The centre of the earth

Question

I'm a real beginner here (first post and first foray into math since high-school, trying to catch up), so I'm going to try my best to explain my problem in mathematical terms then follow up with an intuitive explanation. Thanks in advance!

---

Maths:

Given two points $A$ and $B$ on a sphere, where the coordinates and the angle of the normal of the tangential plane against the sphere of which are known, what is the centre of that sphere?

I'm not sure how best to represent such an angle, it seems it would vary based on the problem/context; perhaps somebody could advise? Let's say I start off with a spherical coordinate without the radius.

Intuitive:

If I am a person stood on a perfectly spherical planet, I can feel the direction of gravity. If I walk a known distance in a known direction, I can feel the pull of gravity pulling from a different absolute direction (assuming I possess an absolute orientation reference!). How can I calculate the centre of that planet and my position relative to it?

Thanks very much. Please excuse my weak expression of my problem.

Answer 1

Your problem is:

Let $S\subset\mathbb{R}^3$ be a sphere (remark: this will in fact work for an $(n-1)$-sphere in $\mathbb{R}^n$), and assume we are given two distinct points $x,y\in S$ not antipodal to each other and the normal vectors to the sphere at these points, $n_x,n_y$. What is the center (and radius) of the $S$?

The solution, as mentioned in the comments, is to take the straight lines defined by the normal vectors and intersect them. The lines can be parametrized by $$\{x + tn_x\mid t\in\mathbb{R}\},\qquad\{y + sn_y\mid t\in\mathbb{R}\}.$$ Their intersection is the solution of $$x + tn_x = y + sn_y.$$ This is an equation in the $3$ coordinates. Write it down and find $t$ (and $s$), then insert in the formula for the line through $x$ to find the center.

Remark: If the points are antipodal, then the lines coincide. In that case, the center is the point given by $c=\frac{x+y}{2}$.