# Scale model solar system in interstellar space - will it work?

Now that FTL travel is here there are all sorts of family outings we can go on.

One of these is a scale model solar system lovingly made by hobbyist Barty Slartfast.

In full scale

Sedna is three times farther away from Earth than Pluto, making it the most distant observable object known in the solar system. It is 143.73 billion km from the Sun, thus giving the Solar System a diameter of 287.46 billion km. It is 143.73 billion km from the Sun, thus giving the Solar System a diameter of 287.46 billion km. https://www.universetoday.com/15585/diameter-of-the-solar-system/

The scale model

Barty scaled it down by a billion. He made sure that all distances and diameters are in proportion to the real thing

The masses of the solar bodies are in proportion to those of the real ones. (but not necessarily scaled by the same factor as the distances)

To be determined

We've divided distances by a billion, how much should we scale down the masses of the mini-planets to make it all work (or doesn't it matter because we can also adjust orbital velocity)?

The planetary spins can be adjusted if this is necessary. He doesn't intend to have atmospheres, liquids or even tidal effects. Just solid bodies.

Question

Ignoring gravitational disturbances from tourists, can we make such a scaled down solar system that will be held together by gravity once the initial velocities and all other parameters have been set correctly?

Bonus question

It would be nice for tourists if the Little Earth year could be about an hour. Can this be done simply by adjusting the masses and orbital velocities?

• Does he want the model Sun to actually be a functioning fusion reactor, or is he fine with sticking a bright lightbulb in there? – Roger Nov 28 '18 at 20:58
• A light bulb will be fine. He doesn't want to be concerned with the evolution of the solar system or solar winds. Just a working gravitational model of its 'present' configuration. – chasly from UK Nov 28 '18 at 21:04
• You also might be interested in some of the terrestrial scale models of the solar system -- several at your proposed scale! See en.wikipedia.org/wiki/… – Roger Nov 28 '18 at 22:34
• @Roger - For a moment I was wondering how they could possibly be working models! – chasly from UK Nov 28 '18 at 23:11

Okay. My initial answer is completely wrong. Took a friend with a stronger grasp of Newtonian mechanics to set my head right. Holding densities the same, in a place where spacetime is nearly completely flat, proportional reduction across the board will work absolutely fine.

This does mean that the masses are scaled by the cube of the distance scaling, but that actually results in all the appropriate values being neatly cancelled. Kepler's 3rd law can be expressed as $$T = \sqrt{\frac{R^3}{M}}$$, and since $$R$$ is being decreased by $${10^{-9}}$$ and $${M}$$ by $$10^{-27}$$, it all comes out even.

The problem with the visibility of the elements of the solar system, given the ratio between their diameters and the vast distances between them, remains a barrier to this being an easily-taken-in tourist attraction.

(For the bonus, in order to speed up the earth while keeping its orbital distance correct, you'd have to radically increase the mass of the sun, which would have a similar effect on the entire system.)

Preserving my original answer so others can learn from my mistakes!

The scale of your scale model is dependent on which properties are being scaled. Gravity is a result of the product of the two masses and inversely proportional to the square of the distance between them. If all dimensions are being reduced by the same constant, the masses of all the components of the scale model will be reduced by the cube of the reduction.

By means of explanation:

$$F=G\frac{{m_1}{m_2}}{{r^2}}$$

is the equation for gravitation. Assuming density is held constant, $$m_1$$ and $$m_2$$ both have their masses reduced by the cube, because

$$V_{sphere}=4/3 {\pi} {r^3}$$

So the reduction in mass, assuming, again, that density is consistent, is $$1*{10^{-9}}^{-3}$$.

Meanwhile, the orbital velocity equation is

$$v = \sqrt{\frac{GM}{R}}$$

$$G$$ is the gravitational constant, $$M$$ is the mass of the system (in the case where one object is substantially more massive than the other, generally just the "central" object of the orbit), and $$R$$ is the orbital radius. $$R$$ has decreased by a factor of a billion, while $$M$$ has decreased by a factor of an octillion. Even though v also needs to decrease a billionfold, accounting for the root, a factor of a billion is still missing.

This means that everything in the scale model will have to be ten thousand times as dense to compensate... and given that Earth is currently 5.51g/cm^3, you're running into special material problems. The densest material on earth, Osmium, tops out at ~23g/cm^3, which is still orders of magnitude less dense than our rocky planets have to be.

Assuming you have access to some manner of degenerate matter that has the appropriate density, you still can't arbitrarily reduce the orbital period of any given component of the Mini System(tm). Orbital period is defined by the masses of both elements of the system and the distance between them, so you'll be screwing up your scale model by dramatically reducing the orbit diameter of the Earth.

On top of all this, your model earth, at one-billionth scale, would be a marble 1.2cm in diameter. Not so much fun for the kiddies when it's lost in the ~150m between it and the beach-ball-sized super-dense sun.

It's a hobby project, but even if you have access to strange materials from which to assemble your planets, it's not going to be much to look at.

Edit: Just noticed an important detail in your bonus question - orbital velocity is a function of mass and distance, and cannot be adjusted independent of them.

Edit the second: It should also be noted that you'd also need a way to keep the matter in whatever state hits the correct density for your project. "Naturally occuring" degenerate matter is in dead stars where the crushing gravity keeps it in what is ordinarily a deeply unnatural state. With that in mind, if you can manage that trick, you could presumably manipulate everything in the Mini System to behave however you want. But then it wouldn't be hanging together "naturally".

• I'm still digesting your answer. In the meantime. Barty has agreed, on your advice, to make each celestial body glow with a bright light to make it visible. Each will have its own colour. – chasly from UK Nov 28 '18 at 21:23
• This question is tagged hard-science, but this answer isn't. A hard science would demonstrate the equations that you allude to, and prove that you are right. This doesn't answer a hard-science question. – kingledion Nov 28 '18 at 21:56
• Then vote it down. Demonstrating how it would work would require hard science. There are a variety of reasons as to why it doesn't work, and most of them are elementary school math (like the radii of the planets relative to their orbits from the perspective of anyone coming to visit the attraction). – jdunlop Nov 28 '18 at 21:58
• Unnecessarily snarky on my part. Equations do make things better, and my mass estimates were off by an enormous factor. – jdunlop Nov 28 '18 at 22:22
• Perhaps I should have omitted the tourist attraction part. I did it mainly to add colour. What I'm really interested in is whether a tiny solar system is possible at all. My bad. – chasly from UK Nov 28 '18 at 23:15