Iterative Tangents

A curiosity journal of math, physics, programming, astronomy, and more.

Torus world

An Earth-sized torus

In the previous post I speculated about some characteristics of the horizon on the outer equator of a toroidal world. To make the discussion a bit more concrete, let's pick a size for our hypothetical toroid. Personally, I like the look of a torus with a 4-to-1 ratio of major radius to minor radius:

Let's give our torus a major radius equal to Earth's radius. How does the surface area compare to Earth's?

StartLayout 1st Row 1st Column upper A Subscript e a r t h 2nd Column equals 4 pi upper R Subscript e a r t h Superscript 2 EndLayout

Solving for Rearth:

StartLayout 1st Row 1st Column upper R Subscript e a r t h 2nd Column equals StartRoot StartFraction upper A Subscript e a r t h Baseline Over 4 pi EndFraction EndRoot EndLayout

The surface area of a torus is

StartLayout 1st Row 1st Column upper A Subscript t o r u s 2nd Column equals 4 pi upper R r EndLayout

We've chosen r = R/4 and R = Rearth, so

StartLayout 1st Row 1st Column upper A Subscript t o r u s 2nd Column equals 4 pi upper R Subscript e a r t h Baseline StartFraction upper R Subscript e a r t h Baseline Over 4 EndFraction 2nd Row 1st Column Blank 2nd Column equals pi upper R Subscript e a r t h Superscript 2 Baseline 3rd Row 1st Column Blank 2nd Column equals pi StartFraction upper A Subscript e a r t h Baseline Over 4 pi EndFraction 4th Row 1st Column Blank 2nd Column equals StartFraction upper A Subscript e a r t h Baseline Over 4 EndFraction EndLayout

Our toroidal world's surface area is only a quarter Earth's, which makes sense given how much top and bottom have to be cut off a sphere to make a torus like the one above.

How does volume compare?

StartLayout 1st Row 1st Column upper V Subscript e a r t h 2nd Column equals four thirds pi upper R Subscript e a r t h Superscript 3 Baseline 2nd Row 1st Column upper R Subscript e a r t h 2nd Column equals RootIndex 3 StartRoot StartFraction 3 Over 4 pi EndFraction upper V Subscript e a r t h Baseline EndRoot 3rd Row 1st Column upper V Subscript t o r u s 2nd Column equals 2 pi squared upper R r squared 4th Row 1st Column Blank 2nd Column equals 2 pi squared upper R Subscript e a r t h Baseline left-parenthesis StartFraction upper R Subscript e a r t h Baseline Over 4 EndFraction right-parenthesis squared 5th Row 1st Column Blank 2nd Column equals 2 pi squared upper R Subscript e a r t h Superscript 3 Baseline one sixteenth 6th Row 1st Column Blank 2nd Column equals 2 pi squared StartFraction 3 Over 4 pi EndFraction upper V Subscript e a r t h Baseline one sixteenth 7th Row 1st Column Blank 2nd Column equals StartFraction 3 pi Over 32 EndFraction upper V Subscript e a r t h EndLayout

Our toroid's volume would be less than a third Earth's.

Part of the reason to calculate volume is so we can calculate mass and gravitational pull. Of course, talking about the gravity of a toroidal world is an exercise in absurdity, since on the inner equator gravitational attraction is toward the center of mass in the (empty) middle of the toroid, meaning objects would be pulled off the inner surface. In a more full-fledged hypothetical, we'd have to imagine a countering force that keeps objects on the inner surface, but that force would also potentially lessen gravitational pull on the outer equator. (Edit: see this correction.) For now I'll ignore all that and only calculate the gravitational pull on the outer equator.

Giving our torus an Earth-like density, its mass is

StartLayout 1st Row 1st Column upper M Subscript t o r u s 2nd Column equals StartFraction 3 pi Over 32 EndFraction upper V Subscript e a r t h Baseline normal rho Subscript e a r t h Baseline 2nd Row 1st Column Blank 2nd Column equals StartFraction 3 pi Over 32 EndFraction upper M Subscript e a r t h EndLayout

Gravitational acceleration on Earth's surface is

StartLayout 1st Row 1st Column a Subscript e a r t h 2nd Column equals StartFraction upper G upper M Subscript e a r t h Baseline Over upper R Subscript e a r t h Superscript 2 Baseline EndFraction EndLayout

So gravitational acceleration on our torus's outer equator is

StartLayout 1st Row 1st Column a Subscript t o r u s normal bar o u t e r normal bar e q u a t o r 2nd Column equals StartStartFraction upper G StartFraction 3 pi Over 32 EndFraction upper M Subscript e a r t h Baseline OverOver left-parenthesis upper R Subscript e a r t h Baseline plus StartFraction upper R Subscript e a r t h Baseline Over 4 EndFraction right-parenthesis squared EndEndFraction 2nd Row 1st Column Blank 2nd Column equals StartFraction 3 pi upper G upper M Subscript e a r t h Baseline Over 32 left-parenthesis five fourths upper R Subscript e a r t h Baseline right-parenthesis squared EndFraction 3rd Row 1st Column Blank 2nd Column equals StartFraction 3 dot 16 dot pi upper G upper M Subscript e a r t h Baseline Over 32 dot 25 dot upper R Subscript e a r t h Superscript 2 Baseline EndFraction 4th Row 1st Column Blank 2nd Column equals StartFraction 3 pi Over 50 EndFraction a Subscript e a r t h EndLayout

Due to the smaller mass and further distance from the center of gravity, downward pull on the outer equator would be less than 20% of Earth's. If the residents of this world are familiar with Newton's law of gravitation (perhaps from observations of neighboring spherical planets and their moons), and they have some expectation of their planet's Earth-like density, the lack of gravity would be very surprising. Most likely residents would speculate that they lived on a hollow sphere, in this case not an unreasonable conclusion.

However, such a theory would be quickly undermined by expeditions off the outer equator, which we'll discuss in a subsequent post.