## 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. Points A and B have the same longitudes but different latitudes. Point A (66
^{o }N, 40^{0}E) B, Point B ( 25^{o}N ,40^{0}E)A. Find the angle subtended at the center of the Earth by arc AB if A is (25^{0}N,40^{0}E) and B is (66^{o}s ,40^{0}E)

2

AOB =BOP- AOP

AOB =66^{0}– 25^{o}_{
}=41^{0}

**Exercise 1**

In questions 1 to 4 consider the town and cities indicated an d the answer the questions that follow.

- Which of the following towns and cities lie on the same meridian?

- Tabora (5
^{0}s, 33^{0}E) Dar-es salaam (7^{0}s, 39^{o}E) Mbeya (9^{0 }s,33^{0}E) Chahe Chahe (5^{ 0}s, 40^{0 }E), Tanga ( 5^{0}S, 39^{0}E) Moshi ( 3^{0 }s, 37^{0 }E) , Zanzibar ( 6^{0 }S, 40^{0 }E),Mwanza ( 3^{0 }S, 33^{0 }E) Morogoro (7^{0 }S, 38^{0 }E), Nakuru ( 0^{0 , }36^{0 }E) , Kampala (0^{0},33^{0}E) , Gulu ( 3^{0}N,32^{0}E)

- Which town and cities have latitudes like that of;
- a) Moshi?

Mwanza

- b) Chahe Chahe?

Tanga and Tabora

- Which towns and cities have longitudes like that of
- Mbeya? Tabora , Mwanza and Kampala.
- Dar-Es-Salaam? Tanga
- 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** **A**Mwanza (3^{0 }S, 33^{0}E)

** B **Mbeya (9^{0 }S, 33^{0 }E)

SOB = BOA- AOS

SOB= 9^{0 }– 3^{0 }

SOB = 6^{0}

- c. X Nakuru ( 0
^{0}, 36^{0 }E) - Y Kampala ( 0
^{0 }, 33^{0 }E) - TOX = XOY –TOY
- TOX= 36
^{0 }– 33^{0 } - TOX = 3
^{0}

**CASE**** 2**

Point A and B are on the same latitude but different longitudes.

** Examples**

- Find the angle subtended at the center of the Earth by arc XY if X is Nakuru (0
^{0 , }36^{0 }E) and Y is Kampala ( 0^{0 }, 33^{0}E)

XOY = XOP- YOP

= 36^{0 }– 33^{0}

= 3^{0}

- Two towns A and B are on the equator.The longitude of A is 35
^{0 }E and that of B is 72^{0 }W. Find the angle subtended by the arc AB at the center of the Earth.Solution

AOB = AOP + POB

= 35^{0 }+ 72^{0}

= 107^{0}

- Two towns A and B in Africa are located on the Equator. The longitude of A is 10
^{0 }E and that of B is 42^{0 }E . Find the angle subtended by the arc AB at the center of the Earth.Solution

AOB = SOB – SOA

= 42^{0 }– 10^{0}

= 32^{0}

The angle formed by arc AB is 32^{0}

**Exercise 2**

- 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 = ( 0^{0 }, 30^{0 }W)

Latitude of A = ( 50^{0 }N, 0^{0 })

There after draw a figure similar to that of question 1 to illustrate the position of point H (60^{0}S,45^{0}E).

- P and Q are towns on latitude 0
^{0 }. if the longitude of P is 116^{0 }E and that of Q is 105^{0 }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;

1^{0 }= 60 minutes

1^{0 }= 60 nautical miles

For a great circle, angle at the center if the Earth is 360^{0 }.

Length of a great circle.

1^{0 }= 60 nautical miles

360^{0 }= ?

360^{0 x} 60^{0 }= 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**

- If the latitude of Nakuru is O
^{0 }, find the distance ( length) in nautical miles from this town to the North Pole.

** Solution**.

From O^{0 }to North Pole , the angle is 90^{0}

1^{0 }= 60^{0 }nautical miles.

90^{0} = ?

90^{0} x 60^{0 }= 5400 nautical miles

1 nautical mile = 1.852 km

5400 nautical mile = ? km

Distance = 1. 852 x 5400

= 10000.8km

- calculate the distance of the prime meridian from south to North pole in
- nautical mile
- kilometers.

**Solution**

- 1
^{0 }= 60 nautical miles.

180^{0 }= ? nautical miles

180^{0 }x 60^{0} = 10800 nautical miles.

.^{.}. The distance of prime meridian from south to north pole is 10800Nm

- 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

- Calculate the distance of the equator from east to West in Nautical Miles.

1^{0 }= 60nautical miles.

180^{o }?

180^{0 }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 latitudeθº 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)

.^{.}.θ^{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.**

- Calculate the circumference of a small circle in kilometers along the parallel of latitude 10
^{0 }S.

Soln

C = 2πR cos θ

= 2x 3.14x 6370x cos 10^{0}

=39395.54528km

- Calculate the length of the parallel of the latitude through Bombay If Bombay is located 19
^{0 }N, 73^{0 }E

C= 2πR cos θ

= 2×3.14x6370xcos19^{0}

=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**

- The city of Kampala lies along the equator. Calculate the distance in kilometers from the city of Kampala to the South Pole2. How far is B from A if A is 0
^{0 },0^{0}and B is 0^{0}, 180^{0 }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 70^{0 }N**Solution**. 1

1^{0}= 60 nautical miles.

90^{0 }= ?

60^{0 }x90^{0 }= 5400 nautical miles.

In kilometers .

5400nm x 1.852km/nm= 10000.8km

2.

1^{0 }= 60nm

180^{0 }= ?

= 180^{0} x 60^{0 }= 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

r = Rcos θ

= 6370 x 0.3420

=2178.4 km

**Distance between points along the same meridian.**

A ( 60^{0 }N , 30^{0 }E)

B (20^{0 }N , 30^{0 }E)

K=20^{0}

∝= 60^{0}

^{
=600 – 200
}

1^{0 }= 60 nautical miles

40 ^{0 }= ? Nm

40^{0 }x 60 ^{0 }= 2400 nautical miles

**Examples**

- 1. Find the distance between A ( 30
^{0 }N, 139^{0 }E) and B ( 45^{0 }N , 139^{0}E) in

- Nautical miles.
- 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}

=15^{0}

=15^{0 }x 60 nm/ ^{0} = 900 nautical Mile

**OR**

Distance AB = πRθ/ 180^{0}

= 1666.8 km

1nautical mile = 1.852 km

= 900 nautical miles

- 2. Find the distance in kilometers between A (9
^{0 }S, 33^{0 }E) and B ( 8^{0 }S, 33^{0 }E)

Solution

The points have the same longitudes but their latitudes differ.

The difference in latitude is

=9^{0 }– 8^{0}

= 1^{0 }

1^{0 = }60 nm

The distance is 60nm

- 3. Find the distance in nautical miles on the same meridian with latitude
- 10
^{0 }N, 35^{0 }N - 20
^{0 }N, 42^{0 }S

**Solution**

- 35
^{0 – }10^{0 }= 25^{0 }

1 ^{0 }= 60 nm

25^{0 }=?

60×25= 1500 nm

- 20
^{0 }+ 42^{0 }= 62^{0 }

1^{0 }= 60 nm

62^{0 }=?

60x 62 = 3720 nm

- 4. Find the distance AB In nautical miles between each of the following pairs of places.

- A (18
^{0 }N, 12^{0 }E)

B (65^{0 }N, 12^{0 }E)

(65-18) ^{0}= 47^{0}

1^{0} = 60 nautical miles

47^{0}=?

60 x 47^{0}= 2820 nautical miles

- A( 31
^{0 }S, 76^{0 }W) and B ( 22^{0 }N , 76^{0 }W)

31^{0 }S + 22^{0 }N = 53 ^{0 }

1^{0 }= 60 nm

53^{0 }=?

=3180nm

5. Find the distance in kilometers between;

- Tanga (5
^{0 }S, 39^{0 }E) and Addis Ababa (9^{0 }N, 39^{0}E)

**Solution**:

(9+5) = 14^{0}

1^{0} = 60 nm

14^{0 }= ?

=840 nautical miles

1nm = 1.852 km

840nm =?

Distance = 1555.68 km

- Mbeya (9
^{0 }S, 33^{0}E) and Tabora ( 5^{0}S, 33^{0}E)

**Soln**.

(9-5) ^{0} = 4^{0}

1^{0 }= 60 nm

4^{0 }=?

= 240nm

=240 nmx 1.852km/nm

=444.48 km

- A ship sails northwards to Tanga (5
^{0 }S, 39^{0 }E)at an average speed of 12 nm/ hr. If the ships starting point is Dar Es salaam (7^{0}S, 39^{0}E) at 12:00 noon , when will it reach Tanga?

**Solution**

(7-5) ^{0}= 2^{0}

1^{0}= 60 nm

2^{0 }= ?

Distance = 120 nm

Velocity = 12nm/hr

Time = 10 hours

It will reach tanga at 10:00 pm

- A plane flying at 595 km/hr leaves dar-es-salaam (7
^{0 }s , 39^{0 }E) at 8:00 am. When will it arrive at Addis Ababa (9^{0 }N, 39^{0 }E)?

Solution

9^{0 }N + 7^{0} S = 16^{0 }

1^{0 }= 60 nm

16^{0 }=?

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 0^{0 }, 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.

360^{0} = 2πr

θ = ?

**Example.**

- A ship is streaming in a western direction from Q and P. if the position of P is ( 40
^{0 }S, 178^{0}E) and that of Q ( 40^{0 }S, 172^{0}E). how far does the ship move from Q to P?

**Solution**

Difference in longitude = 178-172= 6^{0}

**Exercise**

- Two points on latitude 50
^{0}N lie on longitudes 35^{0}E and 40^{0}W. what is the distance between them in nautical miles.2. An airplane flies westwards along the parallel of latitude 20^{0 }N from town A on longitude 40^{0 }E to town B on a longitude 10^{0}W. find the distance between the two towns in kilometers.3. An aeroplane flies from Tabora ( 5

^{0}S, 33^{0}E) to Tanga ( 5^{0}S, 39^{0}E) 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 (7

^{0 }S , 38^{0}E) and that of Dar-Es-Salaam is ( 7^{0}S, 39^{0}E). find the distance between them In kilometers.

5. A ship after sailing for 864 nautical miles eastwards find that her longitude was altered by 30^{0}. What parallel of the latitude is the ship sailing?

6. An aeroplane takes off from B (55^{0 }S, 33^{0}E) to C ( 55^{0}S, 39^{0}E) 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 20^{0 }S for a distance of 1440km. find the latitude of the point it reaches.

Solution

1.

35^{0}E + 40^{0}W = 75^{0}

1^{0 }= 60 cos θ

75^{0}= ?

60^{0}x 75^{0}x Cos 45º= 2892.6 nm

2.

10^{0}W +40 ^{0 }E = 50^{0}

1^{0 }= 60 cos 20^{0}

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

6^{0}=?

=6^{0}x 60 xcos 5^{0}

=359nm

Velocity = 332 km /hr

Distance = 664.86 km

Time = 2 hrs

Arrival time = 5:00 pm

**4.**

Longitude difference = 39^{0}– 38^{0}= 1^{0}

= 110.29Km

L = 864 nm =1600.128km

Cos θ = 0.4799952

θ = 61º

(39-33) ^{0}= 6^{0}

1 ^{0} = 60 cos θ nm

6^{0}= ?

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

1 nm = 1.852 km

? = 1440 km

= 777.54 nm

1^{0} = 60 nm

? = 777.54 nm

=12.95 ^{0}

The latitude it reaches will be 12.95 ^{0}