PHYSICS s c i s y h p / y m . u d e . k i r t a m . s m k . w w w
CHAPTER 7
CHAPTER 7: Gravitation (2 Hours)
PHYSICS
CHAPTER 7
In this chapter, we learns about 7.1 Gravitational force and field strength 7.2 Gravitational potential 7.3 Satellite motion in a circular orbit
PHYSICS
CHAPTER 7
7.1 Gravitational Force and Field Strength 7.1.1 Newton’s law of gravitation 7.1.2 Gravitational Field 7.1.3 Gravitational force and field strength
PHYSICS CHAPTER 7 Learning Outcome: s c i s y h p / y m . u d e . k i r t a m . s m k . w w w
7.1 Newton’s law of gravitation (1 hour) At the end of this chapter, students should be able to:
State and use the Newton’s law of gravitation,
PHYSICS
CHAPTER 7
7.1.1 Newton’s law of gravitation
States that a magnitude of an attractive force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. them . OR mathematically,
F g m1m2 m1m2 F g 2 r
and
F g
1 2
r
where
m1 , m2 : masses of pa p articl rticle 1 and 2
r : distance be betwee tween pa p artic rticle 1 and 2 G : Universal gravitatio nal nal Constant 6.67 x10 11 N m 2 kg 2
PHYSICS
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The statement can also be shown by using the Figure 7.1.
m1
m2
F 21
F 12 r
Figure 7.1
F 21 F 12 F g G
m1m2 r 2
where
F 21 : Gravitatio nal nal force by pa p articl rticle 2 on pa p articl rticle 1
F 12 : Gravitatio nal nal force by pa p articl rticle 1 on pa p articl rticle 2 Simulation 7.1
PHYSICS
CHAPTER 7
Figures 7.2a and 7.2b show the gravitational force, F g varies with the distance, r .
F g
F g
gradient Gm1m2
0
r 0
1 r 2
Figure 7.2b Figure 7.2a 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.
PHYSICS
CHAPTER 7
Example 7.1 : A spaceship of mass mass 9000 kg travels travels from the Earth Earth to the Moon 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=6.001024 kg, the mass of the Moon, mM=7.351022 kg and the universal gravitational constant,
G=6.67 10
11
N m2 kg 2)
PHYSICS
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Solution :
mE 6.00 10 24 kg; mM 7.35 10 22 kg; ms 900 0 kg; r EM 3.84 108 m
mE
m F Es s F Ms
mM
r EM x
x r EM
F Es 2FMs GmE ms GmM ms 2 2 x r EM x 2 2 mE x
Given
r EM x
2
x
2
3.84 10
8
x
2
2mM 6.00 10
24
27.35 10 22 3.32 108
PHYSICS
CHAPTER 7
Example 7.2 :
C
50 g
6 cm 3.2 kg
8 cm
A
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 7.3. 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 = 6.6710
11
N m2 kg 2)
PHYSICS
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Solution : a.
mA 3.2 kg; mB 2.5 kg; mC 50 10 3 kg r BC 6 10 2 m; r AC 10 10 2 m C
F A 10 10
2
θ
sin sin θ 0.6 cos θ 0.8
m
F B
6 102 m
θ
A
8 10- 2 m
B
The magnitude of the forces on mC,
F A
GmA mC r AC
2
6.67 10 3.250 10 10 10
F A 1.07 10 9 N
11
3
2 2
PHYSICS
CHAPTER 7 mA 3.2 kg; mB 2.5 kg; mC 50 10 3 kg r BC 6 10 2 m; r AC 10 10 2 m
Solution :
F B
GmB mC r BC
2
6.67 10 2.550 10 6 10 11
3
2 2
F B 2.32 10 9 N Force
F A
F B
x-component (N)
y-component (N)
F A sin sin θ F A cos θ 9 9 0.6 1.07 10 1.07 10 0.8 6.42 1010 8.56 1010 F B 0 2.32 109
PHYSICS
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Solution :
F 6.42 10
F x 8.56 10 10 N 10
y
.96 10 9 N 2.32 10 9 2.96
The magnitude of the nett force is
F F F 2
2
x
y
8.56 10 2.96 .96 10 10 2
9 2
F 3.08 .08 109 N
9 F 2 .96 10 y 1 1 θ t an t an 10 F x 8 .56 .5 6 10 θ 73.9 (254 from +x axis anticlockwise)
and its direction is
PHYSICS
CHAPTER 7
Solution : b. By using the Newton’s second law of motion, thus
F m a C
3.08 10 9 50 10 3 a
a
3.08 10 9 50 10 3
a 6.16 10 8 m s 2 and the direction of the acceleration in the same direction of the nett force on the mC i.e. 254 from +x axis anticlockwis anticlockwise. e.
PHYSICS
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7.1 7. 1.2 Gra ravi vita tati tion ona al Fie Fielld
is defined as a region of space surrounding a body that has the property of mass where the attractive force 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 For spherical objects (such as the Earth) the field is radial as shown in Figure 7.4.
M
Figure 7.4
PHYSICS
Note:
CHAPTER 7
The gravitational field in small region near the Earth’s surface are uniform and can be drawn parallel to each other as other as shown in Figure 7.5.
Figure 7.5 The field lines indicate two things:
The arrows arrows – – the direction of the field
The spacing spacing – – the strength of the field
The gravitational field is a conservative field in which the work done in moving a body from one point to another is independen independentt of the path taken.
N e w
PHYSICS
CHAPTER 7
Exercise 7.1 : Given G = 6.6710
11
N m2 kg
2
1. Four identical identical masses masses of 800 800 kg each each are place placed d at the corners 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 Three 5.0 5.0 kg sphe spheres res are are locat located ed in the xy plane as shown in Figure 7.6.Calculate the magnitude of the the nett gravitati gravitational onal force force on the sphere at the origin due to the other two spheres. ANS. : 2.1 10
8
N
Figure 7.6
PHYSICS
CHAPTER 7
Exercise 7.1 : 3.
Figure 7.7 In Figure 8.7, 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
PHYSICS CHAPTER 7 Learning Outcome: s c i s y h p / y m . u d e . k i r t a m . s m k . w w w
7.1.3 Gravitational force and field strength At the end of this chapter, students should be able to: Define gravitational field strength as gravitational force per unit mass,
Derive and use the equation equation for gravitational gravitational field strength.
Sketch a graph of a g against r and explain the change in a g with altitude and depth from the surface of the earth.
PHYSICS
CHAPTER 7
7.1.3 Gravitational field strength, ag is defined as the gravitational force per unit mass of a body (test mass) placed at a point. OR
where F g
: Gravitatio nal force ag : Gravitatio nal field strength
m : mass of a body (test mass)
It is a vector quantity. quantity. The S.I. unit of the gravitational field strength is N kg 1 or m or m s 2.
PHYSICS
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It is also known as gravitational acceleration (the free-fall acceleration).. acceleration) Its direction is in the same direction of the gravitational force. force. Another formula for the gravitational gravitational field strength at a point is given by
ag
F g m
and
F g
GMm r 2
where
M : mass of the point point mass
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.
a g
M
Figure 7.8
r
GM 2
r
PHYSICS
CHAPTER 7
The gravitational field in the small region near the Earth’s R) are uniform surface( r uniform where its strength is 9.81 m s 2 and its direction can be shown shown by using using the Figure 7.9.
a g g
GM R
2
Figure 7.9 where
R : radius of the Earth
g : gravitatio nal accelerati on 9.81 m s 2
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 = 6.6710
M = 6.00
11
N m2 kg 2, mass of the Earth,
1024 kg and radius of the Earth, R = 6.40 106 m)
Solution : a.
g
R
r R 6.40 10 m; a g g 6
r M
The gravitational field strength is
g
GM R 2
6.67 10 6.00 10 6.40 10 11
24
6 2
g 9.77 N kg 1 OR 9.77 m s 2 (Towards the centre of the Earth)
PHYSICS
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Solution : b.
a g h
r
r R h 6 3 6.40 10 350 10 r 6.75 10 6 m The gravitational field strength is given by
R
M
ag
GM r 2
6.67 10 6.00 10 6.75 10 11
24
6 2
ag 8.78 m s 2 (Towards the centre of the Earth)
PHYSICS
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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 1.5 R 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
g
GM R 2
9.81 N kg 1
The distance of point C from the Earth’s centre is
r R 1.5 R 2.5 R
PHYSICS
CHAPTER 7
Solution : a. Thus the gravitational field strength at point C is given by
a g
GM r C
2
a g
GM
2.5R
2
1 GM 2 6.25 R 1 9.81 1.57 N kg 1 a g 6.25 b. Given
m 2.5 kg
The weight of the rock is
(Towards the centre of the Earth)
W ma g
1.57 2.5 W 3.93 N (Towards the centre of the Earth)
PHYSICS
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Example 7.5 :
5 km
B A
Figure 7.10
Figure 8.10 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.
PHYSICS
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Solution :r 5 10
A
3
m; M A 4M B
a g 1 C a g 2 r x a g
At point C,
0 a g 1 a g 2 nett
GM A
r x
2
4 M B
5 10
3
r
x
2
GM B
M B
x 2
x 2
x 1.67 103 m
x
B
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7.1.4 7.1 .4 Variat Variation ion of gra gravit vitati ationa onall fiel field d stre strengt ngth h on the distance from the centre of the Earth Outside the Earth (
) r > R
Figure 8.11 shows a test mass which is outside the Earth and at a distance r from the centre.
M
r
R
Figure 8.11 The gravitational field strength outside the Earth is
ag
GM r 2
PHYSICS
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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
ag
GM R 2
g 9.81 m s 2
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 ' : the mass of spherical portion portion of the Earth of radius, r
Figure 7.13
The gravitational field strength inside the Earth is given by
ag
GM ' 2
r
PHYSICS
CHAPTER 7
By assuming the Earth is a solid sphere and constant density,, hence density
M ' M
r r M R R
M '
V ' V
M '
r 3 R
3
4 3
3
3
4 3
3
3
M
Therefore the gravitational field strength inside the Earth is
r 3 G 3 M R ag 2 r
ag
GM R
3
r
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 ag
GM R
2
g
ag r 0
R
ag
1 2
r
r Figure 7.14
PHYSICS CHAPTER 7 Learning Outcome: s c i s y h p / y m . u d e . k i r t a m . s m k . w w w
7.2 Gravitational potential (½ hour) At the end of this chapter, students should be able to: Define gravitational potential in a gravitational field. Derive and use the formulae,
Sketch the variation of gravitational potential, V with distance, r from the centre of the earth.
PHYSICS 7.2
CHAPTER 7 Gravitational potential
7.2. 7. 2.1 1 Wo Work rk do done ne by the the ext exter erna nall forc force e
Consider an external force, F is required to bring a test
m dr
F g
At the distance r 2 from the centre of the Earth,
F F g
F
mass, m from r 1 to r 2 , as shown in Figure 7.18.
The work done by the external force through the small displacement
dr is
dW Fdr cos cos 0 dW F g dr
r 1 r 2
M
Figure 7.18
PHYSICS
CHAPTER 7
Therefore the the work done by the external external force to bring test mass, m from r 1 to r 2 is
r 2
GMm
r 1 r 2
r 2
W
dW
r 1
F g dr and F g GMm 2
r
W GMm
r 2
dr 1 2
dr
r r 2 1 W GMm r r 1 r 1
where
r 1 : initial distance r 2 : final distance
PHYSICS 7.2.2
CHAPTER 7 Gravitational potential, V
at a point is defined as the work done by an external force in bringing a test mass from from infinity to a point per unit the test mass. mass. OR mathematically, V is written as: where
m : mass of the test mass
V : gravitatio nal pot potent ential at a poin point
W : work done in bri bringing a test mass from infinity t o a poin point
It is a scalar quantity. quantity . Its dimension is given by
W V m
V
2
ML T M
V L T 2
2
2
PHYSICS
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The S.I unit for gravitational potential is m2 s 2 or J or J kg 1. Another formula for the gravitational gravitational potential at a point is given by
1 1 and W GMm GMm V m r 1 r 2 GMm GMm 1 1 where r 1 and r 2 r V m r 1 r 2 GMm 1 1 V m r W
where
r : distance between betw een the point point and the point point mass, M
PHYSICS
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The gravitational potential potential difference between point A and B (V AB) in the Earth’s gravitational field is defined as the work done in bringing a test mass from point B to point A per unit the test mass. OR mathematically, mathematically, V AB is written as:
where
W BA : work done in bri bringing nging the test mass from point point B to point point A.
V A : gravitatio nal potential potential at point point A V B : gravitatio nal potential potential at point point B
PHYSICS
CHAPTER 7
Figure 7.19 shows two points A and B at a distance r A and r B from the centre of the Earth respectively in the Earth’s gravitational field.
A
r A
B
r B M
Figure 7.19
The gravitational potential difference between the points A and B is given by
V AB V A V B
GM GM V AB r A r B
PHYSICS
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The gravitational potential difference between point B and A in the Earth’s Earth’s gravitational field is given by
The variation of gravitational potential, V when the test mass, m move away from the Earth’s surface surface is illustrated illustrated by the graph graph in Figure 7.20.
V 0
r
R
V
1 r
Figure 7.20
GM R
Note: The Gravitational potential at infinity is zero. V 0
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Example 7.7 : 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 7 1. 5.4510 J kg
a. Calculate the satellite’s distance from the Earth’s surface. b. Determine the satellite’s mass. (Given G = 6.6710
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 : F g 19 10
3
N; V 5.45 10 7 J kg 1
F g
R
h
r
PHYSICS
CHAPTER 7 1 3 7 1 9 1 0 N; V 5 . 4 5 1 0 J k g g
Solution : F
a. By using the formulae of gravitational potential, thus
V
5.45 10
GM r
6.67 10 5.98 10 11
7
r 7.32 10 6 m
24
r
Therefore the satellite’s distance from the Earth’s surface is
r h R 6 6 7.32 10 h 6.38 10 h 9.4 10 5 m
PHYSICS
CHAPTER 7 3 7 1 1 9 1 0 N; V 5 . 4 5 1 0 J k g g
Solution : F
b. From the Newton’s law of gravitation, hence
F g 19 10
GMm GM m r 2
6.67 10 5.98 10 m 7.32 10 11
3
m 2552 kg
24
6 2
PHYSICS CHAPTER 7 Learning Outcome: s c i s y h p / y m . u d e . k i r t a m . s m k . w w w
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,
period,
PHYSICS
CHAPTER 7
7.3
Satellite motion in a circular orbit
7.3. 7. 3.1 1 Ta Tang ngen enti tial al (lin (linea ear/ r/or orbi bita tal) l) velo veloci city ty,, v Consider a satellite of mass, m travelling around the Earth of mass, M , radius, R, in a circular orbit of radius, r with constant tangential (orbital) speed, v as shown in Figure 7.22.
Figure 7.22
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The centripetal force, F c is contributed by the gravitational force of attraction, F g exerted on the satellite by the Earth.
F g F c mac 2 GMm mv 2
r
r
Hence the tangential velocity, v is given by
r : distance of the satellite from the centre of the Earth M : mass of the Earth G : universal gravitatio nal constant
where
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For a satellite close to the Earth’s surface, surface , 2 and GM gR
r R
Therefore
The relationship between tangential velocity and angular velocity is
Hence , the period, T of the satellite orbits around the Earth is given by
PHYSICS 7.3. 7. 3.2 2
CHAPTER 7 Sync Sy nchr hron onou ous s (Geo (Geost stat atio iona nary ry)) Sate Satell llit ite e
Figure 8.23 shows a synchronous (geostationary) satellite which stays above the same point on the equator of the Earth.
Figure 8.23 The satellite have the following characteristics: It revolves in the same direction as the Earth. It rotates with the same period of rotation as that of the Earth (24 hours). It moves directly above the equator. The centre of a synchronous satellite orbit is at the centre of the Earth. It is used as a communication satellite.
PHYSICS
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Example 7.12 : The weight of a satellite in a circular orbit round the Earth is half of its weight on the surface of the Earth. If the mass of the satellite is 800 kg, determine a. the altitude of the satellite, b. the speed of the satellite in the orbit, (Given G = 6.6710
11
N m2 kg 2, mass of the Earth,
M = 6.00 1024 kg, and radius of the Earth , R = 6.40 106 m)
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Solution : a. The satellite orbits the Earth in the circular path, thus
b. The speed of the satellite is given by
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Example 7.13 : The radius of the Moon’s orbit around the Earth is 3.8 108 m and the period of the orbit is 27.3 days. The masses of the Earth and Moon are 6.0 1024 kg and 7.4 1022 kg respectively. Calculate the total energy of the Moon in the orbit. Solution : r 8.50 10 m; The period of the satellite is 6
m 120 120 kg; kg; g 9.50 m s 2
T 3.53600 T 12600 s
The tangential speed of the satellite is 2 r
v
v
T 6 2 8.50 10
12600 v 4.24 103 m s 1
PHYSICS
CHAPTER 7 2 6 Solution : r 8.50 10 m; m 120 120 kg; kg; g 9.50 m s
A satellite orbits the planet in the circular path, thus thus
F g F c 2 GMm mv
2
r
v 2
v 2
4.24 10
3 2
r GM
and
r 2 gR
GM gR
r 9.50R 2
8.50 106 6 R 4.0110 m
2
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Exercise 7.2 : Given G = 6.6710
11
N m2 kg
2
1. A rocket is launched launched verticall vertically y from the surface 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 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
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Exercise 7.2 : 3. A geostationary satellite of mass 2400 2400 kg is placed 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
PHYSICS s c i s y h p / y m . u d e . k i r t a m . s m k . w w w
CHAPTER 7
THE END… Next Chapter… CHAPTER 8 : Simple Harmonic Motion