrary. The gravitational field of extended bodies is actually the sum of the separate field from each infinitesimal bit of mass. It just happened that you can often take the results as if all the mass was at the CoM. –  dmckee Oct 16 '12 at 21:50
    
The second paragraph above mentioned the two pebbles settle down permanently at a distance of 0.09m. So I wondered whether neighbours then begin to experience a stronger 'pull'. –  Everyone Oct 17 '12 at 9:14

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