Distances Along Great Circles & Small Circles - SS3 Mathematics Lesson Note

DISTANCES ALONG GREAT CIRCLES

Recall, great circles are the lines of longitude and the equator whilst small circles are the other parallels of latitude.

The shortest distance $D$ between two points along a great circle on the earth surface is given by the length of the arc subtending an angle at the center of the earth. Thus, $D=\frac{\theta }{360}\times 2\pi R$, where $R$ is the radius of the earth and $\theta$ is the angular difference between the two points.

Example 3 Find the distance between this pair of points on the earth surface: $A(30{}^{\circ }N,40{}^{\circ }W)$ and $B(30{}^{\circ }S,40{}^{\circ }W)$. Take $R=6,400km$

Solution

$$Angulardifference=30{}^{\circ }+30{}^{\circ }=60{}^{\circ }$$

$$Distancebetweenthepoints=\frac{\theta }{360}\times 2\pi R$$

$$Distancebetweenthepoints=\frac{60}{360}\times \frac{2}{1}\times \frac{22}{7}\times \frac{6400}{1}=6,704.76km$$

DISTANCES ALONG SMALL CIRCLES

Recall as one moves from the equator along the parallels of latitude to the north and south poles, their radii decreases. The equator has the largest radius whilst the north and south poles have effective radii of zero.

The distance along small circles, $d=\frac{\alpha }{360}\times 2\pi R\cos \theta$, where $Rist$he radius of the earth, $\alpha$ is the angular difference between the points and $\theta$ is the common latitude of both points.

Example 4 Find the distance between this pair of points on the earth surface: $P(28{}^{\circ }S,20{}^{\circ }E)$ and $Q(28{}^{\circ }S,60{}^{\circ }E)$. Take $R=6,400km$ and $\pi =3.142$

Solution

Given $P(28{}^{\circ }S,20{}^{\circ }E)$ and $Q(28{}^{\circ }S,60{}^{\circ }E)$:

$$d=\frac{\alpha }{360}\times 2\pi R\cos \theta$$

$$\alpha =60{}^{\circ }-20{}^{\circ }=40{}^{\circ }$$

$$R=6,400km$$

$$\pi =3.142$$

$$\theta =28{}^{\circ }$$

$$d=\frac{40}{360}\times 2\times 3.142\times 6400\times cos28{}^{\circ }=3946km$$