Flatland on a sphere.Find the radius

[Storm Butler]

Note: this is not a H.W. problem. I'm just curious about it.

Imagine you are a flatlander living on a spherical planet. If you don't have any suns or any other satellites to work with how would you go about finding out what the radius of the planet is?

I don't think i know enough spherical trig. to give a good answer.

Would there be a way to trek out a triangle and measuer the angles and then tell from the excess of the angles?

 

[HallsofIvy]

An obvious way would be to move in a "straight line" until you come back to your starting point! The distance you moved would be the circumference of a great circle and the radius of the sphere is that distance divided by 2\pi.

You mention "excess of the angles". Yes, that is a good method. The radius is equal to the square root of (the area of the triangle divided by the excess).

 

[mathwonk]

or compare area of a triangle to the angle sum. there is a formula relating the radius to the angle sum of a spherical triangle of a certain area. ah yes, Halls said this.

 

[Storm Butler]

The problem with walking around would be that you might walk around the circumference of a cap of a circle and not end up going along a great circle.

 

[Michael Redei]

The problem with walking around would be that you might walk around the circumference of a cap of a circle and not end up going along a great circle.

That's not true if you walk really "straight". To do that, you need to move horizontally to a plane that is tangent to your sphere. This is easily done by driving a four-wheeled car. The plane through the four wheels' centres will by such a plane. (You could even ride a tricycle, although your colleagues might start making jokes about you then.)

 

[HallsofIvy]

I said "in a straight line"- i.e. following a geodesic, a great circle on the sphere.

 

[Vargo]

Others mentioned the angle excess formula, so I won't repeat that.

Lets say the sphere is large and you can only access a small part of the sphere. Let's also assume that you have no direct method to measure area. (Can anyone suggest an accurate method to measure area without already knowing R?).

Lets say you can measure lengths and angles. Both these things can be done with fairly pedestrian tools. Let's also assume you can measure them with perfect accuracy. Otherwise any error you make will make it impossible to tell your triangle from a Euclidean triangle (we are assuming we can only access a small part of the sphere).

Well, under these conditions, you can use Spherical Trigonometry. If you have a compass you can make an equilateral triangle (using the same method in Euclid's first proposition). Let s be the measured side length and alpha the measured angle. (The angles are equal by the law of sines). Let x be the cosine of alpha (x is known by calculation!).

Let S = s/R. To determine R, we must determine S. By the second Law of Cosines:
x= -x^2+(1-x^2)\cos(S)
Hence we can calculate
\cos(S) = \frac{x+x^2}{1-x^2}=\frac{x}{1-x}
Taking the inverse cosine we can reconstruct R.

I will point out that in the euclidean case, alpha=pi/3 which impliles that x = 1/2. The formula would then give cos(S) =1, which means R=infinity, which confirms the flatness of the space. (I am only mentioning this as a check on my calculation)

Challenge: If we allow the measurements to have error, is it possible to adjust this calculation so that it can actually distinguish positive curvature from zero curvature? I suspect the answer is no since I have assumed that measurements can only be taken locally and a sphere locally looks like Euclidean space.

 

[Michael Redei]

(Can anyone suggest an accurate method to measure area without already knowing R?).

Erect walls around the area you want to measure, fill this "basin" with water and divide its volume by the height it comes up to.

 

[Storm Butler]

So traveling around a complete loop on the surface of a sphere wouldn't be a geodesic? What would you notice differently physically if you took this path?

Also, someone was saying if you look at Newtonian gravity in flat land (dont consider GR) then you get that the force goes as 1/r. how do you figure that out?