Question #45260
posted on 05/26/2008 3:01 a.m.

Q:

Dear 100 Hour Board,

I'm looking at constructing a scale model of the solar system. If I were to use a standard size tennis ball to represent Jupiter, what would you suggest I use to accurately represent those other heavenly bodies?

- Elliot

A:
Dear Elliot,

You may have to rethink this idea, as you'll see at the end. However, here's what I did: first, I found the ratio of the tennis ball's diameter* to that of Jupiter. Then I multiplied this by the diameter of each of the remaining planets (and Pluto) to get the diameter of the object you would need to represent them. Here are the results, in centimeters, followed by my (rough) suggestion of object:

Sun - 61.8 cm (a beach ball that's 2' across)

Mercury - 0.22 cm (a bead 2 mm across)

Venus - 0.54 cm (a pea)

Earth - 0.57 cm (a slightly bigger pea)

Mars - 0.30 cm (a BB)

Jupiter - 6.35 cm (tennis ball)

Saturn - 5.35 cm (a billiard ball)

Uranus - 2.27 cm (a shooter marble)

Neptune - 2.20 cm (also a shooter marble)

Pluto - 0.11 cm (a very small pebble 1 mm across)

Now why could this be tricky? Well, I decided to apply the same ratio to the Solar System, which I chose to be the diameter out to Pluto's orbit. Your model would have to be 2.62 km across. If you only count out to Neptune's orbit it would be 2.00 km across. If you decided to show all the way out to where the Sun's gravitational influence stops dominating (about a four light-year diameter), you would need 16,800 km. Unfortunately, the Earth is only 12,742 km in diameter. I'd recommend picking either Pluto's or Neptune's orbits as the boundary, personally. Assuming you have access to a vehicle, that could actually be kind of fun.

—Laser Jock

*all diameters taken from Wikipedia

You may have to rethink this idea, as you'll see at the end. However, here's what I did: first, I found the ratio of the tennis ball's diameter* to that of Jupiter. Then I multiplied this by the diameter of each of the remaining planets (and Pluto) to get the diameter of the object you would need to represent them. Here are the results, in centimeters, followed by my (rough) suggestion of object:

Sun - 61.8 cm (a beach ball that's 2' across)

Mercury - 0.22 cm (a bead 2 mm across)

Venus - 0.54 cm (a pea)

Earth - 0.57 cm (a slightly bigger pea)

Mars - 0.30 cm (a BB)

Jupiter - 6.35 cm (tennis ball)

Saturn - 5.35 cm (a billiard ball)

Uranus - 2.27 cm (a shooter marble)

Neptune - 2.20 cm (also a shooter marble)

Pluto - 0.11 cm (a very small pebble 1 mm across)

Now why could this be tricky? Well, I decided to apply the same ratio to the Solar System, which I chose to be the diameter out to Pluto's orbit. Your model would have to be 2.62 km across. If you only count out to Neptune's orbit it would be 2.00 km across. If you decided to show all the way out to where the Sun's gravitational influence stops dominating (about a four light-year diameter), you would need 16,800 km. Unfortunately, the Earth is only 12,742 km in diameter. I'd recommend picking either Pluto's or Neptune's orbits as the boundary, personally. Assuming you have access to a vehicle, that could actually be kind of fun.

—Laser Jock

*all diameters taken from Wikipedia