# Scale model of solar system

If I were to create a perfectly scaled model of our solar system in a 12 ft sphere, would the planets be visible to the naked eye ?

Clarifying question: What do you consider as the outer limit of the bounds of the solar system? The orbit of the outermost full planet (Neptune)? The orbit of the outermost dwarf planet? The Kuiper Belt? The Oort cloud? The heliopause or the bow shock?

What do you envision is embedded in the surface of that 12 foot diameter?

By my basic arithmetic, the diameter of Jupiter (the largest planet) is about 1/32,000 the distance from the Sun to Neptune; 12 feet divided by 32,000 is about .0045 inches, which I believe is about the size of a pinhead, so just barely visible. Not sure if any of the other planets would be visible – they would all be smaller than that.

Also, is the naked eye allowed to get right up next to the planets, or is it outside the sphere?

Good Question, I would say double the distance we have observed or past the furthest known planet.

This is kind of the size range I was thinking.

I don’t really envision anything on an edge of some kind, more like I see it diminishing as it gets further from the sun, and possibly., incoming forces of some kind that would diminish as they approached the center.

I would say from outside the sphere.

Taking Neptune @ 4.47 billion Kms from the Sun is on the outer edge of the 12’ sphere the representation of Earth at 147.17 million Kms would be 4.7 inches off the sphere’s centre and 0.0004 inches in diametre. A pea would be approx 0.2in, as above a pinhead 0.0045in.

So not visible from 12ft away unless it was light emitting or reflecting.

Here is a NASA blog post on the subject. Their scale is an order of magnitude larger (football field (300 feet) instead of 12 feet).

According to their definitions, the farthest out well-known solar body (Pluto) is about 80 yards from the Sun (implying the rest of the solar system beyond that point; i.e., the Kuiper Belt). The Sun itself is 2/3 of an inch in diameter – the size of a dime. And the inner planets are small grains of sand.

Shrink that down by a factor of 25 (ratio of their scenario and yours) and I feel pretty safe the inner planets are not naked-eye visible.

Here is a web site that has an accurate scale model of the solar system. The moon around our planet is one pixel in size, to give you the scope of the model.

The amount of space between the planets is staggering. This web site is the only way I was able to grasp it. I can’t imagine how difficult it is to send a probe or anything else between planets, considering the fact that they are always in motion. It must be like shooting a flea a mile away from you.

https://joshworth.com/dev/pixelspace/pixelspace_solarsystem.html

That really does help in gaining perspective.

Here’s a calculator for our solar system scales:

https://www.exploratorium.edu/ronh/solar_system/

Quick one for fun:
If the Sun were 6 inches in diameter,
Earth would be .05 inches diameter, 54 feet from Sun
Pluto would be .01 inches diameter, 0.4 miles from the Sun

Earth is about 93M miles from the sun, Mars about 130M. So at closest approach both on the same side of the sun, a bit less than 40M miles.

If we made 40M miles a football field, 100yards, then 1M miles would be 300/40 = 7.5 feet. The moon’s orbit is about half a million miles diameter, 240,000miles radius. So draw a circle at one goal line, a bit less than 4 feet diameter, and that’s the moon’s orbit.

Earth is 8,000mi diameter, so at this scale it would be a bit less than 3/4 inch diameter (a big gumball, say) - and keep in mind, earth is the biggest of the rocky planets. By that same scale, Jupiter at 10 times earth’s diameter would be about 7 inches diameter, an inch or so smaller than the size of a volleyball. But the closest we get to Jupiter would be 400M miles, so 10 football fields away. The sun would be 93M miles away, so almost 2-1/2 football fields away, about 700 feet. But the sun is 864,000 miles diameter, so on this scale - about 6 feet diameter, a bit less than a million miles ( 1M = 7.5 feet).

We’ve talked Earth-Mars closest approach - on opposite sides of the sun they are 130M+93M=223M miles apart, or 5.5 football fields, 1650 feet. Instead of 12 feet, our diorama at this scale for Neptune’s orbit would be 2.8billion miles is 70 football fields in radius, or 140 in diameter, which is almost 8 miles diameter across.

= = = =

For the proposed 12 foot sphere of the OP, Jupiter at 0.0045 inches is 1/222 of an inch, much smaller than a pinhead. But divide that in half, so 1/500th inch, since we’re talking radius vs. diameter. I presume the 12 feet would be the diameter-of-orbits of the solar diorama not one radius. Fortunately, 1/500th is twice the thickness of a human hair, so Jupiter at least is presumably somewhat visible. The sun would be 10 times that, so 1/50th of an inch. Good rulers have 1/32 inch divisions, some really good ones have even 1/64ths.

Those are wildly different distances … at least by my assumptions about what you mean by the very imprecise terms “have observed” and “planet”.

There have been a few attempts at building physical models of the solar system at scale.

This project at Tennessee Tech spreads it across the whole campus and has Pluto at pinhead size.

This page at the Exploratorium allows to calculate the sizes and distances of each body.

(A bit off-topic)

So many times I see this error, and I’m seeing it again in this thread: an American football field is not 100 yards; it is 120 yards long.

For instance, Sedna is in the same size class as Pluto, and it aphelion takes it completely beyond the heliopause – in other words, out of plama environment (“solar atmosphere”) of the Sun. Into interstellar space, technically.

Twice that is sucking up a lot of fundamentally empty interstellar vacuum.

The football field as a unit of measurement is usually understood to mean 100 yards. That’s not an error, it’s just a convention.

Here’s a table of planet sizes/orbits with the model size set to Neptune having an average orbital radius of 6’. This puts Pluto’s average distance from the sun at around 8’ which would mean its max distance is closer to 10’, so if Pluto’s entire orbit is supposed to fit into the 12’ sphere everything would have to be about half the size.

At this scale, the sun is a couple hundredths of an inch in diameter, and Jupiter & Saturn are just a couple thou across. These inch/feet measurements just seem wrong for this purpose. Feel free to click the metric button to restore sanity.