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I'm working through a chapter about forces and I was wondering about the definition of gravitational force of which things of mass are brought together. When thinking about the solar-system this "feels" some kind of intuitive to imagine the gravitational idea. But it makes me curious if this would also work in a smaller version:

Would it be possible to build a tiny model of the solar system made by a bowling ball and a tennis ball keeping the proportion in radius and mass of the objects at some place somewhere in the universe? And would be it be further possible to "stick" an extremely tiny object on the bowling ball and it would stay there just by the gravitation of the bowling ball?

Of course I know there would occur problems to install this model since every tool and the person itself would come up with gravitational forces ruining the model. As well as there mustn't be any object of mass somewhere next to my model.

Hope this one doesn't sound silly, since this is a rather elementary question but I was curious.

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  • $\begingroup$ The gravitational force of a bowling ball is infinitesimally small... $\endgroup$ – QuIcKmAtHs Jan 4 '18 at 14:48
  • $\begingroup$ Assuming using only normal matter, consider how the gravitational forces and the mass of objects scale relative to each other. $\endgroup$ – Jon Custer Jan 4 '18 at 14:48
  • $\begingroup$ @XcoderX but the tennis ball is in a relatively small distance and has a small gravitation and mass as well - and that is what I wonder about, is keeping the proportion the only condition for two objects to form such system like the solar system ? $\endgroup$ – LurioTabasco Jan 4 '18 at 15:04
  • $\begingroup$ While the proportion is similar @LurioTabasco, this does not seem productive, as they will not orbit each other due to their weak force of attraction. $\endgroup$ – QuIcKmAtHs Jan 4 '18 at 15:07
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    $\begingroup$ I don't think this is what you are after, but here's a video of a fascinating honest scale model of the solar system. $\endgroup$ – garyp Jan 4 '18 at 16:16
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I presume that you want to build a tiny working scale model, keeping all distances in the same proportions, and keeping the rotation periods the same.

The same equations would apply for the model as for the real Solar System. The force of gravitational attraction provides the centripetal force keeping a planet in circular motion :
$\frac{GMm}{r^2} = mr\omega^2$

If the period of rotation is the same then $\omega$ is also the same. $G$ is a universal constant. The only issue is how masses $M$ and $m$ scale. From the above equation
$M \propto r^3$
ie mass should scale with the cube of distances. This is what happens if you keep the average density the same for the bowling ball as for the Sun. The average density of the Sun is 1.4g/cc which is slightly more than the density of a bowling ball. The average densities of the planets vary from 0.7 g/cc for Saturn to 5.5 g/cc for the Earth. So you would need to find different materials for your Sun and planets, or perhaps make them non-uniform in density, while keeping their radii in correct proportion with other distances.

The forces of gravity between the model Sun and planets would be extremely small, and would be swamped by other forces if it were kept on Earth. But if the model were placed in space far from other influences it would work. Even a model Moon moving around the Earth.

However, you could not use a bowling ball for the Sun and a tennis ball for the Earth. The Sun is 109 times the diameter of the Earth, so if the Sun were a bowling ball the Earth would be about the size of a peppercorn. The distance of Earth from the Sun is 108 times the diameter of the Sun, and Pluto is 40 times as far from the Sun as the Earth is. So on the bowling ball scale the Solar System would have a diameter of about 2km!

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  • $\begingroup$ Thank you for that one, now I have clarification about that one. Especially the last two passages really helped. Can't upvote this one due to low reputation. $\endgroup$ – LurioTabasco Jan 4 '18 at 15:57
  • $\begingroup$ As a curiosity, you can find an accurate comparison between the sizes of the planets and their distances from the sun here: joshworth.com/dev/pixelspace/pixelspace_solarsystem.html $\endgroup$ – GRB Jan 4 '18 at 16:15
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    $\begingroup$ Minor nit: That should be a diameter of about 4 km, not 2, for a 40 AU radius. Yet another minor nit: Pluto's orbit is rather eccentric, swinging out to almost 50 AU. $\endgroup$ – David Hammen Jan 4 '18 at 20:06
  • $\begingroup$ @DavidHammen Solar radius is 1/215 AU, so the radius of Pluto's orbit is 40*215=8600 times larger. The diameter of a bowling ball is about 0.22m so the diameter of the solar system model will be 0.22*8600=1900m approx. $\endgroup$ – sammy gerbil Jan 4 '18 at 20:39
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Newton's law of universal gravitation says that any two bodies exert a nonzero gravitational force on each other. This then means that, if they are isolated enough in an empty region of space where there are no significant contributions from other forces and any gravitational influences from other bodies are uniform and/or can be neglected, then yes, the two bodies will orbit each other.

As pointed out by mmesser, this does come at a cost with respect to scaling, and it is pretty much a foregone conclusion that the period of the orbit will need to be rather long: if the bodies are small, their gravitational interaction is small, and they need to be moving very slowly to remain in the (comparatively) shallow potential well exerted by that gravitational interaction. However, if they're moving slow enough (and they have enough time to do the orbit without e.g. running into a planet or something) then they can indeed orbit each other.

In contrast to mmesser's answer, though, the scaling analysis need not be quite so rigid. If you want the model to be completely to scale, then that places additional constraints, but if all you want is a model system with bodies that orbit each other, then it can be perfectly reasonable to e.g. scale the orbit sizes differently to the bodies' radii or play with their relative densities. This is unlikely to save you in terms of getting this to a feasible experiment (which it isn't: the gravitational interaction between human-scale bodies has been measurable using torsion pendulums and other precision apparatus since about 1800, but getting this to be the dominating force is just not feasible) but as a thought experiment it's useful to have a play around with the different length, mass and time scales.

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  • $\begingroup$ You really helped me out with that one, unfortanetly I can't give any upvotes due to low reputation. $\endgroup$ – LurioTabasco Jan 4 '18 at 15:59
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Some things would change, others would not. To answer this, you need to think about how things change with size. Do they change as $r^3$, $r^2$, $r$, ...?

Suppose you made a model where everything was $1/10^\mathrm{th}$ as big. The Sun and Earth were $1/10^\mathrm{th}$ their diameters, and $1/10^\mathrm{th}$ as far apart.

The volume of a sphere is $$V = 4/3 \pi r^3$$

Since $r$ is $1/10^\mathrm{th}$ as big, V is $1/1000^\mathrm{th}$ as big. So the mass of the Sun and Earth are $1/1000^\mathrm{th}$ as big

The gravitational force between two objects is $$F = GmM/r^2$$

Given the changes in m, M, and r, F is $1/10,000^\mathrm{th}$ as big.

Earth is accelerated by the Sun's gravity. This is $1/10^\mathrm{th}$ as big $$a = GM/r^2$$

Earth's orbit around the Sun is very nearly circular. Without gravity, Earth would move in a straight line. Gravity pulls Earth into a circular motion. The centripital acceleration needed to do that is

$$a = v^2/r$$

So for Earth to orbit the Sun, it would have to move at a speed where

$$v = \sqrt{ar}$$

So v is $1/10^\mathrm{th}$ as big. The distance around the Sun is also $1/10^\mathrm{th}$ as big. A year is the time to orbit the Sun. This would not change.

$$T = 2\pi r/v$$

The force of gravity at the surface of the Earth uses the formulas above. Earth has $1/1000^\mathrm{th}$ the mass, but the surface is $1/10^\mathrm{th}$ as far from the center. You would accelerate $1/10^\mathrm{th}$ as quickly when you fell.

You would have $1/1000^\mathrm{th}$ the mass, so you would weigh $1/10,000^\mathrm{th}$ as much.

The atmosphere is held in place by gravity. You would have a hard time breathing.

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protected by Qmechanic Jan 4 '18 at 16:36

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