chapter 7 matriculation stpm
TRANSCRIPT
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PHYSICS CHAPTER 7
1
CHAPTER 7:Gravitation
(2 Hours)
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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
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PHYSICS CHAPTER 7
7.1 Gravitational Force and Field
Strength7.1.1 Newtons law of gravitation
7.1.2 Gravitational Field
7.1.3 Gravitational force and field strength
3
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At the end of this chapter, students should be able to:
State and use the Newtons law of gravitation,
Learning Outcome:
7.1 Newtons law of gravitation (1 hour)
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7.1.1 Newtons 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.
OR mathematically,
2
1
rFg 21mmFg and
2
21
r
mmFg
2and1particleofmasses:, 21 mm
2and1particlebetweendistance:r2211
kgmNx106.67ConstantnalgravitatioUniversal:
G
where
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The statement can also be shown by using the Figure 7.1.
where
2
211221
r
mmGFFF g
1m 2m
r12F
Figure 7.1
21F
2particleon1particlebyforcenalGravitatio:12F
1particleon2particlebyforcenalGravitatio:21F
Simulation 7.1
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Figures 7.2a and 7.2b show the gravitational force,Fg varieswith the distance, r.
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 andthe forces always act along the line joining the two pointmasses.
gF
r0
gF
2
1
r0
21mGmgradient
Figure 7.2a Figure 7.2b
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A spaceship of mass 9000 kg travels from the Earth to the Moon
along a line that passes through the Earths centre and the Moons
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=6.0010
24 kg, the mass of the
Moon, mM=7.351022 kg and the universal gravitational constant,
G=6.671011 N m2 kg2)
Example 7.1 :
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Solution :
Given
kg;107.35kg;106.00 22M24
E mmm103.84kg;0900 8EMs rm
Em Mmsm
x
EMr
xr EM
EsF
MsF
MsEs F2F
2EM
sM
2
sE 2xr
mGm
x
mGm
ME
2
EM
2
2m
m
xr
x
2224
28
2
107.352
106.00
103.84
x
x
m103.32 8x
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PHYSICS CHAPTER 7
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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 spheres acceleration.
(Given G = 6.671011 N m2 kg2)
Example 7.2 :
Figure 7.3
A B
C
cm8kg3.2 kg2.5
g50
cm6
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Solution :
a.
The magnitude of the forces on mC,
22
311
2
AC
CAA
1010
10503.2106.67
r
mGmF
N101.07 9AF
kg1050kg;.52kg;3.2 3CBA mmm
m1010m;106 2AC2
BC rr
0.6sin 0.8cos
A B
C
m108 2-
m106 2
m1010 2
AF
BF
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Solution :
22
311
2
BC
CBB
106
10502.5106.67
r
mGmF
N10.322 9BF
Force x-component (N) y-component (N)
AF
F cosA F sinA
0.8101.07 910
108.56
0.6101.07 910
106.42
BF
BF09102.32
kg1050kg;.52kg;3.2 3CBA mmm
m1010m;106 2AC2
BC rr
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Solution :
The magnitude of the nett force is
and its direction is
N108.56 10
x
F
22
yx FFF
N10.96210.322106.42 9910 yF
29210 10.962108.56
N10.083 9F
109
11
10.56810.962tantan
x
y
FF
.973 (254 from +x axis anticlockwise)
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Solution :
b. By using the Newtons second law of motion, thus
and the direction of the acceleration in the same direction of the
nett force on the mC i.e. 254 from +x axis anticlockwise.
amF C a39 1050103.08
28 sm10.166 a
3
9
1050
103.08
a
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is defined as a region of space surrounding a body that has
the property of mass where the attractive force isexperienced if a test mass placed in the region.
Field lines are used to show gravitational field around an object
with mass.
Forspherical objects (such as the Earth) the field is radial asshown in Figure 7.4.
7.1.2 Gravitational Field
M
Figure 7.4
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The gravitational field in small region near the Earths surfaceare uniform and can be drawn parallel to each otheras shown
in Figure 7.5.
The field lines indicate two things: The arrows the direction of the field
The spacing the strength of the field
Figure 7.5
The gravitational field is a conservative field in which the work donein moving a body from one point to another is independent of
the path taken.
Note:
New
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Exercise 7.1 :
Given G = 6.671011 N m2 kg2
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. : 8.2103 N; 45
2. Three 5.0 kg spheres are located in thexy plane as shown inFigure 7.6.Calculate the magnitude
of the nett gravitational force
on the sphere at the origin due to
the other two spheres.
ANS. : 2.1108 N
Figure 7.6
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Exercise 7.1 :
3.
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 ofm5 = 250 kg.
ANS. : 1.68102 N; 45
Figure 7.7
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At the end of this chapter, students should be able to:
Define gravitational field strength as gravitational force perunit mass,
Derive and use the equation for gravitational field strength.
Sketch a graph ofag against rand explain the change in agwith altitude and depth from the surface of the earth.
Learning Outcome:
7.1.3 Gravitational force and field strength
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7.1.3 Gravitational field strength,
is defined as the gravitational force per unit massof a body (test mass) placed at a point.
OR
It is a vector quantity.
The S.I. unit of the gravitational field strength is N
kg1
orm s2
.
ga
where
strengthfieldnalGravitatio:ga
forcenalGravitatio:gF
mass)(testbodyaofmass:m
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It is also known as gravitational acceleration (the free-fallacceleration).
Its direction is in the same direction of the gravitational force. Another formula for the gravitational field strength at a point is
given by
m
Fa
g
g and 2g r
GMmF
where
masspointtheofmass:M
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Figure 7.8 shows the direction of the gravitational field strength
on a point S at distance rfrom the centre of the planet.
2r
GMag
r
M
Figure 7.8
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The gravitational field in the small region near the Earths
surface( r R) are uniform where its strength is 9.81 m s2 and
its direction can be shown by using the Figure 7.9.
Figure 7.9
2R
GMgag
Earththeofradius:Rwhere2sm9.81onacceleratinalgravitatio: g
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Determine the Earths gravitational field strength
a. on the surface.
b. at an altitude of 350 km.
(Given G = 6.671011 N m2 kg2, mass of the Earth,
M= 6.00 1024 kg and radius of the Earth,R = 6.40 106 m)
Solution :
a.
Example 7.3 :
R M
gaRr g m;1040.66
262411
21040.6
1000.61067.6
RGMg
The gravitational field strength is
1kgN77.9 g OR 2sm77.9
rg
(Towards the centre of the Earth)
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Solution :
b.
2g rGMa
26
2411
1075.6
106.001067.6
2g sm78.8
a(Towards the centre of the Earth)
R M
hRr 36 103501040.6 m1075.6 6r
ga
hr
The gravitational field strength is given by
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The gravitational field strength on the Earths surface is 9.81 N kg1.
Calculate
a. the gravitational field strength at a point C at distance 1.5R from
the Earths surface whereR is the radius of the Earth.
b. the weight of a rock of mass 2.5 kg at point C.Solution :
a. The gravitational field strength on the Earths surface is
The distance of point C from the Earths centre is
Example 7.4 :
1kgN81.9 g
1
2kgN81.9
R
GMg
RRRr 5.25.1
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Solution :
a. Thus the gravitational field strength at point C is given by
b. Given
The weight of the rock is
N93.3W
2
Cr
GMag 25.2 RGMag
225.6
1
R
GM
gmaW
kg5.2m
57.15.2
1kgN57.181.925.6
1 ga
(Towards the centre of the Earth)
(Towards the centre of the Earth)
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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.
Example 7.5 :
A
B
km5
Figure 7.10
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Solution :
At point C,
BA3 4m;105 MMr
0nett ga
2B
23
B
105
4
x
M
x
M
m10.6713
x
r
A
BC
xr x
2ga
1ga
21 gg aa
2
B
2
A
x
GM
xr
GM
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Outside the Earth ( r> R)
Figure 8.11 shows a test mass which is outside the Earth and at
a distance rfrom the centre.
The gravitational field strength outside the Earth is
7.1.4 Variation of gravitational field strength on the
distance from the centre of the Earth
R
rM
Figure 8.11
2gr
GMa
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On the Earth ( r= R)
Figure 7.12 shows a test mass on the Earths surface.
The gravitational field strength on the Earths surface is
R
rM
Figure 7.12
2
2gsm81.9 g
R
GMa
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R
r
M
'M
Inside the Earth ( r< R)
Figure 7.13 shows a test mass which is inside the Earth and at
distance r from the centre.
The gravitational field strength inside the Earth is given by
Figure 7.13
2g
'
r
GMa
where
portionsphericalofmassthe:'Mradius,ofEarththeof r
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By assuming the Earth is a solid sphere and constant
density, hence
Therefore the gravitational field strength inside the Earth is
V
V
M
M
'' 33
3
34
334'Rr
Rr
MM
M
R
rM
3
3
'
2
3
3
g r
MR
rG
a
rR
GMa
3g
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The variation of gravitational field strength, ag as a function of
distance from the centre of the Earth, ris shown in Figure 7.14.
Figure 7.14
R
ga
r0 R
gR
GMa
2g
ra g2g
1r
a
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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, Vwithdistance, rfrom the centre of the earth.
Learning Outcome:
7.2 Gravitational potential ( hour)
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7.2 Gravitational potential
7.2.1 Work done by the external force
Consider an external force, F
is required to bring a test
mass, m from r1
to r2
,
as shown in Figure 7.18.
At the distance r2 from the
centre of the Earth,
The work done by theexternal force through
the small displacement
dris
m
M
1r2r
F
gF
dr
Figure 7.18
gFF
0cosFdrdW
drFdW g
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Therefore the work done by the external force to bring test
mass, m from r1 to r2 is
2
1
r
rgdrFdW
2
12
r
rdr
r
GMmW
2rGMmFg and
2
1
1r
rr
GMmW
2
12
1rr
drr
GMmW
where
distancefinal:2rdistanceinitial:1r
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at a point is defined as the work done by an external force in
bringing a test mass from infinity to a point per unit thetest mass.
OR mathematically, Vis written as:
It is a scalar quantity.
Its dimension is given by
7.2.2 Gravitational potential, V
where
masstesttheofmass:mpointaatpotentialnalgravitatio:V
masstestabringingindonework:Wpointaoinfinity tfrom
mW
V M
TML 22 V
22TL V
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The S.I unit for gravitational potential is m2 s2 orJ kg1. Another formula for the gravitational potential at a point is given
by
21
11
rrm
GMmV
m
WV and
21
11
rrGMmW
where 1r
and rr 2
rm
GMmV
11
where
pointebetween thdistance:rMmass,pointtheand
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The gravitational potential difference between point A and B
(VAB) in the Earths gravitational field is defined as the work
done in bringing a test mass from point B to point A perunit the test mass.
OR mathematically, VAB is written as:
where
A.pointtoBpointfrom
masstestthebringingindonework:BAW
Apointatpotentialnalgravitatio:AV
Bpointatpotentialnalgravitatio:BV
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Figure 7.19 shows two points A and B at a distance rA and rBfrom the centre of the Earth respectively in the Earths
gravitational field.
M
A
BrA
rB
Figure 7.19
The gravitational potential
difference between the points A
and B is given by
BAAB VVV
BA
ABr
GM
r
GMV
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The gravitational potential difference between point B and A in
the Earths gravitational field is given by
The variation of gravitational potential, Vwhen the test mass, mmove away from the Earths surface is illustrated by the graph
in Figure 7.20.
R
R
GM
r0
V
rV
1
Note:
The Gravitational potential at infinityis zero. 0V
Figure 7.20
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When in orbit, a satellite attracts the Earth with a force of 19 kN
and the satellites gravitational potential due to the Earth is5.45107 J kg1.
a. Calculate the satellites distance from the Earths surface.
b. Determine the satellites mass.
(Given G = 6.67
1011
N m2
kg2
, mass of the Earth,M= 5.981024 kg and radius of the Earth , R = 6.38106 m)
Solution :
Example 7.7 :
R
gF
rh
173 kgJ10455N;1019 .VFg
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Solution :
a. By using the formulae of gravitational potential, thus
Therefore the satellites distance from the Earths surface is
rGMV
m1032.7 6r
66
1038.61032.7
hm104.9 5h
r
.. 24117 10985106761045.5
Rhr
173 kgJ10455N;1019 .VFg
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Solution :
b. From the Newtons law of gravitation, hence
2rGMmFg
kg2552m
262411
3
10327
1098510676
1019
.
m..
173 kgJ10455N;1019 .VFg
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At the end of this chapter, students should be able to:
Explain satellite motion with:
velocity,
period,
Learning Outcome:
7.3 Satellite motion in a circular orbit ( hour)
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7.3 Satellite motion in a circular orbit
7.3.1 Tangential (linear/orbital) velocity,v
Consider a satellite of mass, m travelling around the Earth ofmass,M, radius, R, in a circular orbit of radius, rwith constanttangential (orbital) speed, v as shown in Figure 7.22.
Figure 7.22
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The centripetal force, Fc is contributed by the gravitational force
of attraction,Fgexerted on the satellite by the Earth.
Hence the tangential velocity, vis given by
ccg maFF
r
mv
r
GMm 2
2
where
Earththeofmass:M
fromsatellitetheofdistance:rEarththeofcentrethe
constantnalgravitatiouniversal:G
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For a satellite close to the Earths surface,
Therefore
The relationship between tangential velocity and angular
velocity is
Hence , the period, Tof the satellite orbits around the Earth is
given by
Rr and 2gRGM
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Figure 8.23 shows a synchronous (geostationary) satellite which
stays above the same point on the equator of the Earth.
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 ofthe Earth.
It is used as a communication satellite.
7.3.2 Synchronous (Geostationary) Satellite
Figure 8.23
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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 is800 kg, determine
a. the altitude of the satellite,
b. the speed of the satellite in the orbit,
(GivenG
= 6.671011 N m2 kg2, mass of the Earth,
M= 6.001024 kg, and radius of the Earth , R = 6.40106 m)
Example 7.12 :
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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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The radius of the Moons orbit around the Earth is 3.8 108 m and
the period of the orbit is 27.3 days. The masses of the Earth andMoon are 6.0 1024 kg and 7.4 1022 kg respectively. Calculate
the total energy of the Moon in the orbit.
Solution :
The period of the satellite is
The tangential speed of the satellite is
Example 7.13 :
sm50.9kg;120m;1050.8 26 gmr
T
rv
2
13 sm1024.4 v
36005.3Ts12600T
12600
1050.82 6
v
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Solution :
A satellite orbits the planet in the circular path, thus
cg FF
2
2 r
mv
r
GMm
sm50.9kg;120m;1050.8 26 gmr
r
GMv
2
and2
gRGM
r
gRv
22
62
23
1050.850.91024.4
R
m1001.4 6R
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Exercise 7.2 :
Given G = 6.671011 N m2 kg2
1. A rocket is launched vertically from the surface of the Earthat speed 25 km s-1. Determine its speed when it escapes from
the gravitational field of the Earth.
(Givengon the Earth = 9.81 m s2, radius of the Earth ,
R = 6.38
10
6
m)ANS. : 2.24104 m s12. 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 kg1
. Determinea. 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 s1 ; 1.11104 rad s1
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Exercise 7.2 :
3. A geostationary satellite of mass 2400 kg is placed
35.92 Mm from the Earths surface orbits the Earth along acircular 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. : 7.27105 rad s1; 3.08103 m s1; 0.224 m s2;537 N ; 6.001024 kg
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THE END
Next ChapterCHAPTER 8 :
Simple Harmonic Motion
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