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2 Force and Motion
Chapter 10 Gravitation
6.67  10 11 (7)7
0.218 2
10 Gravitation
=
Practice 10.1 (p.376)
= 6.88  108 N
(b)
Gravitational force on ball
1
B
2
B
= weight of ball
3
C
= 7  9.81
4
B
= 68.7 N
5
A
(c)
GM
1
g = 2  2 (for constant M)
r
r
1
2
2
R2
gY
r
r
= Y = X2 =
1
( R  H )2
gX
rY
2
rX
(R + H)2 =
By Newton’s third law,
gravitational force on Earth = 68.7 N
11
(a)
gX
 R2 = 4R2
gY
(b)
R + H = 2R
H =R
6
A
7
A
= 5.90  103 N kg1
g0
9.81
=
= 1660
g S 5.90  10  3
On the Earth’s surface, the gravitational
field strength due to the Earth is 1660
times that due to the Sun.
GM
M
 2
r2
r
ME
2
2
aE
M r
r
= E = E M2
aM M M
M M rE
2
rM
a=
12
The acceleration due to gravity g is 9.81 m s2
only at positions near the Earth’s surface.
Celestial bodies are far away from the Earth,
13
rM
1
1
1
aE M M
=
= 6
=


81 3.67 4
rE
aM M E
8
Gravitational field strength due to Sun gS
GM
= 2
r
6.67  10 11 (1.99  10 30 )
=
(1.50  1011 ) 2
so g no longer equal to this value.
GMm
F= 2 M
r
 F is halved when M is halved.
F
A
2
2
6370
RE
= 9.81 
= 9.69 m s2
2
(6370  39 ) 2
r
GMm
By F = 2 ,
r
2
11 .2(1.48  10 8  10 3 ) 2
Fr
m=
=
= 1850 kg
GM 6.67  10 11 (1.99  10 30 )
g = g0
9
10
(a)
Assume that the balls are spherical
symmetrical.
Gravitational force =
GMm
r2
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
mass halved
r
0
14
(a)
On Earth’s surface, by W = mg,
W 7080
m=
=
= 721.7 kg
9.81
g
When W = 3540 N,
1
2 Force and Motion
g=
Chapter 10 Gravitation
W 3540
=
= 4.905 m s2
m 721 .7
4
GMm mv 2
=
,
r
r2
GM
v=
r
By
2
By g = g0
r = RE
RE
,
r2
g0
g
= 6370  103
Period =
9.81
4.905
= 9.01  106 m
5
Infinity
W
6
Angular speed
2π
2π
=
=
= 8.25  104 rad s1
T 127  60
(b)
Its weight provides the centripetal force.
Weight = mr  2
= 2350[(1740 + 200)  103] 
Gravitational field strength
R
= g0 E2
r
(8.25  104)2
2
= 0.658 
1150 2
10 000 2
7
= 8.70  103 N kg1
1
B
2
D
3
A
GMm mv 2
=
,
r
r2
GM
1
r = 2  2 (for constant M)
v
v
1
2
rD vD 2 vP
=
= 2
1
rP
vD
2
vP
rD =
vP
vD
2
 rP =
RvP
vD
2
2
2
 2π 
(3.5 10 7 )3 

18  3600 

=
6.67 10 -11
By
2
= 3100 N
GMm
By
= mr  2,
r2
r 3 2
M=
G
 2π 
r3 
T
=  
G
Practice 10.2 (p.385)
2
(a)
r
0
15
B
By
It is 9010 km from the Earth’s centre.
(c)
2 πr
r3
2 πr
=
= 2
v
GM
GM
r
GMm
= mr  2,
r2
1
GM
=

3
r
r3
= 9010 km
(b)
A
2
= 6.04  1024 kg
8
The Earth’s mass is 6.04  1024 kg.
GM E m mv 2
By
=
,
r
r2
GM
r = 2E
v
=
GM E
RE
2

RE
v2
2
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
2 Force and Motion
Chapter 10 Gravitation
Revision exercise 10
2
g0 R E
v2
9.81(6.37 10 6 ) 2
=
3000 2
=
Concept traps (p.388)
1
The acceleration due to gravity decreases
= 4.42  107 m
W
50
Mass of object m = 0 =
= 5.01 kg
g 0 9.81
Weight of object in orbit = centripetal force
mv 2
=
r
5.01(3000 ) 2
=
4.42  10 7
when the distance from the Earth increases. It
equals 9.81 m s2 on the Earth’s surface only.
2
(a)
Gravitational field strength
= g0
RE
r2
Earth in a circular orbit remains constant. The
gravitational force does no work on the
satellite as it is perpendicular to the
displacement of the satellite.
3
Multiple-choice questions (p.388)
2
6370
(6370  900 ) 2
= 7.53 N kg
4
B
5
C
1
GMm M
 2 (for constant m)
r2
r
MP
2
2
WP
M r
r
= P = P E2
WE M E M E rP
2
rE
W=
By mg = mr  ,
2
=
=
g
r
7.53
(6370  900 )10 3
rP = rE
= 1.018  103 rad s1
 1.02  103 rad s1
The angular speed is 1.02  103 rad s1.
2π
2π
Period =
=
= 6170 s
 1.018  10 3
2
(c)
=
T
2
= 9.81 
(b)
F
The energy of a satellite travelling around the
= 1.04 N
9
F
g
=
r
g0
RE
2
r 2 = g RE  1
0
r
r3
r3
r 
2π
T=

T
The angular speed is lower and the
period is longer at a higher orbit.
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
6
C
7
C
M P WE
= RE 2  2 = 2RE

M E WP
GMm mv 2
=
,
r
r2
1
GM
v=

(for constant M)
r
r
1
rY
vY
r
=
= X
1
vX
rY
rX
By
vY = vX
rX
= 2v
rY
3
2 Force and Motion
8
Chapter 10 Gravitation
C
= 5.24  103 
GMm M
 2 (for constant m)
r
r2
MM
2
2
WM
M r
r
= M = M E2
ME
WE
M E rM
2
rE
W=
M M rE
2
M E rM
2
= 4.86  1024 kg
GM
g= 2
r
6.67  10 11 (4.86  10 24 )
=
= 6.52 N kg1
[( 6050  1000 )10 3 ]2
15
(HKALE 2009 Paper 2 Q5)
16
(HKALE 2010 Paper 2 Q12)
17
(HKDSE 2012 Paper 1A Q14)
18
(HKDSE 2013 Paper 1A Q15)
= 3340 N
19
(HKDSE 2014 Paper 1A Q11)
WE 8830
=
= 900 kg
g 0 9.81
Conventional questions (p.390)
WM = WE 
 1 
= 8830  0.107 

 0.532 
9
4
(6050  105)3
3
2
B
m=
20
(a)
3340
W
aM = M =
= 3.71 m s2
900
m
10
A
11
A
12
C
a=
13
X
(X closer to the Moon)
(b)
By F =
GMm
,
r2
1A
X should be closer to the Moon
GMm mv 2
=
,
r
r2
GM
r= 2
v
GM
2π 2
2 πr
v = 2 πGM  1
T=
=
v
v3
v3
v
1
3
TX v X 3 vY
=
= 3
1
TY
vX
3
vY
to make the forces from the bodies equal
vY = vX
3
TX
=v
TY
3
1
= 0.79v
2
1A
in magnitude.
21
(c)
It would move towards the Earth.
(a)
Gravitational field strength
GM
= 2
r
6.67  10 11 (1.99  10 30 )
=
(2.28  10 8  10 3 ) 2
(b)
= 2.55  103 N kg1
GMm mv 2
By
=
,
r
r2
GM
v=
r
C
4
M =   r3
3
4
1A
B
By
14
The Earth is more massive than the
Moon,
1
W
1
=
=
= 1.5 m s2
m slope 20  0
30  0
1A
=
1A
1M
1A
1M
6.67  10 11 (1.99  10 30 )
2.28  10 8  10 3
= 2.41  104 m s1
1A
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
2 Force and Motion
(c)
Chapter 10 Gravitation
The orbital speed is 2.41  104 m s1.
2 πr
Period =
1M
v
2π(2.28 108 10 3 )
=
2.41 10 4
= 5.937  107 s = 687 days
22
(a)
(b)
(b)
The force acting on the satellite by the
planet is always perpendicular to the
(c)
1A
Orbital period = 4  70 = 280 days 1A
2 πr
Orbital speed =
1M
T
2π(2.0 1013 )
=
280  24  3600
displacement of the satellite.
GMm mv 2
(i) By
=
,
r
r2
rv 2
M=
G
(c)
1M
From the figure, when v = 4 km s1,
r = 4  104 km
4 10 4 10 3 (4 10 3 ) 2
M=
6.67 10 11
= 5.194  106 m s1
= 9.60  1024 kg
 5.19  10 m s
GMm mv 2
By
=
,
r
r2
rv 2
M=
G
2.0 1013 (5.194 10 6 ) 2
=
6.67 10 11
6
1A
1
1A
1A
The mass is 9.60  10 kg.
24
1M
= 8.09  1036 kg
1M
(ii)
1A
The black hole’s mass is 8.09  10 kg.
36
23
(a)
Weight = mg
1A
By Newton’s second law, F = ma
1A
(v halved)
ma = mg
a =g
(b)
Acceleration due to gravity
GM
= 2
r
6.67  10 11 (7.35  10 22 )
=
(1740  10 3 ) 2
= 1.62 m s2
(c)
Take downwards as positive.
1
By s = ut + at 2,
2
1
1.2 = 0 + (1.62)t 2
2
t = 1.22 s
24
(a)
1A
(a)
v
F
1M
1A
(i)
1M
(b)
(Tangential direction)
1A
and radius
1A
1A
(ii) (Towards Earth’s centre)
1A
2
mv 2
RE
(i) By mg =
and g = g0 2 ,
r
r
1M + 1M
1A
The planet’s mass
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
25
1A
g0
RE
r2
2
2
=
v
r
5
2 Force and Motion
Chapter 10 Gravitation
v =
g 0 RE
r
2
(d)
F/N
724
9.81(6370  10 3 ) 2
=
(6370  7000 )10 3
= 5460 m s1
1A
0
1
t1
The linear speed is 5460 m s .
(ii) Weight
= mg
= mg0
RE
r2
2
27
6370 2
= 1000(9.81)
(6370  7000 ) 2
= 2230 N
26
(a)
(a)
1A
gravitational force acting per unit mass
gM = gE
1A
(Smooth curve)
1
s = ut + at 2
2
1
= 0 + (9.81)12
2
1A
A
C
1M
O
For AOB being very small,
2
distance travelled by object in 1 s
M E rM
2
= AC  AB
2
1A
2
AO + AB = BO
2
(6370  10 ) + AB
3 2
1A
2
= (6370  103 + 4.91)2
1A
AB = 7910 m
surface of Mars is 0.4g.
Mass of Opportunity
W
1810
= E =
= 184.5 kg
g E 9.81
1A
B
1A
The gravitational field strength on the
(c)
1M
The speed of the object is 7910 m s1.
mv 2
By mg =
,
1M
r
v = gr
= 9.81(6370103 )
Weight on Mars
= 7910 m s1
= mg
The speed found by Newton’s law of
= 184.5  0.4  9.81
1A
universal gravitation is the same as the
answer to (b).
6
1M
M M rE
= 0.4g
= 724 N
(F increasing)
(b)
 1 
= g   22
 10 
(c)
1A
The distance fallen is 4.91 m.
in a gravitational field is the
(b)
(Correct labels with units)
= 4.905 m  4.91 m
The gravitational field strength at a point
at that point.
GM
M
g= 2  2
r
r
MM
2
2
gM
M r
r
= M = M E2
ME
gE
M E rM
2
rE
t/s
1A
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
2 Force and Motion
Chapter 10 Gravitation
2
28
(a)
W = mg = mg0
R0
r2
h = 4810  3400 = 1410 km
(i)
man
200
mg 0
= 3400
= 4810 km
W
100
r = R0
(b)
(c)
1M
weight mg0
1A
Earth
The height is much smaller than the
planet’s radius.
The centripetal acceleration of the
1A
satellite is g0 while that of the man is
mg 0  N
equal to
< g0.
1A
m
2π
By a = r  2 and T =
,

1
2π
T=

(for constant r)
1A
a
a
r
Therefore, the change in g is
negligible.
(ii)
1A
Gain in PE = loss in KE
1
mgh = mv2
2
2
32
v
g=
=
= 5 m s2
2h 2 ( 0 . 9 )
m=
(c)
W 200
=
= 40 kg
5
g
(a)
1M
1M
1A
 The period of the satellite is shorter.
30
(a)
Period = 24  3600 = 86 400 s
(b)
Angular speed of the men
2π
2π
=
=
= 7.27  105 rad s1 1M
T 86 400
By F = mr  2,
Centripetal force needed by X
= 60(6370  103)(7.27  105)2
1A
= 2.02 N
By mg = mr  2,
Centripetal force needed by Y
=
1A
1M
The mass of the planet is 8.67  1023 kg.
=
1M
1A
= 60[(6370  103) sin 40](7.27  105)2
g
r
= 1.30 N
9.81
6370  10 3
1A
(c)
=1.24  103 rad s1
2π
Period of satellite =

weight mg
1M
Earth
2π
=
1.24  10 3
= 5060 s
(b)
aT
1A
The mass of the object is 40 kg.
GMm
By F = W = 2 ,
1M
r
Wr 2
M=
Gm
200 (3400  10 3 ) 2
=
6.67  10 11 (40 )
= 8.67  1023 kg
29
normal
reaction N
1A
1M
2.02 = 60(9.81)  N
N = 587 N
= 24  3600
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
normal
reaction N
Centripetal force = mg  N
Period of man
= 86 400 s
X
1A
The scale’s reading is 587 N.
1A
7
2 Force and Motion
Chapter 10 Gravitation
(d)
(c)
The mass of Jupiter is 1.97  1027 kg.
GMm mv 2
By
=
,
r
r2
orbital speed
=
GM
r
=
6.67 10 11 (1.97 10 27 )
128 000 10 3
plane of orbit
= 3.20  104 m s1
(Passing through Y and Earth’s centre)
1A
31
(a)
GMm mv 2
2 πr
By
=
and T =
,
2
r
v
r
GMm m  2πr 
= 

r T 
r2
T2 =
32
(HKALE 2005 Paper 1 Q6)
33
(a)
1M
Speed is a scalar (is described by
magnitude only)
(b)
4π 2 r 3
GM
1M
2.39
9.41
38.3
208
r3 / 1025 m3
7.52
30.2
123
664
by both magnitude and direction).
1
(i) By s = (u + v)t,
2
1
 4.26 
3.6 = (u  0)

2
 2 
u = 3.4 m s1

200
180
projectile is 3.4 m s1.
vu
(ii) a =
t
0  3.4
=
2.13
= 1.60 m s2
1M
1A
1M
1M
1A
the Moon is 1.60 m s2.

(iii) The other time
100
200
300
400
500
600
r 3 / 1025 m3
(Correct labels with units)
1A
(Data points correct)
1A
(A correct straight line)
4π 2
Slope =
GM
10
(60  0)10
4π 2
=
25
(200  0)10
6.67 10 11 M
1A
M = 1.97  1027 kg
8
1M
The acceleration due to gravity on

(b)
1A
The initial vertical velocity of the
T 2 / 1010 s2
0
1A
while velocity is a vector (is described
2
T2 / 1010 s2
160
140
120
100
80
60
40
20
1A
= 4.26  0.90
1M
= 3.36 s
1A
(iv)
1M
1A
(Correct diagram)
3  1A
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
2 Force and Motion
Chapter 10 Gravitation
(v) This method is not valid on the
from the Moon has a larger magnitude
Earth
1A
because there is air resistance on
force points towards the Moon.
the Earth and the motion of the
At the point of intersection, the forces of
projectile will be affected.
(c)
(i)
than that from the Earth. The resultant
1A
gravity from the two bodies have the
Resultant initial velocity
same magnitude and their resultant is
= 2.0 2  3.4 2
zero.
= 3.94 m s1
3.4
tan  =
2.0
the force of gravity from the Earth has a
larger magnitude than that from the
1A
Moon. The resultant force points
The resultant initial velocity of the
1
projectile is 3.94 m s at 59.5
(e)
1A
the Earth and the Moon are in the
both values)
opposite direction as its motion.
1M
35
1A
Experiment questions (p.394)
acting on it during its flight, its
36
(a)
8
1M
Moment = Fd
GMm
2.1  1010 = 2  d
r
G (0.05 )(1.5)
(0.1)
=
0.045 2
The calculated value agrees with the
1A
The Moon’s mass is much less that the
Earth’s.
(b)
3.45  10 km
1A
The resultant field at this point is zero. /
The fields from the Earth and the Moon
balance each other at this point.
1A
Before reaching the point of intersection
1M
G = 5.67  1011 N m2 kg2 1A
6.67  5.67
Percentage error =
 100%
6.67
1A
5
1M
The two forces form a couple.
1M
= 0.04 N kg1
graph.
Let F the force between the large ball
and the small ball.
GMm
F= 2
r
1A
r = 1.0  10 km = 1.0  10 m
GM
g= 2
r
6.67 10 11 (6.0 10 24 )
=
(1.0 10 8 ) 2
5
1A
(HKDSE Practice Paper 2012 Paper 1B Q4)
As there is no horizontal net force
same as the vehicle.
(d)
It would require more energy
(Correct calculating method for
horizontal velocity remains the
(c)
1A
because both the forces of gravity from
moving Moon vehicle.
(b)
towards the Earth.
above the horizontal.
(ii) The projectile will land on the
(a)
1A
After passing the point of intersection,
1A
 = 59.5
34
1A
= 15.0%
(c)
Air current
(d)
(i)
1A
1A
Let T be the tension in the string, M
be the mass of X and m be the mass
of Y.
Consider the vertical direction.
indicated in Figure q, the force of gravity
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
9
2 Force and Motion
Chapter 10 Gravitation
T cos  = mg
(1)
1A
Consider the horizontal direction.
GMm
T sin  = 2
(2)
1A
r
(2)  (1),
GM
tan  = 2
gr
1A
The angle  is independent of m, so
he is incorrect.
1A
(ii) No,
1A
this is because the angle  is very
small ( 109 degrees).
1A
The error is very large in measuring
such a small angle.
1A
Physics in article (p.395)
37
(a)
Period = 24  3600 = 86 400 s
(b)
The centripetal acceleration is equal to
1A
the acceleration due to gravity.
r 2 = g
1M
2
R
 2π 
r   = g0 E2
T
r
 
2
1M
2
r =3
=3
g 0 RE T 2
4π 2
9.81(6370 10 3 ) 2 86 400 2
4π 2
= 4.222  107 m
Height above Earth’s surface
= 4.222  107  6370  103
= 3.59  107 m
(c)
1A
A geostationary satellite appears at a
fixed position in the sky.
1A
Satellite antennas on the ground need not
move and can point in a fixed direction
in communicating with the satellite. 1A
10
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015