This unit carry information of Acceleration Due to the Gravity (g), Satellite and Planetary Motion and Gravitational Field, Potential Energy, Kinetic Energy and Total energy of the satellite. in each section, there is an example so as you could be able to manipulate those equations that are associated with this unit. Also, there is problem practice so as to straighten the understanding of this module.

1. GRAVITATION
Peter Huruma Mammba
Department of General Studies
DODOMA POLYTECHNIC OF ENERGY AND EARTH RESOURCES MANAGEMENT (MADINI INSTITUTE) –DODOMA
peter.huruma2011@gmail.com

2. GRAVITATION
I. Acceleration Due to the Gravity
II. Satellite and Planetary Motion
III.Gravitational Field, Energy and Potential

4. Acceleration Due to the Gravity (g)
• Is the acceleration that results in an object due to
earth's gravity.
• If air resistance is neglected, all bodies regardless of
size and mass fall with the same acceleration under
gravity alone. (i.e. 9.8 m/s2)

5. The following points may be noted about the g
The value of g varies slightly from place to place on the
earth's surface. However it is normally sufficiently accurate
to use a value of 9.8
The acceleration due to the gravity always acts downward
towards the center of the earth.

6. The following points may be noted about the g
Like all accelerations, the acceleration due to the gravity is
vector.

7. Newton’s Law of
Universal Gravitation
States that every body in the universe attracts every
other body with a force which directly proportional to
the product of their masses and inversely proportional
to the square of distance between their centers.

8. Gravitation
• Every object with mass attracts every other object with mass.
• Newton realized that the force of attraction between two
massive objects:
• Increases as the mass of the objects increases.
• Decreases as the distance between the objects increases.

9. Law of Universal Gravitation
FG = G
• G = Gravitational Constant= 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2
• M1 and M2 = the mass of two bodies
• r = the distance between them
M1M2
r2

10. Example 1
Estimate the gravitational force between a 55 Kg woman and a
75 Kg man when they are 1.6 m apart.
Solution
𝐹 = 𝐺
𝑚1 𝑚2
𝑟2 = 6.67x10−11
𝑥
55 𝑥 75
1.6 2 = 10−7
𝑁

11. Example 2
• Two 8 Kg spherical lead balls are placed with their
centers 50 cm apart. What is the magnitude of the
gravitational force each exerts on the other.

12. Solution
• Here 𝑚1= 𝑚2 = 8 Kg; G = 6.67𝑥10−11
𝑁𝑚2
𝐾𝑔−2
; r = 0.5 m
𝐹 = 𝐺
𝑚1 𝑚2
𝑟2
𝐹 = 6.67𝑥10−11
𝑥
8𝑥8
(0.5)2
∴F= 1.71𝑥10−8

13. Example 3
• Assuming the orbit of the earth about the sun to be
circular (it is actually slightly elliptical) with radius
1.5𝑥1011
𝑚, find the mass of the sun. The earth
revolves around the sun in 3.15𝑥107
𝑠𝑒𝑐𝑜𝑛𝑑𝑠.

14. SOLUTION
Centripetal force = Gravitational force
Or
𝑚 𝑒 𝑣2
𝑅
=
𝑚 𝑒 𝑚 𝑠
𝑅2
𝑚 𝑠=
𝑅𝑣2
𝐺
v= speed of earth in its orbit around the sun
G = 6.67𝑥10−11
𝑁𝑚2
𝐾𝑔−2

15. SOLUTION…
2𝜋𝑅
𝑇
=
2𝜋𝑥 (1.5𝑥1011
)
3.15𝑥107
= 3𝑥104
𝑚𝑠−1
𝑚 𝑠=
(3𝑥104)2 𝑥1.5𝑥1011
6.67𝑥10−11
∴ 𝒎 𝒔= 𝟐𝒙𝟏𝟎 𝟑𝟎
𝑲𝒈

16. Expression for the Acceleration Due to Gravity
• According to the law of the gravitation, the force of attraction
acting on the body due to earth is given by;
𝐹 = 𝐺
𝑚1 𝑚2
𝑟2 …………………….. i
The attractive force which the earth exert on the object is
simply weight of the object i.e.
W= mg……………………..……..ii

17. From equation (i) and (ii),
mg =
𝐺𝑀 𝐸 𝑚
𝑅 𝐸
2
g =
𝑮𝑴 𝑬
𝑹 𝑬
𝟐
∴The equation for the expression for acceleration due to the
gravity on the surface of the earth shows that the value of g does
not depend on the mass m of the body.

18. Differences between G and g
G g
G is the universal gravitational Constant g is the acceleration due to the gravity
G = 6.67 𝑥 10−11 𝑁𝑚2 𝐾𝑔−2 Approximately g = 9.8 𝑚𝑠2, value of g on
the earth varies from one place to another
Constant through the universe Change every place on the earth. Example
on the mon the value of g = 1
6 𝑡ℎ of that of
the earth surface

19. • Using the law of universal gravitation and the
measured value of the acceleration due to gravity,
we can find,
Mass of Earth
Density of Earth

20. Mass of Earth
• From equation, g =
𝑮𝑴 𝑬
𝑹 𝑬
𝟐 then; 𝑴 𝑬 =
𝑮g
𝑹 𝑬
𝟐
where ; G = 6.67𝑥10−11
𝑁𝑚2
𝐾𝑔−2
, g = 9.8 𝑚𝑠−2
,
𝑹 𝑬=6.37𝑥106
𝑚
∴Mass of earth, 𝑴 𝑬=
9.8𝑥 6.37𝑥106 2
6.67𝑥10−11 = 6𝑥1024
𝐾𝑔

21. Density of Earth
• Let 𝜌 be the average density of the earth. Earth is the sphere of radius 𝑅 𝐸.
Mass of Earth,𝑀 𝐸 = Volume (𝑉𝐸) x Density (𝜌)
4
3
𝜋𝑅 𝐸
3
𝜌
𝑀 𝐸 =
4
3
𝜋𝑅 𝐸
3
𝜌 ………i
But 𝑴 𝑬 =
𝑮𝑹 𝑬
𝟐
g
……….ii

22. • Substitute the equation (i) into equation (ii);
𝑮𝑹 𝑬
𝟐
g
=
4
3
𝜋𝑅 𝐸
3
𝜌
𝜌 =
𝟑g
𝟒𝜋𝑮𝑹 𝑬
G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2, g = 9.8 𝑚𝑠−2, 𝑹 𝑬=6.37𝑥106 𝑚, g = 9.8 𝑚𝑠−2
𝜌 =
3 𝑥 9.8
𝟒𝜋 𝑥 6.37𝑥106 𝑥6.67𝑥10−11
∴ 𝝆 = 𝟓. 𝟓 𝒙𝟏𝟎 𝟑
𝑲𝒈 𝒎−𝟑

23. Example 4
What will be the acceleration due to the gravity on the surface
of the moon if its radius is 1/5th the radius of the earth and its
mass 1/80th of the mass of the earth?
SOLUTION
The acceleration due to the gravity on the surface of the earth is given by;
𝑔 𝐸 =
𝐺𝑀 𝐸
𝑅 𝐸
2

24. SOLUTION
The acceleration due to the gravity on the surface of the moon is given by;
𝑔 𝑚 =
𝐺𝑀 𝑚
𝑅 𝑚
2
𝑔 𝑚
𝑔 𝐸
=
𝑀 𝑚
𝑀 𝐸
𝑥
𝑅 𝐸
𝑅 𝑚
2
, Now
𝑀 𝑚
𝑀 𝐸
=
1
80
and
𝑅 𝐸
𝑅 𝑚
= 4
𝑔 𝑚
𝑔 𝐸
=
1
80
𝑥
4
1
2
=
1
5
∴ 𝑔 𝑚 =
𝑔 𝐸
5

25. Problems for Practice 1
• Two 20 Kg spree are placed with their centers 50 cm apart. What is the
magnitude of gravitational force exerts on the other? (1.07 𝑥 107
𝑁)
• A spree of mass 40 Kg is attracted by another of mass 15 Kg when their
centers are 0.2 m apart, with a force of 9.8 𝑥 10−7 𝑁. Calculate the constant
of gravitational. (6.53 𝑥 10−11
𝑁𝑚2
𝐾𝑔−2
)

26. Problems for Practice …
• Assuming the mean density of the earth is 5500 𝐾𝑔𝑚−3, that G is
6.67𝑥 10−11 𝑁𝑚2 𝐾𝑔−2 and that radius of the earth is 6400 Km, Find
the value for the acceleration of free fall at the earth’s surface.
(9.9 𝑚𝑠−2)
• The Acceleration due to the gravity at the moon’s surface is 1.67 𝑚𝑠−2
.
If the radius of the moon is 1.74 𝑥 106 𝑚 and 6.67𝑥 10−11 𝑁𝑚2 𝐾𝑔−2,
calculate the mass of the moon. (7.58 𝑥 1022 𝐾𝑔)

27. Satellite and
Planetary Motion

28. Natural and Artificial Satellite
• Natural Satellite is a heavenly body revolves around a planet
in a close and stable orbit.
• For examples, moon is the natural satellite of the earth,
• It goes round the earth in about 27.3 days in a nearly circular
orbit of radius 3.84 x 10^5 Km.

29. Earth satellite
• Artificial Satellite; is a man-made satellite that orbits around
the earth or some other heavenly body.
• For example, the communication satellite are used routinely
to transmit information around the globe.

30. Projection of a Satellite
• Why it is necessary to have at least a two stages rocket to
launch a satellite?
• A rocket with at least two stages is required to launch a
satellite because;
The first stage is used to carry the satellite up to the desired
height.
In the second stage, rocket is turned horizontally (through 90
degrees) and the satellite is fired with the proper horizontal
velocity to perform circular motion around the earth.

31. Orbital velocity of the satellite
• Is the velocity required to put a satellite into a given
orbit.
• It is also known as Critical velocity (Vc) of
the satellite

32. Critical Velocity of the Satellite
• Let
M = Mass pf the earth
m = Mass of the satellite ( )
h = Height of the satellite above the earth
r = R + h (the distance of the satellite form the center
of the earth)
V - Velocity of projection in a horizontal direction
𝑉𝐶 - Critical velocity
𝑉𝐸- Escape velocity

33. Critical Velocity of the Satellite
• The centripetal force necessary for a circular motion of the satellite
around the Earth is proved by the gravitational force of attraction
between the Earth and the satellite.
Centripetal force = Gravitational force
𝑚𝑉𝑐
2
𝑟
=
𝐺𝑀𝑚
𝑟2
𝑉𝑐
2
=
𝐺𝑀
𝑟
𝑉𝑐 =
𝐺𝑀
𝑟

34. Critical Velocity of the Satellite
• Factors on which Critical velocity of a satellite depends on;
1. Mass of the planet
2. Radius of the planet
3. Height of the planet
NB. Critical velocity is not dependent on the mass of the satellite

35. Critical Velocity of the Satellite
• But we know that
𝑔ℎ =
𝐺M
(R + h)2 =
𝐺M
𝑟2
GM = 𝑔ℎ(R + h)2
…………………………………….(i)
Substitute the equation (i) to the equation of the critical velocity. We get
𝑉𝑐 =
𝑔ℎ(R + h)2
R + h
𝑉𝑐 = 𝑔ℎ(R + h)

36. Special Case of Critical Velocity of the Satellite
• When the satellite orbits very
close to the surface of the earth,
h ≅ 0. Therefore, equation of the
critical velocity will be;
• Now
g = 9.8 𝑚𝑠−2
, 𝑹 =6.37𝑥106
𝑚
𝑉𝑐 = 9.8 𝑥 6.4 𝑥106
≈ 8 𝑥 103
𝑚𝑠−1
∴≈ 8 𝐾𝑚𝑠−1
Thus the orbital speed of the satellite
close to the earth’s surface is about
8 𝐾𝑚𝑠−1
𝑉𝑐 = gR

37. Time Period of the Satellite
•Is the time taken by the satellite to complete one
revolution around the earth
•It is denoted by T
T =
𝑪𝒊𝒓𝒄𝒖𝒎𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒐𝒓𝒃𝒊𝒕
𝑶𝒓𝒃𝒊𝒕𝒂𝒍 𝑺𝒑𝒆𝒆𝒅

38. Time Period of the Satellite
𝑇 =
2𝜋(𝑅+ℎ)
𝑉𝑐
…………………… (i)
But; 𝑉𝑐 =
𝐺𝑀
R+h
……………….. (ii)
Putting the value of 𝑉𝑐 given by equation (ii) into equation (i), we have,
𝑇 = 2𝜋 𝑥
(𝑅 + ℎ)3
𝐺𝑀

39. Special Case of Time Period of the Satellite
• When the satellite orbits very close to the
surface of the earth, h ≅ 0. therefore,
equation of the time period will be;
𝑇 = 2𝜋 𝑥
𝑅3
𝐺𝑀
………………… (i)
But; g𝑅2
= 𝐺𝑀 ……………….. (ii)
Substitute the value of GM given by
equation (ii) into equation (i), we have,
• Now
• g = 9.8 𝑚𝑠−2, 𝑹 =6.37𝑥106 𝑚
• 𝑇 = 2𝜋 𝑥
6.37𝑥106
9.8
= 5075 s = 84 Minutes
∴ Thus the orbital speed of the satellite
revolving very near to the earth’s surface is
about 8 𝐾𝑚𝑠−1
and its period of revolution is
nearly 84 minutes.
𝑻 = 𝟐𝝅 𝒙
𝑹
g

40. Example 5
• Calculate the velocity required
for a satellite moving in a
circular orbit 200 Km above the
earth’s surface. Given that
radius of the earth is 6380 Km
and mass of the earth =
5.98 𝑥 1024 𝐾𝑔
SOLUTION
G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 , 𝑹 𝑬 = 6.37𝑥106 𝑚 , h
=200 𝑥 103 𝑚, M = 5.98 𝑥 1024 𝐾𝑔
𝑉𝑐 =
𝐺𝑀
𝑅+ℎ
𝑉𝑐 =
6.67𝑥10−11 𝑥 5.98 𝑥 1024
6.37𝑥106+200 𝑥 103
∴ = 7.79 𝑥 103 𝑚𝑠−1

41. Example 6
•The period of moon around
the earth is 27.3 days and
the radius of its orbit is
3.9 𝑥 105
𝐾𝑚. If G =
6.67𝑥10−11
𝑁𝑚2
𝐾𝑔−2
,
find the mass of the earth
SOLUTION
T = 27.3 days = 27.3 𝑥 24 𝑥 60 𝑥 60 𝑠,
R+h = 3.9 𝑥 108
𝑚,
𝑇 = 2𝜋 𝑥
(𝑅 + ℎ)3
𝐺𝑀
𝑀 =
4𝜋2
𝑅 + ℎ 3
𝐺𝑇2
𝑀 =
4𝜋2
3.9 𝑥 10 3
6.67𝑥10−11 𝑥 27.3 𝑥 24 𝑥 60 𝑥 60 2
∴ 𝑀 = 6.31 𝑥 1024
𝐾𝑔

42. Problems for Practice 2
• A satellite is revolving in a circular orbit at a distance of 2620 Km from the
surface of the earth. Calculate the orbital velocity and the period of revolution
of the satellite. Radius of the earth = 63802 Km, mass of the earth =
6 𝑥 1024
𝐾𝑔 and G = 6.67𝑥10−11
𝑁𝑚2
𝐾𝑔−2
. (6.67 𝑲𝒎𝒔−𝟏
; 2.35 Hours)
• An earth’s satellite makes a circle around the earth in 90 minutes. Calculate the
height of the satellite above the surface of the earth. Given that radius of the
earth = 6400 Km and g = 9.8 𝑚𝑠−2. (268 Km)

43. Uses of Artificial Satellite
• They are used to learn about the atmosphere near the earth.
• They are used to forecast weather.
• They are used to study radiation from the sun and the outer space.
• They are used to receive and transmit various radio and television signals.
• They are used to know the exact shape and dimensions of the earth.
• Space fights are possible due to artificial satellite.

44. Escape Velocity

45. Escape velocity
• Is the minimum velocity which it is to be projected so
that it just overcome the gravitational pull of the earth
(or any other planet)

46. • In order to project a body with escape
velocity, we give kinetic energy to it. Let
us calculate this Energy.
• Suppose a body of mass m is projected upward with
escape velocity . When the body is at point P at a
distance x from the center of the earth, the gravitational
force of attraction exerted by the earth on the body is
𝐹 = 𝐺
𝑀𝑚
𝑥2
In moving a small distance ∆𝑥 against this gravitational
force, the small work done at the expense of the KE of
the body is given by;
∆𝑊 = 𝐹∆𝑥 = 𝐺
𝑀𝑚
𝑥2 ∆𝑥
P
Fig. PH
x
Q
∆𝑥
R
M = Mass of the earth
R = Radius of the earth

47. • Total work done (W) in moving the body from earth’s surface
(where x=R) to infinity (where 𝑥 = ∞) is given by;
𝑊 =
𝑅
∞
𝐺𝑀𝑚
𝑥2
𝑑𝑥 = 𝐺𝑀𝑚
𝑅
∞
𝑥−2
𝑑𝑥
= 𝐺𝑀𝑚 −
1
𝑥 𝑅
∞
=
𝐺𝑀𝑚
𝑅
∴ 𝑊 =
𝐺𝑀𝑚
𝑅

48. • If the body is to be able to do this amount of work (and so
escape), it needs to have at least this amount of KE at the
moment it is projected. Therefore, escape velocity is given by;
1
2
𝑚𝑉𝑒
2
=
𝐺𝑀𝑚
𝑅
𝑽 𝒆 =
𝟐𝑮𝑴
𝑹

49. The escape velocity can be written in other
equivalent form as shown below;
• The acceleration due to the gravity
of the surface of the earth is;
g =
𝐺𝑀
𝑅2 or 𝐺𝑀 =g𝑅2
Substitute the equation above to the
general equation of escape velocity;
𝑉𝑒 =
𝟐g 𝑅2
𝑹
∴ 𝑉𝑒 = 𝟐g𝑹

50. Example 7
• What is the escape
velocity for a rocket on
the surface of the
Marks? Mass of Marks
= 6.58 𝑥 1023
𝐾𝑔 and
radius of Mars =
3.38 𝑥 106
𝑚.
Solution
𝑽 𝒆 =
𝟐𝑮𝑴 𝒎
𝑹 𝒎
𝑴 𝒎= 6.58 𝑥 1023 𝐾𝑔, 𝑹𝒎 = 3.38 𝑥 106 𝑚, G
= 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2
𝑽 𝒆 =
𝟐 𝑥 6.67𝑥10−11 𝑥 6.58 𝑥 1023
3.38 𝑥 106
∴ 𝑽 𝒆 = 5.1 𝑥 103 𝑚𝑠−1

51. Example 8
• Find the velocity of
escape at the moon
given that its radius is
1.7 𝑥 106
𝑚 and the
value of g at its surface
is 1.63 𝑚𝑠−2
.
SOLUTION
𝑉𝑒 = 𝟐g𝑹
Here; g = 1.63 𝑚𝑠−2, R = 1.7 𝑥 106 𝑚
𝑉𝑒 = 𝟐 𝑥 1.63 𝑥1.7 𝑥 106
𝑉𝑒 = 2.354 𝑥 103
𝑚𝑠−1

52. Kepler’s First Law
• The path of each planet about the sun is an ellipse with sun at
the one focus of the ellipse.

53. What is an ellipse?
2 foci
An ellipse is a
geometric shape with
2 foci instead of 1
central focus, as in a
circle. The sun is at
one focus with
nothing at the other
focus.
FIRST LAW OF PLANETARY MOTION

54. An ellipse also has…
…a major axis …and a minor axis
Semi-major axis
Perihelion Aphelion
Perihelion: When Mars or any another planet is
closest to the sun.
Aphelion: When Mars or any other planet is
farthest from the sun.

55. Kepler’s Second Law
• Each planet moves in such a way that an imaginary line
drawn from the sun to the planet sweeps out equal areas in
equal periods of time.

56. Kepler also found that Mars changed speed as it orbited around
the sun: faster when closer to the sun, slower when farther from
the sun…
A B
But, areas A and B, swept out by a
line from the sun to Mars, were
equal over the same amount of
time.
SECOND LAW OF PLANETARY
MOTION

57. Kepler’s Third Law
• The square of the period of any planet (time needed for one revolution
about the sun) is directly proportional to the cube of the plane’s average
distance from the sun.
𝑇2
∝ 𝑟3
Or
𝑇1
2
𝑇2
2 =
𝑟1
3
𝑟2
3
∴
𝑇2
𝑟3
= 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

58. Derivation of Newton’s Law of Gravitation
from Kepler's Third Law
• Newton was able to show that Kepler's law could be derived
mathematically from his law of universal gravitation and his
laws of motion.
• He used Kepler’s third law as evidence in favor of his law of universal gravitation.

59. Derivation continue…
• Consider a planet of mass m
revolving around the sun of mass
M in a circular orbit of radius r.
• Suppose 𝑣 is the orbital speed of
the planet and T is its time period.
• In time, T, the planet travels a
distance 2𝜋𝑟.
𝑇 =
2𝜋𝑟
𝑣
or 𝑣 =
2𝜋𝑟
𝑇
• The centripetal force F required to
keep the planet in the circular orbit
is;
𝐹 =
𝑚𝑣2
𝑟
=
This centripetal force is provided by
the sun’s gravitational force on the
planet.

60. • According to Kepler’s third law
𝑇2
𝑟3 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (k) or 𝑇2 = 𝑘𝑟3
Putting 𝑇2 = 𝑘𝑟3 in the derive
centripetal force, we have ;
𝐹 =
4𝜋2
𝑘
𝑥
𝑚
𝑟2
• From Universal
F = G
𝑀𝑚
𝑟2

62. Gravitational Field due to a Material of a Body
• Is the space around the body in which any other mas experiences a
force of attraction.
E.g. the gravitational field of the earth

63. Gravitational Field Strength
• Is defined as the force per unit mass acting on a test mass
placed at that point.
𝐸 =
𝐹
𝑚
Its SI unit is 𝑁𝐾𝑔−1

64. Gravitational Field Strength Due to the Earth
• Since, 𝐸 =
𝐹
𝑚
…………………….. (i)
and, 𝐹 =
𝐺𝑀𝑚
𝑅2 ……………………….. (ii)
Substitute equation (ii) into equation
(i), we have;
𝐸 =
𝐺𝑀
𝑅2
• The field strength at any point in a
gravitational field is equal to the
gravitational acceleration of any
mass placed at that point.
• i.e.
g =
𝑮𝑴 𝑬
𝑹 𝑬
𝟐 = 𝐸 𝐸 =
𝐺𝑴 𝑬
𝑹 𝑬
𝟐

65. Gravitational Potential Energy
• The gravitational P.E of a body at a point in the
gravitational field is defined as the amount of work done in
bringing the body from infinity to that point.

66. Gravitation P.E…
• Suppose a body of mass m is situated outside the
earth at Point A at a distance r from the center of
the earth. It’s Gravitational P.E (𝑼 𝑨) is W.D
• Suppose at any instant the body is at point B at a
distance x from the center of the earth. The
gravitational force exerted by the earth on the body
at B is
𝐹 =
𝐺𝑀𝑚
𝑥2
Consider the earth to be a
spherical of radius R and
Mass M
R
o
c
B
dx
x
a
r

67. Gravitation P.E…
• Small amount of work done when
the body moves from B to C is;
𝑑𝑊 = 𝐹𝑑𝑥 =
𝐺𝑀𝑚
𝑥2
𝑑𝑥
• Total work done by the gravitational
force when the body of mass m at a
distance r from the center of the earth
is;
𝑊 =
∞
𝑟
𝐺𝑀𝑚
𝑥2
𝑑𝑥 = 𝐺𝑀𝑚
∞
𝑟
1
𝑥2
𝑑𝑥
𝑊 = 𝐺𝑀𝑚
𝑥−1
−1 ∞
𝑟
= −𝐺𝑀𝑚
1
𝑟
−
1
∞
𝑊 = −
𝐺𝑀𝑚
𝑟
∴ Therefore, gravitational potential energy (𝑼 𝑨) of a body of mass m at
a distance r from the center of the earth is;
𝑼 𝑨 = −
𝑮𝑴𝒎
𝒓

68. Kinetic Energy of the Satellite
• Suppose a satellite of mas m moves round the earth in a circular orbit at the
height h above the surface of the earth. The radius of the orbit of the
satellite is (R + h). If v is the speed of the satellite in the orbit, then,
KE of the satellite = 1
2 𝑚𝑉𝑐
2
‘……………………………(i)
The orbital velocity of the satellite is
𝑉𝑐 =
𝐺𝑀
𝑟
………………………………………. (ii)

69. Kinetic Energy of the Satellite
• Suppose a satellite of mas m moves round
the earth of mas M in a circular orbit at
the height h above the surface of the earth.
• The radius of the orbit of the satellite is
(R + h). If v is the speed of the satellite in
the orbit, then,
K.E of the satellite = 1
2 𝑚𝑉𝑐
2
……(i)
The orbital velocity of the satellite
is;
𝑉𝑐 =
𝐺𝑀
(R + h)
……………. (ii)
• Substitute equation (i) into
equation (ii), we get;
∴ 𝐾. 𝐸 =
𝐺𝑀𝑚
2(R + h)

70. Total Energy of the Satellite
• The satellite in orbit around the earth has both K.E and Potential
Energy(𝑼 𝑨).
Total Energy, E = K.E + 𝑼 𝑨
=
𝐺𝑀𝑚
2(R + h)
+ −
𝑮𝑴𝒎
R + h
∴ E = -
𝐺𝑀𝑚
2(R + h)

71. Example 9
• A spaceship is stationed on Mars. How much energy must be
expend on the spaceship to rocket it out of the solar system?
Mass of the spaceship = 1000Kg, mass of the sun =
2𝑥1030
𝐾𝑔, Mass of mars = 6.4𝑥1023
𝐾𝑔, radius of mars =
3395 Km. radius of orbit of mars = 2.28𝑥108
𝐾𝑚, G =
6.67𝑥10−11
𝑁𝑚2
𝐾𝑔−2
.

72. Solution
Data given;
Mass of the spaceship = 1000Kg, M
ass of the sun = 2𝑥1030 𝐾𝑔,
Mass of mars = 6.4𝑥1023 𝐾𝑔,
radius of mars = 3395 Km,
radius of orbit of Mars = 2.28𝑥108 𝐾𝑚,
 G = 6.67𝑥10−11
𝑁𝑚2
𝐾𝑔−2
,
E = ?.
E = -
𝐺𝑀𝑚
2(R + h)
E = -
6.67𝑥10−11 𝑥 2𝑥1030 𝑥1000
2 x (2.28𝑥1011+3.395 𝑥 106)
E = −2.9 𝑥 1011 𝐽
∴ The energy extended on the spaceship =
𝟐. 𝟗 𝒙 𝟏𝟎 𝟏𝟏
𝑱