Distance between two cities on Earth

Question

Barcelona (Spain) has the coordinates (approx): $\theta = 2^\circ$, $\phi = 41^\circ$, and New York has the coordinates: $\theta = −74^\circ$, $\phi = 41^\circ$. Notice that both cities lie on the same latitude, which makes the calculations easier. For your calculations, use Earth’s radius $R = 6378$ km, and give all your result in kilometers.

1. On most maps, the naive ‘straight line’ between Barcelona and New York lies on the line of latitude $\phi = 41^\circ$. What is the distance between the two cities along this line?

2. What is the actual shortest distance between those two cities?

   Hint: Notice that it is determined by the angle Barcelona–Earth’s center–New York. Use the fact that the angle $\psi$ between two unit vectors $nA$ and $nB$ satisfies $\cos \psi = nA \cdot nB$.

   Hint: Observe that both curves are parts of circles. By using this, you can avoid integration altogether.

For the first part, I'm trying to take the Euclidean distance. \begin{eqnarray} x &=& R\cos\theta\cos\phi \\ y &=& R\cos\theta\sin\phi \\ z &=& R\sin\theta, \end{eqnarray} Using this, I get coordinates for NY and Barcelona.

New York: $(x, y, z) = (2620.58, 421.01, 5799)$

Barcelona: $(x, y, z) = (-1081.34, -173.72, 6283.26)$

If I try to find the distance by the equation

\begin{eqnarray} \sqrt{x^2 + y^2 + z^2} \end{eqnarray}

I get 3780 km.

Am I doing this right? I feel a bit lost because this "naive" distance differs so greatly from the real one (ie 6157 km)

The Euler-Lagrange equation is this: \begin{eqnarray}ϑ˙^2 cos ϕ sin ϕ/ \sqrt{ϕ˙^2 + ϑ˙^2 cos^2 ϕ} + d/dt ϕ˙^2 \sqrt{ϕ˙^2 + ϑ˙^2 cos^2 ϕ} = 0 \end{eqnarray}

Answer 1

Be careful. All angles are measured in radians. In accord with the question $\phi$ means the latitude ($=\frac{\pi}{2}-\text{polar angle}$) and $\theta$ means the longitude (azimuth angle).

If one considers the Earth being a ball, the shortest path is: $$ d=R\arccos(\sin\phi_1\sin\phi_2+\cos\phi_1\cos\phi_2\cos(\theta_2-\theta_1)), $$ the argument of $\arccos$ being the scalar product of unit vectors directed from the Earth center to the points on the surface.

The "naive path" for $\phi_1=\phi_2\equiv\phi$ has length: $$ d^*=R(\theta_2-\theta_1)\cos\phi. $$

In the last expression $0\le\theta_2-\theta_1\le\pi$ is assumed. Generally $\arccos\cos(\theta_2-\theta_1)$ can be used instead.

Answer 2

Use that chord length and the Earth's radius to determine the arc length around the curved path.

Answer 3

Using the distance equation you will get the "straight line distance" if you bored a hole through the earth.

The radius of the $41^\circ$ parallel is $R\cos 41^\circ$

And the distance a person would travel if they followed this arc would be this radius times the longitude traveled in radians: $(R\cos 41)(\frac {76 \pi}{180})$

The great circle path...

$a\cdot b = \|a\|\|b\| \cos \psi\\ (R\cos 74^\circ\cos 41^\circ, R\cos 74^\circ\sin 41^\circ, R\sin 41^\circ)\cdot(R\cos -2^\circ\cos 41^\circ, R\sin -2^\circ\cos 41^\circ, R\sin 41^\circ) = R^2(\cos 76^\circ\cos^2 41^\circ + \sin^2 41^\circ)\\ \psi = \arccos (\cos 76\sin^2 41 + \cos^2 41)\\ D = R\psi$

Answer 4

Cut the Earth along its center $O$ and the two cities $N$ and $B$ to get a circle with the radius $R$ and the angle $\measuredangle BON=76^\circ$.

a) The length of the chord $BN$ is: $$BN=\sqrt{NO^2+BO^2-2NO\cdot BO\cdot \cos{\measuredangle BON}}=\\ \sqrt{2\cdot 6378^2-2\cdot 6378^2\cdot \cos{76^\circ}}\approx 7853.38 \ \text{km}.$$ Note: The Cosine Theorem is applied in the triangle $\Delta BON$.

b) The length of the arc $BN$ is: $$l_{BN}=2\pi R\cdot \frac{\alpha^\circ}{360^\circ}=2\pi\cdot 6378\cdot \frac{76^\circ}{360^\circ}\approx 8460.1 \ \text{km}.$$