Spherical Geometry

Let A = ($\theta_{A}, \psi_{A}$) be a point on the earth at latitude $\psi_{A}$ and longitude $\theta_{A}$. Let B = ($\theta_{B}, \psi_{B}$) be another point on the earth. Let R be the radius of the earth. Prove that the distance |AB| between A and B is given by $\cos(\frac{|AB|}{R}) = \sin\psi_{A}\sin\psi_{B} + \cos\psi_{A}\cos\psi_{B}\cos(\theta_{B}-\theta_{A})$

Question

Spherical Geometry

Let A = ($\theta_{A}, \psi_{A}$) be a point on the earth at latitude $\psi_{A}$ and longitude $\theta_{A}$. Let B = ($\theta_{B}, \psi_{B}$) be another point on the earth. Let R be the radius of the earth. Prove that the distance |AB| between A and B is given by $\cos(\frac{|AB|}{R}) = \sin\psi_{A}\sin\psi_{B} + \cos\psi_{A}\cos\psi_{B}\cos(\theta_{B}-\theta_{A})$

astrophysics · Answer

@ganeshie8

jhannybean · Answer

@ikram002p @zzr0ck3r

ikram002p · Answer

this need a long type -,- (if u wanna prove an exist theorem)
and i have to go now, unless u wanna it directly

anonymous · Answer

If you don't have the time it's okay. It seems like that the problem is asking to prove the law of cosines but with a twist to it. Unless it's simpler than that.

anonymous · Answer

So, it looks really similar to:
$\cos(c) = \cos(a)\cos(b) + \sin(a)\sin(b)\cos(C)$
But of course the cosines and sines are flipflopped and the above formula assumed unit sphere. So knowing how to adapt the formula and then see how things got switched up like they did.

ganeshie8 · Answer

$$d^2 = (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2$$

change to spherical coordinates and simplify ?

anonymous · Answer

So represent all the points like $(\rho, \theta, \phi$) and see if I can get something?

ganeshie8 · Answer

Exactly! Also $\rho $ = constant since we are "on" the sphere

ganeshie8 · Answer

we may first work the formula in regular spherical coordinates
changing them to latitude/longitude in the end shouldn't be hard

ganeshie8 · Answer

No wait, that wont work. that formula still gives the direct distance between those two points

anonymous · Answer

Oh. Lol, okay.

anonymous · Answer

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