PHYSICS
CHAPTER 7
CHAPTER 7:
Gravitation
(2 Hours)
1
PHYSICS
CHAPTER 7
Learning Outcome:
7.1 Gravitational Force and Field Strength(1 hour)
At the end of this chapter, students should be able to:

State and use the Newton’s law of gravitation,
m1m2
Fg  G 2
r

Define gravitational field strength as gravitational force per unit mass
ag 

F
m
Derive and use gravitational field strength,
ag  G

M
r2
Sketch a graph of ag against r and explain the change in ag with altitude
and depth from the surface of the earth.
2
PHYSICS
CHAPTER 7
7.1 Newton’s law of gravitation
7.1.1
Newton’s law of gravitation
 States that a magnitude of an …………………..between two point masses
is directly ………………....... to the product of their masses and inversely
proportional to the square of the …………………between them.
OR mathematically,
Fg  m1m2
m1m2
Fg  2
r
and
where
1
Fg  2
r
m1m2
Fg  G 2
r
Fg : Gravitational force
m1 , m2 : masses of particle1 and 2
r : distance between particle1 and 2
G : Universal gravitational Constant  6.67 x1011 N m2 kg2
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PHYSICS

CHAPTER 7
The statement can also be shown by using the Figure 7.1.
m1
m2

F21

F12
r
Figure 7.1


m1m2
F21  F12  Fg  G 2
r
where
F21 : Gravitational force by particle 2 on particle 1

F12 : Gravitational force by particle1 on particle 2
4
PHYSICS

CHAPTER 7
Figures 7.2a and 7.2b show the gravitational force, Fg varies with the
distance, r.
Fg
Fg
gradient  Gm1m2
0

Figure 7.2a
r
0
Figure 7.2b
1
r2
Notes:
 Every spherical object with constant density can be reduced to a
point mass at the centre of the sphere.
 The gravitational forces always attractive in nature and the forces
always act along the line joining the two point masses.
5
PHYSICS
CHAPTER 7
Example 7.1 :
A spaceship of mass 9000 kg travels from the Earth to the Moon along
a line that passes through the Earth’s centre and the Moon’s centre.
The average distance separating Earth and the Moon is 384,000 km.
Determine the distance of the spaceship from the Earth at which the
gravitational force due to the Earth twice the magnitude of the
gravitational force due to the Moon.
(Given the mass of the Earth, mE=6.001024 kg, the mass of the
Moon, mM=7.351022 kg and the universal gravitational constant,
G=6.671011 N m2 kg2)
6
PHYSICS
CHAPTER 7
Solution :
mE  6.00 10 24 kg; mM  7.35 10 22 kg;
ms  900 0 kg; rEM  3.84 108 m
mE
 m 
FEs s FMs
x
rEM  x
mM
rEM
Given
FEs  2FMs
7
PHYSICS
CHAPTER 7
C
Example 7.2 :
50 g
6 cm
3.2 kg
A
8 cm
2.5 kg
B
Figure 7.3
Two spheres of masses 3.2 kg and 2.5 kg respectively are fixed at points A
and B as shown in Figure 8.3. If a 50 g sphere is placed at point C,
determine
a. the resultant force acting on it.
b. the magnitude of the sphere’s acceleration.
(Given G = 6.671011 N m2 kg2)
8
PHYSICS
Solution :
a.
CHAPTER 7
mA  3.2 kg; mB  2.5 kg; mC  50 10 3 kg
rBC  6 10 2 m; rAC  10 10 2 m
C
 θ
sin θ  0.6
FA
cos θ  0.8

10  10 2 m
FB
2
6  10
m
θ
A
8  10 - 2 m
B
The magnitude of the forces on mC,
9
PHYSICS
Solution :
CHAPTER 7
mA  3.2 kg; mB  2.5 kg; mC  50 10 3 kg
rBC  6 10 2 m; rAC  10 10 2 m
GmB mC
FB 

2
rBC
Force

FA

FB
x-component (N)
 FA cos θ
0
y-component (N)
 FA sin θ
 FB
10
PHYSICS
Solution :
CHAPTER 7
F
F
x

y

The magnitude of the nett force is
 F   F    F 
2
2
x
and its direction is
  Fy
θ  t an 
F
x

θ
1
y




11
PHYSICS
CHAPTER 7
Solution :
b. By using the Newton’s second law of motion, thus
F  m a
C
and the direction of the acceleration in the …………………..of the nett force
on the mC i.e………………………………………
12
PHYSICS
CHAPTER 7
Exercise 7.1 :
Given G = 6.671011 N m2 kg2
1. Four identical masses of 800 kg each are placed at the corners of a square
whose side length is 10.0 cm. Determine the nett gravitational force on one
of the masses, due to the other three.
ANS. : 8.2103 N; 45
2. Three 5.0 kg spheres are located in the xy plane as shown in Figure
7.4.Calculate the magnitude
of the nett gravitational force
on the sphere at the origin due to
the other two spheres.
ANS. : 2.1108 N
Figure 7.4
13
PHYSICS
CHAPTER 7
Exercise 7.1 :
3.
Figure 7.5
In Figure 7.5, four spheres form the corners of a square whose side is 2.0
cm long. Calculate the magnitude and direction of the nett gravitational
force on a central sphere with mass of m5 = 250 kg.
ANS. :
1.68102 N; 45
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PHYSICS
CHAPTER 7
7.1.2 Gravitational Field

is defined as a ……………………………surrounding a body that has the
property of mass where the …………………….is experienced if a test
mass placed in the region.


Field lines are used to show gravitational field around an object with mass.
For spherical objects (such as the Earth) the field is radial as shown in
Figure 7.6.
M
Figure 7.6
15
PHYSICS

CHAPTER 7
The gravitational field in small region near the Earth’s surface are ……………..and
can be drawn ……………………..to each other as shown in Figure 7.7.
Figure 7.7

The field lines indicate two things:
 The arrows – the ………………. of the field
 The spacing – the ………………. of the field
Note:
The gravitational field is a conservative field in which the work done in
moving a body from one point to another is independent of the path
taken.
16
PHYSICS
CHAPTER 7
Gravitational field strength,


ag
is defined as the ……………………per ………………of a body (test
mass) placed at a point.

OR
Fg

ag 
m
where



Fg : Gravitational force
ag : Gravitational field strength
m : mass of a body (test mass)
It is also known as gravitational acceleration (the free-fall
acceleration).
It is a ………………………..
The S.I. unit of the gravitational field strength is ……………..or …………...
17
PHYSICS
CHAPTER 7
7.1.3 Gravitational Acceleration


Its direction is in the ……………………..of the gravitational force.
Another formula for the gravitational field strength at a point is given by
ag 
ag 
Fg
and
m
GMm
Fg  2
r
1  GMm 


m  r2 
GM
ag  2
r
where
M : mass of t hepoint mass
r : distancebetween test mass and pointmass
18
PHYSICS

CHAPTER 7
Figure 7.8 shows the direction of the gravitational field strength on a point S
at distance r from the centre of the planet.
GM
ag  2
r
M
Figure 7.8
r
19
PHYSICS

CHAPTER 7
The gravitational field in the small region near the Earth’s surface( r R)
are uniform where its strength is 9.81 m s2 and its direction can be shown
by using the Figure 7.9.
GM
ag  g  2
R
Figure 7.9
where
R : radius of the Earth
g : gravitational acceleration  9.81m s2
20
PHYSICS
CHAPTER 7
Example 7.3 :
Determine the Earth’s gravitational field strength
a. on the surface.
b. at an altitude of 350 km.
(Given G = 6.671011 N m2 kg2, mass of the Earth,
M = 6.00  1024 kg and radius of the Earth, R = 6.40  106 m)
Solution :
a.

g
r
M
R
r  R  6.40 10 6 m; ag  g
The gravitational field strength is
GM
g 2 
R
21
PHYSICS
Solution :
b.
CHAPTER 7

ag h
r  Rh
r
 6.40  10 6  350  10 3
r  6.75  10 6 m
The gravitational field strength is given by
R M
GM
ag  2
r
22
PHYSICS
CHAPTER 7
Example 7.4 :
The gravitational field strength on the Earth’s surface is 9.81 N kg1.
Calculate
a. the gravitational field strength at a point C at distance 1.5R from
the Earth’s surface where R is the radius of the Earth.
b. the weight of a rock of mass 2.5 kg at point C.
Solution :
g  9.81 N kg 1
a. The gravitational field strength on the Earth’s surface is
The distance of point C from the Earth’s centre is
23
PHYSICS
CHAPTER 7
Solution :
a. Thus the gravitational field strength at point C is given by
ag 
GM
rC
2
b. Given m  2.5 kg
The weight of the rock is
ag 
GM
2.5R 2
W  ma g
24
PHYSICS
CHAPTER 7
Example 7.5 :
5 km
B
A
Figure 7.10
Figure 7.10 shows an object A at a distance of 5 km from the object B. The
mass A is four times of the mass B. Determine the location of a point on the line
joining both objects from B at which the nett gravitational field strength is zero.
25
PHYSICS
Solution :
CHAPTER 7
r  5 103 m; M A  4M B
A


a g1 C a g 2
rx
At point C,
a 
g nett
r
B
x
0
a g1  a g 2
26
PHYSICS
CHAPTER 7
7.1.4 Variation of gravitational field strength on the
distance from the centre of the Earth
Outside the Earth ( r > R)
 Figure 7.11 shows a test mass which is outside the Earth and at a distance
r from the centre.
M
r
R

Figure 7.11
The gravitational field strength outside the Earth is
GM
ag  2
r
1
ag  2
r
27
PHYSICS
CHAPTER 7
On the Earth ( r = R)
 Figure 7.12 shows a test mass on the Earth’s surface.
M
r
R
Figure 7.12

The gravitational field strength on the Earth’s surface is
GM
ag  2  g  9.81 m s 2
R
28
PHYSICS
CHAPTER 7
Inside the Earth ( r < R)

Figure 7.13 shows a test mass which is inside the Earth and at distance r
from the centre.
M
M'
where
r
R
M ' : themass of sphericalportion
of theEarthof radius, r
Figure 7.13

The gravitational field strength inside the Earth is given by
GM '
ag  2
r
29
PHYSICS

CHAPTER 7
By assuming the Earth is a solid sphere and constant density, hence
M ' V '

M
V





M '  43 r 3
r3
 4 3  3
M  3 R
R
3
r
M ' 3 M
R
Therefore the gravitational field strength inside the Earth is
 r3 
G 3 M 
R


ag 
r2
GM
ag  3 r
R
ag  r
30
PHYSICS

CHAPTER 7
The variation of gravitational field strength, ag as a function of distance
from the centre of the Earth, r is shown in Figure 7.14.
R
ag
GM
ag  2  g
R
ag  r
0
1
ag  2
r
r
R
Figure 7.14
31
PHYSICS
Learning Outcome:
CHAPTER 7
7.2 Gravitational potential (1 hour)
At the end of this chapter, students should be able to:
 Define gravitational potential in a gravitational field.

Derive and use the formulae,
GM
V 
r

Sketch the variation of gravitational potential, V with distance, r
from the centre of the earth.
32
PHYSICS
7.2.1

CHAPTER 7
Gravitational potential, V
at a point is defined as the ………………...by an external force in bringing
a test mass from infinity to a point per unit the test mass.
OR
mathematically, V is written as:
W
V
m
where
m : mass of the test mass
V : gravitatio nal potential at a point
W : work done in bringing a test mass
from infinity t o a point

It is a ……………………...
33
PHYSICS


CHAPTER 7
The S.I unit for gravitational potential is ……………or ……………….
Another formula for the gravitational potential at a point is given by
W
V
m
and
GMm  1 1 
  
V
m  r1 r2 
GMm  1 1 
V
  
m  r
GM
V 
r
where
1 1
W  GMm  
 r1 r2 
where
r1  
and r2
r
r : distancebetween the point
and thepointmass, M
34
PHYSICS

CHAPTER 7
The gravitational potential difference between point A and B (VAB) in
the Earth’s gravitational field is defined as the ………………..in bringing
a test mass from point B to point A per unit the test mass.
OR
mathematically, VAB is written as:
WBA
VAB 
 VA -VB
m
where
WBA : work done in bringing thetest mass
frompointB to pointA.
VA : gravitational potentialat pointA
VB : gravitational potentialat pointB
35
PHYSICS

CHAPTER 7
Figure 7.15 shows two points A and B at a distance rA and rB from the
centre of the Earth respectively in the Earth’s gravitational field.

A
rA
B
rB
M
Figure 7.15
The gravitational potential
difference between the points A
and B is given by
VAB  VA  VB
 GM   GM 
   

VAB   
 rA   rB 
1 1
VAB  GM   
 rB rA 
36
PHYSICS

CHAPTER 7
The gravitational potential difference between point B and A in the Earth’s
gravitational field is given by
VBA

WAB
 VB  VA 
m
The variation of gravitational potential, V when the test mass, m move
away from the Earth’s surface is illustrated by the graph in Figure 7.16.
V
0
GM

R
R
r
1
V
r
Figure 7.16
Note:

The Gravitational potential at infinity
is zero.
V  0
37
PHYSICS
CHAPTER 7
Example 7.6 :
When in orbit, a satellite attracts the Earth with a force of 19 kN and the
satellite’s gravitational potential due to the Earth is 5.45107 J kg1.
a. Calculate the satellite’s distance from the Earth’s surface.
b. Determine the satellite’s mass.
(Given G = 6.671011 N m2 kg2, mass of the Earth,
M = 5.981024 kg and radius of the Earth , R = 6.38106 m)
Solution :
Fg  19 103 N; V  5.45 10 7 J kg 1

Fg
h
r
R
38
PHYSICS
Solution :
CHAPTER 7
Fg  19 103 N; V  5.45 10 7 J kg 1
a. By using the formulae of gravitational potential, thus:
GM
V 
r
Therefore the satellite’s distance from the Earth’s surface is:
r hR
39
PHYSICS
Solution :
CHAPTER 7
Fg  19 103 N; V  5.45 10 7 J kg 1
b. From the Newton’s law of gravitation, hence:
GMm
Fg  2
r
40
PHYSICS
CHAPTER 7
Example 7.7 :
The gravitational potential at the surface of a planet of radius R is 12.8 MJ
kg1. Determine the work done in overcoming the gravitational force when a
space probe of mass 1000 kg is lifted to a height of 2R from the surface of the
planet.
Solution :
m  1000 kg; r1  R
R
M
r1 m
h  2R
r2
On the surface of the planet, the gravitational potential is
GM
V 
r1
41
PHYSICS
CHAPTER 7
Solution :
m  1000 kg; r1  R r2  R  h  3R
The final distance of the space probe from the centre of the Earth is
The work done required is given by
1 1
W  GMm  
 r1 r2 
42
PHYSICS
CHAPTER 7
Example 7.8 :
The Moon has a mass of 7.351022 kg and a radius of 1740 km.
a. Determine the gravitational potential at its surface.
b. A probe of mass 100 kg is dropped from a height 1 km onto the
Moon’s surface. Calculate its change in gravitational potential
energy.
c. If all the gravitational potential energy lost is converted to kinetic
energy, calculate the speed at which the probe hits the surface.
(Given G = 6.671011 N m2 kg2)
Solution :
M  7.35 1022 kg; R  1.74 106 m
a. The gravitational potential on the moon’s surface is
GM
V 

R
43
PHYSICS
Solution :
b. Given
CHAPTER 7
M  7.35 1022 kg; R  1.74 106 m
6
m  100 kg; r2  R  1.74 10 m
R
M
r2
h  1.0010 m
3
m
r1
r1  R  h 
Hence the change in the gravitational potential energy is
ΔU  U f  U i
 GMm   GMm 
1 1
   
  GMm  
ΔU   
r2  
r1 

 r1 r2 
44
PHYSICS
CHAPTER 7
Solution :
c. Given
Gravitational potential energy lost = kinetic energy
The speed of the probe when hit the moon’s surface is given by
ΔU  K
1 2
ΔU  mv
2
45
PHYSICS
CHAPTER 7
Learning Outcome:
7.3 Satellite motion in a circular orbit (½ hour)
At the end of this chapter, students should be able to:

Explain satellite motion with:

velocity,
GM
v
r

period,
r3
T  2
GM
46
PHYSICS
7.3
CHAPTER 7
Satellite motion in a circular orbit
7.3.1 Tangential (linear/orbital) velocity, v
 Consider a satellite of mass, m travelling around the Earth of mass, M,
radius, R, in a circular orbit of radius, r with …………………………….speed,
v as shown in Figure 8.22.
Figure 8.22
47
PHYSICS

CHAPTER 7
The centripetal force, Fc is contributed by the gravitational force of
attraction, Fg exerted on the satellite by the Earth.
Fg  Fc  mac
GMm mv

2
r
r
2
Hence the tangential velocity, v is given by
GM
v
r
where r : distance of the satellite from
the centre of the Earth
M : mass of the Earth
G : universal gravitatio nal constant
48
PHYSICS

CHAPTER 7
For a satellite close to the Earth’s surface,
and
rR
Therefore

GM  gR
2
v  gR
The relationship between tangential velocity and angular velocity is
2r
v  r 
T
Hence , the period, T of the satellite orbits around the Earth is given by
2r
GM

T
r
r3
T  2
GM
49
PHYSICS
7.3.2

CHAPTER 7
Synchronous (Geostationary) Satellite
Figure 7.17 shows a synchronous (geostationary) satellite which stays
above the same point on the equator of the Earth.
Figure 7.17


The satellite have the following characteristics:
 It revolves in the ……………………………as the Earth.
 It rotates with the …………………….of rotation as that of the Earth (24
hours).
 It moves directly above the ……………………...
 The centre of a synchronous satellite orbit is at the centre of the Earth.
It is used as a ……………………………...
50
PHYSICS
CHAPTER 7
Exercise 7.2 :
Given G = 6.671011 N m2 kg2
1. A rocket is launched vertically from the surface of the Earth at speed 25
km s-1. Determine its speed when it escapes from the gravitational field of
the Earth.
(Given g on the Earth = 9.81 m s2, radius of the Earth ,
R = 6.38  106 m)
ANS. : 2.24104 m s1
2. A satellite revolves round the Earth in a circular orbit whose radius is five
times that of the radius of the Earth. The gravitational field strength at the
surface of the Earth is 9.81 N kg1. Determine
a. the tangential speed of the satellite in the orbit,
b. the angular frequency of the satellite.
(Given radius of the Earth , R = 6.38  106 m)
ANS. : 3538 m s1 ; 1.11104 rad s1
51
PHYSICS
CHAPTER 7
Exercise 7.2 :
3. A geostationary satellite of mass 2400 kg is placed 35.92 Mm from
the Earth’s surface orbits the Earth along a circular path.
Determine
a. the angular velocity of the satellite,
b. the tangential speed of the satellite,
c. the acceleration of the satellite,
d. the force of attraction between the Earth and the satellite,
e. the mass of the Earth.
(Given radius of the Earth , R = 6.38  106 m)
ANS. : 7.27105 rad s1; 3.08103 m s1; 0.224 m s2;
537 N ; 6.001024 kg
52
PHYSICS
CHAPTER 8
THE END…
Next Chapter…
CHAPTER 8 :
Rotational of A Rigid Body
53