# Distance between points on Earth

So the problem is this: Assuming the surface of the Earth is a sphere of circumference 40, 000 kilometers, estimate the distance between Philadelphia and Paris.

I'm uncertain how to do this problem. I haven't done these kinds of question before, but besides doing some geometry of the earth how am I supposed to solve this? How do you go about using the latitude and longitude coordinates?

• Are you familiar with spherical trigonometry? Commented Oct 9, 2017 at 23:16
• Not really. Ive studied trigonometry, but never learned about spherical trig.
– user482578
Commented Oct 9, 2017 at 23:21

One approach using more familiar methods might be:

• calculate the radius of the earth using the $2\pi r$ expression for the circumference
• longitude $\lambda$ and latitude $\phi$ on a sphere radius $r$ correspond to the coordinates $\left(r \cos(\phi) \cos(\lambda), r \cos(\phi) \sin(\lambda), r \sin(\phi)\right)$ with the origin at the centre of the sphere (remember that W is negative $\lambda$)
• you can find the angle between two points using the dot product, dividing by the product of their magnitudes (i.e. by $r^2$) and then taking the arccosine
• you can find the corresponding great circle arc length distance by multiplying the angle (in radians) by the radius, i.e. $r \theta$

Apparently the real world (non-spherical) answer is just under $6000$km so if you are not close to that then there may be an error

• Well I did the problem best i could and am short 1030km!
– user482578
Commented Oct 10, 2017 at 1:15
• Your problem may be taking the angles in degrees but using a computer function in radians. So for example $r \cos(\phi) \cos(\lambda)$ for Paris should be $6366.2 \times 0.65794 \times 0.99916 \approx 4185$ not $6366.2 \times 0.1613 \times -0.70328 \approx -722.2$. To get from degrees to radians, multiply by $\frac{\pi}{180}$ Commented Oct 10, 2017 at 7:35

Two methods are listed here for obtaining the great-circle distance on a sphere:

1. distance $=R*\cos^{-1}(\cos(\phi_1)\cos(\phi_2)\cos(\lambda_1-\lambda_2)+\sin(\phi_1)\sin(\phi_2))$
2. distance $=2R\,\tan^{-1}(\frac{\sqrt a}{\sqrt {1-a}})$, where $a=\{[1-\cos(\phi_2-\phi_1)]+\cos(\phi_1)\cos(\phi_2)[1-\cos(\lambda_2-\lambda_1)]\}/2$

where $R$ is radius of the earth, $\lambda$ the longitude and $\phi$ the latitude.

The second one is Haversine formula and is faster than the first one if you write both formula into program subroutines.