Form 3 Mathematics – THE EARTH AS A SPHERE

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 (66^o N, 40⁰E) B, Point B ( 25^oN ,40⁰E)A. Find the angle subtended at the center of the Earth by arc AB if A is (25⁰N,40⁰E) and B is (66^os ,40⁰E)

2

AOB =BOP- AOP

AOB =66⁰– 25^o
=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;
3. a) Moshi?

Mwanza

1. b) Chahe Chahe?

Tanga and Tabora

3. Which towns and cities have longitudes like that of
4. Mbeya? Tabora , Mwanza and Kampala.
5. Dar-Es-Salaam? Tanga
6. 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 KampalaSolution
7. 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)
6. Y Kampala ( 0⁰ , 33⁰ E)
7. TOX = XOY –TOY
8. TOX= 36⁰ – 33 ⁰
9. 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 (0^{0 ,} 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 = ?
360^{0 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
3. nautical mile
4. kilometers.

Solution

1. 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

1. 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 = 60nautical miles.

180^o ?

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 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.

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 Pole2. 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

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

1. Nautical miles.
2. 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

1^{0 =} 60 nm

The distance is 60nm

3. 3. Find the distance in nautical miles on the same meridian with latitude
4. 10⁰ N, 35⁰ N
5. 20⁰ N, 42⁰ S

Solution

1. 35^{0 –} 10 ⁰ = 25⁰

1 ⁰ = 60 nm

25⁰ =?

60×25= 1500 nm

1. 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

L = 864 nm =1600.128km

Cos θ = 0.4799952
θ = 61º

(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

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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