Latitude, longitude and distances on the earth · Longitude & latitude

The earth is treated as a sphere with centre $O$ and radius $R$ (WAEC usually gives $R = 6400$ km). Every place has a latitude (how far north or south of the equator) and a longitude (how far east or west of the Greenwich meridian).

The equator (gold) and the other parallels of latitude run east–west; the meridians run from pole to pole. The Greenwich meridian (orange) is longitude 0°.

The circles on the globe

A great circle has its centre at the centre of the earth, so its radius is $R$ . The equator is one; so is every meridian (a circle of longitude, through both poles).
A circle of latitude (a parallel) is smaller, except the equator. Its centre is on the earth’s axis, and the higher the latitude, the smaller it is.

Latitude θ: the angle at the centre O between OP and the equator, measured north or south (0° to 90°).

Longitude φ: seen from above the North Pole, the angle east or west of the Greenwich meridian (0° to 180°).

Latitude and longitude are angles at the centre of the earth, so distances along these circles are arcs: $\frac{\text{angle}}{360} \times 2\pi \times \text{radius}$ . You need arcs and cosine.

Differences in latitude and longitude

Same side (both north, or both east): subtract. $70^\circ$ N and $55^\circ$ N differ by $15^\circ$ .
Opposite sides (one north and one south, or one east and one west): add. $25^\circ$ S and $17^\circ$ N differ by $42^\circ$ .

More: differences in latitude and longitude

JAMB 1993 · UME · Q38The angle between latitudes $30^\circ$ S and $13^\circ$ N is

The radius of a circle of latitude

Distances on the earthChange the latitude

3200 kmradius of latitude 60°, R cos 60°1676 kmalong latitude 60°: 30/360 × 2π × 3200

latitude θ = 60°Ndifference in longitude = 30°

The circle of latitude 60° is smaller than the equator. Its radius r is the same length as the side next to the 60° angle in the right-angled triangle below it: r = R cos 60° = 6400 × 0.5 = 3200 km. Going east or west through 30° of longitude covers 30/360 of that circle.

In the side view, $P$ is a point on latitude $\theta$ . Drop a line from $P$ straight down to the equator plane. That makes a right-angled triangle with hypotenuse $OP = R$ and the angle $\theta$ at the centre $O$ , so the side along the equator is $R\cos\theta$ . The radius of the circle of latitude (gold, at the top) is exactly the same length:

r = R\cos\theta

So:

\text{along a meridian: } \frac{\text{difference in latitude}}{360} \times 2\pi R

\begin{gathered} \text{along a parallel of latitude } \theta\text{:} \\ \frac{\text{difference in longitude}}{360} \times 2\pi R\cos\theta \end{gathered}

More: distances along a parallel of latitude

Worked example · WAEC 2013

WAEC 2013 · Paper 2 · Q12

An aeroplane flies due north from a town $T$ on the equator at a speed of $950\text{ km}$ per hour for 4 hours to another town $P$ . It then flies eastwards to town $Q$ on longitude $65^\circ$ E. If the longitude of $T$ is $15^\circ$ E, (i) represent this information in a diagram; (ii) calculate the: (I) latitude of $P$ , correct to the nearest degree; (II) distance between $P$ and $Q$ , correct to 4 significant figures. $\left[\text{Take }\pi = \frac{22}{7}, \text{ radius of the earth} = 6400\text{ km}\right]$

Draw it
North from $T$ on the equator is along the meridian $15^\circ$ E, up to $P$ . Then east from $P$ is along $P$ ‘s circle of latitude, to $Q$ on $65^\circ$ E.
Think first. Which way does each part of the flight go: along a meridian or along a parallel?
(I) The latitude of P
$TP = 950 \times 4 = 3800$ km. So $\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 6400 = 3800$ , which gives $\theta \approx 34.0^\circ$ . $P$ is at latitude $34^\circ$ N.
Think first. How far does the plane fly north? That distance is an arc of a great circle.
(II) The distance PQ
The difference in longitude is $65^\circ - 15^\circ = 50^\circ$ , along latitude $34^\circ$ N:
$\begin{aligned} &\frac{50}{360} \times 2 \times \frac{22}{7} \times 6400\cos 34^\circ \\ &\approx 4632\text{ km} \end{aligned}$
Think first. What is the difference in longitude? Which radius?

Distance along the latitude vs straight through

Two places on the same latitude can be joined three ways, and questions ask for different ones:

the arc along their circle of latitude (the formula above);
the chord, a straight line through the earth: $2r\sin\frac{\Delta}{2}$ , where $r = R\cos\theta$ and $\Delta$ is the difference in longitude;
the angle that chord subtends at the centre of the earth, found from the chord and $R$ .

Worked example · WAEC 2016

WAEC 2016 · Paper 2 · Q7 (b)

$X(60^\circ\text{N}, 12^\circ\text{E})$ and $Y(60^\circ\text{N}, 42^\circ\text{E})$ are points on the earth's surface.

Taking $\pi = 3.142$ and the radius of the earth $= 6400\text{ km}$ , calculate the: (i) length of the chord $XY$ , correct to the nearest 10 km; (ii) angle that the chord $XY$ subtends at the centre of the earth, correct to one decimal place; (iii) distance between $X$ and $Y$ along their common latitude.

The circle of latitude
Radius $= 6400\cos 60^\circ = 3200$ km. Difference in longitude $= 42^\circ - 12^\circ = 30^\circ$ .
(i) The chord XY
$2 \times 3200 \times \sin 15^\circ \approx 1656$ km, which is 1660 km to the nearest 10 km.
Think first. The chord of a circle of radius 3200 that subtends $30^\circ$ at its centre.
(ii) The angle at the centre of the earth
$2 \times 6400 \times \sin\frac{\alpha}{2} = 1656.4$ , so $\sin\frac{\alpha}{2} \approx 0.1294$ , $\frac{\alpha}{2} \approx 7.44^\circ$ and $\alpha \approx 14.9^\circ$ .
Think first. Now the same chord in a circle of radius 6400.
(iii) Along the latitude
$\frac{30}{360} \times 2 \times 3.142 \times 3200 \approx 1675.7$ km.

Your turn

JAMB 1985 · UME · Q46

Longitude & latitude

Two points $X$ and $Y$ , both on latitude $60^\circ$ S, have longitudes $147^\circ$ E and $153^\circ$ W respectively. Find, to the nearest kilometre, the distance between $X$ and $Y$ measured along the parallel of latitude. (Take $2\pi R = 4 \times 10^4$ km, where $R$ is the radius of the earth.)

Worked solution (try it first)

$X$ is east and $Y$ is west, so the longitude difference one way is $147^\circ + 153^\circ = 300^\circ$ .
The short way round is $360^\circ - 300^\circ = 60^\circ$ .
The parallel of latitude $60^\circ$ has radius $R\cos60^\circ$ , so its length is $2\pi R\cos60^\circ = 40\,000 \times 0.5$
$= 20\,000$ km.
The arc for $60^\circ$ is $\frac{60}{360}$ of that: $\frac16 \times 20\,000 \approx 3333$ km, option E.

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