## THE EARTH AS A SPHERE

The earth surface is very close to being a sphere. Consider a sphere representing the shape of the earth as below.
NS is the Axis of the earth in which the earth rotates once a day.
O is the center of the earth.
The radius of the Earth is 6370 Km.
GREAT CIRCLES.
Is that which is formed on the surface of the earth by a plane passing through the center of the Earth. Its radius is equal to the radius of the Earth.
Examples;

ARB and ANB.
The equator is also a great circle.
NDBS, NGS and NPRS all are meridians. (Longitude).
EQUATOR: Is the line that circles the Earth midway between the North and the South Pole.
PRIME MERIDIAN (: is a meridian (longitude) which passes through Greenwich, England.
SMALL CIRCLES : is one formed on
the surface of the Earth by a plane that cuts through the Earth but does not pass through the centre of the Earth. Eg. ARB.
LATITUDE: The angular distance of a place north to south of the earth’s equator or of a celestial object north or south of the celestial equator. It is measured on the meridian of the point.
Example 1.

CASE 1
1. 1. Points A and B have the same longitudes but different latitudes. Point A (66o N, 400E) B, Point B ( 25oN ,400E)A. Find the angle subtended at the center of the Earth by arc AB if A is (250N,400E) and B is (66os ,400E)
2
AOB =BOP- AOP
AOB =660– 25o
=410
Exercise 1
In questions 1 to 4 consider the town and cities indicated an d the answer the questions that follow.
1. Which of the following towns and cities lie on the same meridian?
•Tabora (5 0s, 33 0E) Dar-es salaam (7 0s, 39 oE) Mbeya (90 s,330E) Chahe Chahe (5 0s, 40 0 E), Tanga ( 50S, 390E) Moshi ( 30 s, 370 E) , Zanzibar ( 60 S, 40 0 E),Mwanza ( 30 S, 330 E) Morogoro (70 S, 380 E), Nakuru ( 0 0 , 360 E) , Kampala (00,330E) , Gulu ( 30N,320E)
2. Which town and cities have latitudes like that of;
a. a) Moshi?
Mwanza
b. b) Chahe Chahe?
Tanga and Tabora
3. Which towns and cities have longitudes like that of
a. Mbeya? Tabora , Mwanza and Kampala.
b. Dar-Es-Salaam? Tanga
4. Find the angle subtended at the center of the Earth by arc AB if A is Mwanza and B is Mbeya.
5. Find the angle subtended at the centre of the Earth by arc XY if X is Nakuru and Y is Kampala

Solution
1. A Mwanza (30 S, 330 E)
B Mbeya (90 S, 330 E)
SOB = BOA- AOS
SOB= 90 – 30
SOB = 60
5. c. X Nakuru ( 00 , 360 E)
d. Y Kampala ( 00 , 330 E)
e. TOX = XOY –TOY
f. TOX= 360 – 33 0
g. TOX = 30
CASE 2
Point A and B are on the same latitude but different longitudes.
Examples
1. Find the angle subtended at the center of the Earth by arc XY if X is Nakuru (00 , 360 E) and Y is Kampala ( 00 , 330 E)
XOY = XOP- YOP
= 360 – 330
= 30
2. Two towns A and B are on the equator.The longitude of A is 350 E and that of B is 720 W. Find the angle subtended by the arc AB at the center of the Earth.

Solution
AOB = AOP + POB
= 350 + 720
= 1070
3. Two towns A and B in Africa are located on the Equator. The longitude of A is 100 E and that of B is 420 E . Find the angle subtended by the arc AB at the center of the Earth.

Solution
AOB = SOB – SOA
= 420 – 100
= 320

The angle formed by arc AB is 320
Exercise 2
1. In the figure below, if the center through N,G,S is the prime meridian , the center of the Earth and the Equator passes through B and G , the longitude and latitude of A.
Longitude of A = ( 00 , 300 W)
Latitude of A = ( 500 N, 00 )
There after draw a figure similar to that of question 1 to illustrate the position of point H (600S,450E).
2. P and Q are towns on latitude 00 . if the longitude of P is 1160 E and that of Q is 1050 W, find the angle subtended by the arc connecting the two places at the center of the Earth . Draw a figure to illustrate their positions.
LENGTH OF A GREAT CIRCLE.
Distance on the surface of the Earth are usually expressed in nautical miles or in kilometers. A nautical mile is the length of an arc of a great circle that subtends an angle of 1 minute at the center of the Earth.
The length of arc AB is 1 nautical mile;
10 = 60 minutes
10 = 60 nautical miles
For a great circle, angle at the center if the Earth is 3600 .
Length of a great circle.
10 = 60 nautical miles
3600 = ?
3600 x 600 = 21600 nautical miles.
Therefore, equator and all meridians are great circles , distance (length) is equal to 21600 nautical miles.

In kilometers.
1 nautical mile = 1.852 Km
21600 nautical miles = ?
21600x 1.852 = 40003.2 km
OR
We can use the following formula to find the distance.( length)
C= 2π r
Where “r” is the radius of the Earth.
r= 6370
π=3.14
C= 2×3.14×6370
C= 40003.6 km
Example
1. If the latitude of Nakuru is O0 , find the distance ( length) in nautical miles from this town to the North Pole.
Solution.
From O0 to North Pole , the angle is 900
10 = 600 nautical miles.
900 = ?
900 x 600 = 5400 nautical miles
1 nautical mile = 1.852 km
5400 nautical mile = ? km
Distance = 1. 852 x 5400
= 10000.8km
2. calculate the distance of the prime meridian from south to North pole in
a. nautical mile
b. kilometers.
Solution
a. 10 = 60 nautical miles.
1800 = ? nautical miles

1800 x 600 = 10800 nautical miles.
... The distance of prime meridian from south to north pole is 10800Nm

b. 1
nautical mile = 1.852 km
10800 Nautical miles =?km
10800x 1.852= 20001.6 km
... The distance of prime meridian from south to north pole is 20001.6 km
3. Calculate the distance of the equator from east to West in Nautical Miles.
10 = 60 nautical miles.
180o ?
1800 x 60 0= 10800 nautical miles.
... The distance of the equator from East to West in Nm is given by 10800Nm
Length of small circle.

Let P be any point on the surface of the earth through this point a small circle is drawn with parallel of latitude θº as shown above. The radius of the earth as R and the radius of the parallel latitude (r) are both perpendicular to the polar axis.
Note: SP is parallel to OQ (Both are perpendicular to NS)
..º = OPS
Then we have

From trigonometrical ratios
r= R cos θ where R is the radius of the earth and r is the radius of the small circle of latitude θ.
... Distance of parallel of latitude θ = 2πr
= 2πRcos θ
Example 1.
1. Calculate the circumference of a small circle in kilometers along the parallel of latitude 100 S.
Soln
C = 2πR cos θ
= 2x 3.14x 6370x cos 100
=39395.54528km
2. Calculate the length of the parallel of the latitude through Bombay If Bombay is located 190 N, 730 E
C= 2πR cos θ
= 2×3.14x6370xcos190
=37823.40km
In nautical miles.
3782km/1.852km/miles = 20423 nautical miles.
Exercise
In the questions below , take the radius of the Earth , R= 6370km and π = 3.14
1. The city of Kampala lies along the equator. Calculate the distance in kilometers from the city of Kampala to the South Pole

2. How far is B from A if A is 00 ,00 and B is 00, 1800 E.

3. What is the latitude of a point P north of the Equator if the length of the parallel of the latitude through p is 28287 kilometers.( give your answer to the nearest degree.
4. What is the radius of a small circle parallel to the equator along latitude 700 N
Solution
1. 10= 60 nautical miles.
900 = ?
600 x900 = 5400 nautical miles.
In kilometers .
5400nm x 1.852km/nm= 10000.8km
2.
10 = 60nm
1800 = ?
= 1800 x 600 = 10800 nm
1nm = 1.852 km
10800nm =?
10800x 1.852= 20001.6 km
3.
C= 2πR cos θ
28287= 2 x 3.14x 6370 x Cos θ
Cos θ = 0.7071
θ= 45º
4.
r = Rcos θ
= 6370 x 0.3420
=2178.4 km
Distance between points along the same meridian.
A ( 600 N , 300 E)
B (200 N , 300 E)
K=200
= 600

= 600 – 200
10 = 60 nautical miles
40 0 = ? Nm
400 x 60 0 = 2400 nautical miles

Examples
1. 1. Find the distance between A ( 300 N, 1390 E) and B ( 450 N , 1390 E) in
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Nautical miles.
1. Kilometers.
Solution
Since A and B have the same longitudes , they are on the same meridian. The difference between their latitudes is
= (45-30)0
=150
=150 x 60 nm/ 0 = 900 nautical Mile
OR
Distance AB = πRθ/ 1800
= 1666.8 km
1nautical mile = 1.852 km
= 900 nautical miles
2. 2. Find the distance in kilometers between A (90 S, 330 E) and B ( 80 S, 330 E)

Solution
The points have the same longitudes but their latitudes differ.
The difference in latitude is
=90 – 80
= 10
10 = 60 nm
The distance is 60nm
3. 3. Find the distance in nautical miles on the same meridian with latitude
a. 100 N, 350 N
b. 200 N, 420 S
Solution
a. 350 – 10 0 = 250
1 0 = 60 nm
250 =?
60×25= 1500 nm
b. 200 + 420 = 620
10 = 60 nm
620 =?
60x 62 = 3720 nm
4. 4. Find the distance AB In nautical miles between each of the following pairs of places.
1. A (180 N, 120 E)
B (650 N, 120 E)
(65-18) 0= 470
10 = 60 nautical miles
470=?
60 x 470= 2820 nautical miles
1. A( 310 S, 760 W) and B ( 220 N , 760 W)
310 S + 220 N = 53 0
10 = 60 nm
530 =?
=3180nm

5. Find the distance in kilometers between;
1. Tanga (50 S, 390 E) and Addis Ababa (90 N, 39 0E)
Solution:
(9+5) = 140
10 = 60 nm
140 = ?
=840 nautical miles
1nm = 1.852 km
840nm =?
Distance = 1555.68 km
1. Mbeya (90 S, 330E) and Tabora ( 50S, 330E)
Soln.
(9-5) 0 = 40
10 = 60 nm
40 =?
= 240nm
=240 nmx 1.852km/nm
=444.48 km

6. A ship sails northwards to Tanga (50 S, 390 E)at an average speed of 12 nm/ hr. If the ships starting point is Dar Es salaam (70S, 390E) at 12:00 noon , when will it reach Tanga?

Solution
(7-5) 0= 20
10= 60 nm
20 = ?
Distance = 120 nm
Velocity = 12nm/hr
Time = 10 hours
It will reach tanga at 10:00 pm

7. A plane flying at 595 km/hr leaves dar-es-salaam (70 s , 390 E) at 8:00 am. When will it arrive at Addis Ababa (90 N, 390 E)?

Solution
90 N + 70 S = 160
10 = 60 nm
160 =?
Distance = 960 nm=1777.92km

Time = 2.988 hours
It will arrive at 11: 00 am
DISTANCE BETWEEN POINTS ALONG THE PARALLEL OF LATITUDES.
Consider the figure below, points A and B are two points having the same latitude 00 , since both lie on the parallel of latitude but they are different in their longitudes i.e. That point A is on a different longitude from that of point B. the difference between their longitudes is θ.
3600 = 2πr
θ = ?
Example.
1. A ship is streaming in a western direction from Q and P. if the position of P is ( 400 S, 1780E) and that of Q ( 400 S, 1720E). how far does the ship move from Q to P?

Solution
Difference in longitude = 178-172= 60

Exercise
1. Two points on latitude 50 0 N lie on longitudes 350E and 40 0W. what is the distance between them in nautical miles.

2. An airplane flies westwards along the parallel of latitude 200 N from town A on longitude 400 E to town B on a longitude 100W. find the distance between the two towns in kilometers.

3. An aeroplane flies from Tabora ( 50S, 330E) to Tanga ( 50S, 390E) at 332 kilometers per hour along a parallel of latitude. If it leaves Tabora at 3 pm, find the arrival time at Tanga airport?

4. The location of Morogoro is (70 S , 380E) and that of Dar-Es-Salaam is ( 70S, 390E). find the distance between them In kilometers.
5. A ship after sailing for 864 nautical miles eastwards find that her longitude was altered by 300. What parallel of the latitude is the ship sailing?
6. An aeroplane takes off from B (550 S, 330E) to C ( 550 S, 390E) at a speed of 332 km/ hr . if it leaves B at 3:00 pm , at what time will it arrive at C airport?
7. A ship sails due North from latitude 200 S for a distance of 1440km. find the latitude of the point it reaches.

Solution

1.
350E + 400W = 750
10 = 60 cos θ
750= ?
600x 750x Cos 45º= 2892.6 nm

2.
100W +40 0 E = 500
10 = 60 cos 200
50 0=?
= 2819.1 nm
1nm =1.852 km
2819.1nm= ?
Distance = 5221km

3.
(39-33)0 = 6 0
1 0 = 60 0 cos θ
60=?
=60x 60 xcos 50
=359nm
Velocity = 332 km /hr
Distance = 664.86 km
Time = 2 hrs
Arrival time = 5:00 pm
4.
Longitude difference = 390– 380= 10

= 110.29Km
5.

L = 864 nm =1600.128km

Cos θ = 0.4799952
θ = 61º
6.
(39-33) 0= 60
1 0 = 60 cos θ nm
60= ?
Where θ = 55 0
= 358.632nm
But 1 nm = 1.852 km
359nm =?
Distance =665 km
Time = 2 hours
3:00 pm + 2 hours = 5:00 pm
7.
1 nm = 1.852 km
? = 1440 km
= 777.54 nm
10 = 60 nm
? = 777.54 nm
=12.95 0
The latitude it reaches will be 12.95 0

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