Download gravitation - Physics Unit

GRAVITATION
SYNOPSIS
Gravitation is one of the four classes of interactions found in nature. In this chapter, we
shall investigate the basic law that governs gravitational interactions. This law is
universal : gravity acts in the same fundamental way between the earth and our body,
between the sun and the planet and among other planets. We’ll apply the law of
gravitation and gravitational field strength together with the concept of escape velocity to
phenomena such as the variation of weight with altitude, the orbits of satellites around
earth and the orbits of planets around the sun.
OBJECTIVES
At the end of this topic, students should be able to :
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Mm
.
r2
Define gravitational field strength as gravitational force per unit mass.
Derive and use the equation for gravitational field strength.
Explain the change of g with altitude, latitude and depth from the surface of the
earth.
Understand concept of weightlessness.
Define gravitational potential at the point in the gravitational. Derive and use the
GM
formula, V 
.
r
GMm
Use the potential energy formula, U 
. For situation close to the surface of
r
the earth, U  mgh where h << r.
Elaborate graphically variation of gravitational field strength and gravitational
potential with distance from the centre of the earth.
Derive and use formula for escape velocity,
GM
Explain satellite motion with velocity, v 
and period, T 
and use the
r
related equations.
State and use the Newton’s Law of gravitation, F  G
State Kepler’s Laws and use Kepler’s third Law,
1
T
3
2r 2
GM
.
8.1 NEWTON’S LAW OF GRAVITATION
In 1687, Sir Isaac Newton published his work on the universal law of gravitation, which
state that :
“ Every particle in the universe attract every other particle with a force that is
directly proportional to the product of their masses and inversely propotional to
the squre of the distance between their centres”.
If the particles have masses m1 dan m2 and are separated by the distance, r the magnitude
of gravitational force between them is :
F 
or
where
F=G
G
m1 m 2
r2
m1 m 2
…..(8.1)
r2
= Constant of universal gravitation
= 6.673 x10-11Nm2kg-2
m1
F12
F21
r
2
m2
Example 1 ( Thinking Physic 1)
A student drops a ball and consider :
“Why the ball falls to the ground as opposed to the situation in which the ball stays
stationary the Earth moves up to meet it.”
The student comes up with the following explanation :
“The Earth is much more massive than the ball, so the Earth pulls much harder on the
ball than the ball pulls on the Earth. Thurs, the ball while the Earth remains stationary”.
What do you think about this wxplanation ?
Explanation :
According to Newton’s Universal Law of Gravity, the force between the ball and the
earth depends on the product of their masses, so both forces, that of the ball on the Earth,
and that of the Earth on the ball, are equal in magnitude. This follows also, of course,
from Newton’s third law. The ball has large motion compared to the Earth because
according to Newto’s second law, the force gives a much greater acceleration to the small
mass of the ball.
Example 2 :
Calculate the gravitational attraction force between bodies of mass 3.0kg and 2.0kg
placed with their centres 50cm apart.
Solution :
From equation
F=G
m1 m 2
r2
= (6.67 x10-11)
(3)( 2)
(0.5) 2
= 1.6 x10-9 N
This is a very small force. To get an appreciable force one or both of the objects must be
very large. Our weight is the result of the gravitational attraction force from the Earth.
3
Example 3 :
Two particles of mass 0.2kg and 0.3kg are placed 0.15m apart. A third particle of mass
0.050kg is placed between them on the line joining the first two particles. Calculate :
a) the gravitational force acting on the third particle if it is placed 0.05m from the
0.3kg mass.
b) where along the line it should be placed for no gravitational force to be exerted on
it.
Solution :
a)
M1 = 0.2kg
M2 = 0.3kg
m = 0.05kg
r1 = 0.1m
r2 = 0.05m
0.15m
F1 = G
mM 1
r1
F2 = G
2
mM 2
r2
 (0.05)(0.2) 

= (6.67 x10-11) 
2

 (0.1)

-11
= 6.7 x10 N
Thus, the net force, F is given by :
F = F2 – F1
= 33.5 x10-11 N (towards M2)
4
2
 (0.05)(0.3) 

= (6.67 x10-11) 
2 
 (0.05) 
= 40.2 x10-11 N
b)
Suppose mass m is x from M1 and thurs (0.15 – x) from M2.
mM
mM 2
Then F1 = G 2 1 and
F2 = G
x
(0.15  x ) 2
For no gravititional force to act on mass m :
F1 = F2
Thurs
G
mM 1
x2
=G
mM 2
(0.15  x ) 2
0 .3
0.2
=
2
(0.15  x ) 2
x
Taking squre roots and cross multiplying gives :
2 (0.15  x ) = 3x
This gives x = 0.067m (from M1)
5
Exercise 1 :
m2
1.
0.4m
m1
0.3m
m3
Figure 8.1
Three 0.3kg billiard balls are placed on a table at the corners of right tringles, as
shown from overhead in figure 8.1. Find the net gravitational force on the ball
designed as m1 due to the forces exerted on it by the other two balls. [answer F =
7.65 x10-11N,  = 29.3o]
2.
Use the gravititional force law to find an approximate value for the mass of the
Earth. [ answer : ME = 5.98 x1024kg ]
3.
Calculate the gravitational attractive force between two particle, each of mass
20kg, placed 1.0m apart.
4.
Consider the Earth as a sphere of radius 6.4 x106m and mass 6.0 x1024kg. Find the
gravititional force on a mass of 5.0kg placed on the surface of the Earth. (Assume
the Earth can be replaced by a point mass acting at its centre). Compare this with
the weight of a 5.0kg mass on Earth.
5.
Two small sphere of mass 4.0kg and M kg are placed 80cm apart. If the
gravititional force is zero at a point 20cm from the 4kg mass along the line between
the two masses, calculate the value of M.
6.
The mass of the Earth is 6.0 x1024kg and that of the moon is 7.4 x1022kg. If the
distance between their centre is 3.8 x10m, calculate at what point on the line joining
their centres there is no gravititional forces. Neglect the effect of other planets and
the sun.
6
8.2 GRAVITATIONAL FIELD STRENGTH
8.2.1 Definition
A gravitational field at a point in space equals the gravitational force experienced by a
test mass placed at the point divided by the test mass.
g
F
m
8.2.2 Derivation and application of the equation, g = GM/r2
A body of mass m at a place on the earth’s surface where the acceleration of free- fall is g
experiences a force, F = mg ( i.e its weight ) because of its attraction by the earth.
m
F = mg
r
Figure 8.2.1
Assuming the earth behaves as if its whole mass m were concertrate at its center, then by
the law of gravitational, we can also say that F is the gravitational pull of the earth on the
body, so
F
GMm
r2
where r is the radius of the earth
mg 
g
GMm
r2
------ 8..2.1
GM
r2
7
We can arrage this equation and find the mass of the erth, ME,
2
gR E
ME 
G
Example 4 :
a) The falling baseball is close enough to the Earth so that the distance of seperation
between the center of the ball and the center of the Earth can be taken as the radius of the
Earth, 6.38 X 106 m. Thus, the mass of the Earth is,
Solution:
(9.81m / s 2 )(6.38x106 ) 2
Me 
 5.98  1024 kg
11
2
2
6.67 x10 Nm / kg
8.2.3 Variation of g with altitude (height)
a) On the earth’s surface ( r=R )
where R = radius of the earth
GM E
2
RE
b) Above the earth’s surface ( r > R )
g
From the equation
GMm
(r  R )
r2
Above the earth’s surface, the acceleration of free-fall varies universely as the
square of the distance, r from the center of the earth. (i.e.it decreases with height ).
mg 
8
Example 5 :
a) What is the effective value of g at a height of 1000 km above the earth’s surface?
That is, what is the acceleration due to gravity of objects allowed to fall freely at this
altitude?
Solution:
Given ME is 6x1024 kg
re replaced by r = 6.38 x 106 + 1000 x 103
= 7380 x 103 m
= 7.38 x 106 m
g=
(6.67 x10 11 Nm2 / kg 2 )(6x10 24 kg)
(7.38x10 6 m) 2
= 54.23 x 106
b) Assuming that the earth is a uniform sphere of radius 6.4x106 m and mass 6.0x1024
kg, find the gravitation field strengh g at a point
(i)
on the surface
(ii)
at height 0.5 times its radius above the earth
Solution:
a)
g
=
GM
r2
(6.7 x10 11)(6x10 24 )
(6.4 x10 6 ) 2
= 9.8 ms-2
b)
r1 = 1.5r
g1 r 2
r2


g r12 (1.5r ) 2
=
(6.4 x10 6 ) 2
(1.5x 6.4x10 6 )
= 0.444
g1 = 0.444 x 9.8 = 4.35 ms-2
9
8.2.4 Variation of g with depth ( r < R )
Mass of the earth, m = V
4

=  r 3 
 3.

From the equation g 
(1)
GM
(2)
r2
Insert (1) into (2),
4

G r 3 
3

=
2
r
4 
g = G  r 
3 
------- 3
Equation (3) shows that g decreases porpotionally with depth, r.
r=R
r<R
g=
r>R
GM
r2
g
1
r2
r
Figure 8.2.3 : illuatrates both thr relation of g with altitude and depths
10
8.2.6 Variation of g with latitude
g3
r3
o
r2
 r1
g2
g1
R
Fc
Fg
Figure 8.2.4
r1 > r2 > r3
g1 < g2 < g3
r = radius
g = gravitational force
The earth is not perfect sphere.
A body of mass m at any point of the earth’s surface ( except at the poles ) must have a
centripetal force acting on it ( i.e. Fc )
This force is supplied by part of the earth’s gravitational attraction for it ( i.e. Fg )
a) A stationary earth
The gravitational pull of the earth on m would be mg where g = acceleration of freefall.
b) The earth under rotation
Because of the earth’s rotation, the observed gravitational pull of the earth on a body
of mass, m equals to mg where go = observed acceleration of free fall.
The centripetal force, Fc = mg - mgo
(1)
c) At the equator
A body of mass, m is moving in a circle of radius, r ( r = radius of the earth ) and has
the same angular velocity,  as the earth,
The centripetal force, Fc = m2r
Compare equation (1) and (2),
mg – mgo = m2r
g – go = 2r
d) At the poles
g = go and  = 0
11
(2)
e) At the latitude  :
A body of mass, m describes a circle of radius r cos . The magnitude of the
centripetal force,
Fc = m2r cos  ( smaller than at the equator )
o
m2r cos 

Q
r kos 
r

P
mg
mgo
mgr

0
R
Figure 8.2.5
The experienced weight, R = m g 2  r 2 4 kos 2   2gr  kos
8.2.7
Weightlessness
a) An Astronaut orbiting the earth
An astronaut orbiting the earth in a space vehicle with its rocket motors off is
said to be weightless. The astronaut is moving with a constant speed along the
orbit but since he is travelling in a circle he has a centripetal acceleration of
the same value as that of the space vehicle and equal to g at that height.
The wall of the vehicle exert no force on the astronaut, he is un supported, the
weight experience disappear s and he floats about ‘ weightless’. Any object
release in the vehicle does not fall.
b) A passenger in a lift
When a lift starts upward the push of the floor on our feet moves downwards
the push of the floor on our feet descreases and we feel lighter until at one
stage where our feet are completely unsupported we will experience
weightlessness.
12
When the lift moves downward :
From Newtons 2nd Law of motion,
mg – R = ma
R = mg – ma
R=m(g–a)
If a = 0, R = mg
If a increases, the weight experience by R decreases
If a = g ( free-fall), R = 0 ( i.e. weightlessness )
When the lift moves upward :
R=m(g+a)
If a increases, the weight experience by R also increases.
i)
ii)
iii)
a=g
N=0
mg N
F  0
w = N = mg
( true weight )
mg
N
mg
Figure 8.2.6 : (i), (ii) and (iii)
F  ma
w’ = N = 0 ( weightless )
Descending with a = g
mg – N = ma
w=N=m(g–a)
Less than true weight
Descending with acceleration a < g
Figure(i) : In a stationary elevator a scale reads true weight. The weight reading
in the reaction force N of the scale on the person.
Figure(ii) : The elevator is descending with an acceleration a < g, the reaction
force and apparent weight are less than the true weight.
Figure(iii) : In the elevator were in free-fall , a = g, the reaction force and
indicated weight would be zero since the scale would be falling as
fast as the person.
Other examples of weightlessness : A ballet dancer and roller skater ‘ flying’ in the air.
13
8.2.8 Gravitational Potential Energy
We know that the earth’s gravitational force on a body of mass m at any point outside the
earth is given normally by,
FG 
Gm E m
r2
where
mE = the mass of the earth
r = the distance of the body from the earth’s center.
FG = earth gravitational force which is a constant in
magnitude and direction.
For problem in which r changes enough that the gravitational force can’t be considered
constant, we need a more general expression for gravitational potential energy.
m
F
r2
curved path
r1
straight path
Me
Figure 8.2.7
To find this expression, we consider a body of mass m outside the earth, and first
compute the work WGrav done by the gravitational force when the body moves directly
away from or toward the center of the earth from r = r1 to r = r2 as in Figure 8.2. This is
given by
WGrav 
r2
 Fr dr
, ------------------ 8.2.5.1
r1
where Fr is the radial component of the gravitational force
the outward from the center of the earth.
14
F , that is the component in
Because F points directly inward toward the center of the earth, Fr is negative. The
magnitude of the gravitational force, by a minus sign ;
Fr 
 Gm E m
r2
,
------- 8.2.5.2
Substituting equation (8.2.5.2) into equation (8.2.5.1), we see that Wgrav is given by
r2
1
1
r2
Wgrav  Gm E mr
Wgrav 
dr
Gm E m Gm E m
,

r2
r1
------- 8.2.5.3
The path doesn’t have to be a straight line, it could also be a curve like the one in Figure
8.2.5. Work depends only on the initial and final values of r, not on the path taken.
From the work-energy theorem,we know that,
W = U2 – U1 ,
------- 8.2.5.4
From equation (8.2.5.3),
Wgrav 
Gm E m Gm E m

r2
r1
when r2 = 
Gm E m
,
r1
Comparing equation (8.2.5.4) and (8.2.5.5)
W0
------- 8.2.5.5
 Gm E m
r1
And r1 = r, U1 = U,
U1 
Therefore U 
 Gm E m
= potential energy
r
15
Change of Potential Energy (P.E) over small distances.
m
U1
U1 
U2 
 Gm E m
r  h 
U2
 Gm E m
r
h
r
r
mE
Figure 8.2.8
Suppose the small mass is moved away from the earth by a distance h, as in Figure 8.2.8.
The change of P.E is given by
  Gm E m    Gm E m 
U  


r
 rh  

1
1 

= Gm E m 


r
r

h


 r  h   r 
= Gm E m

 rr  h  
= Gm E m
h
r r  h 
16
If we assume h is much smaller than r, h < < r (for situation close to the surface of the
earth)
U  Gm E m
= Gm E m
=
Gm E m
r2
h
;h0
r  rh
2
h
r2
h
And we know ,
g
Therefore,
Gm E
r2
U  mgh
So we can only use the ‘mgh’ formula for P.E change if h is much less than r.
This amounts to the same as assuming g does not change significantly over distance h.
Example 6 :
Three masses are in a configuration as shown in Figure below. What is their total
gravitational potential energy ? Given G = 6.7 x 10-11 Nm2 kg-2.
y
m3 = 2kg (0,4m)
x
M1 =1kg , (0,0)
m2 = 2kg ,(3m,0)
17
Solution :
U  U12  U13  U23
 Gm1m 2 Gm1m3 Gm 2 m3


r12
r13
r23
=
  12 12 22 


= 6.67 x1011 
 1.3x1010 J

4
5 
 3
Example 7 :
A satellite of the earth has a mass of 100kg and is at an altitude of 2.00 x 10 6 m. What is
the potential energy of the satellite-earth system. Given mE = 6.0 x 1024 kg, rE = 6.4 x 106
m and G = 6.67 x 10-11 Nm2 kg-2.
Solution
U
 Gm E m
r
=

 6.67 x10 11100 6.0x10 24

6.4x106  2.00x106
=  4.76x109 J
Exercises 2 :
1. The gravitational potential difference between two points is 3.0 x 103 J/kg.
Calculate the work done in moving a mass of 4.0 kg between the two point.
(1.3 x 104 J)
2. The moon has mass 7.7 x 1022 kg and radius 1.7 x 106 m. Calculate the work
needed to completely remove a 1.5 x 103 kg space craff from its surface into outer
space. (4.53 x 109 J)
3. How much energy is required to move a 1000kg mass from the Earth’s surface to
an altitude h = 2RE.(6.25 x 1010 J)
18
8.2.9 Gravitational potential,V
Gravitational potential is the potential energy per unit mass of a small test mass in the
 Gm E
 Gm E m
field. Therefore potential at r is equal to
which equals
.
mr
r
The gravitationalpotential V near a planet of mass M is given by
 GM
r
where r = distance from the planet’s centre.
V
Figure 8.2.5.3 shows how the gravitational potential near a planet or other spherical mass
 GM
varies with distnace from its centre. The potential at the surface is equal to
.
R
The curve is a 1/r curve, not an inverse square curve.So the potential at distance 2R from
the centre is 0.5 x the surface potential, the potential 3R from the centre is 0.33 x the
surface etc.
(Earth’s radius)
0
R
Distance r from Earth’s centre(x 106 m)
10
20
-20
-40
-60
Gravitational potential(MJ/kg)
Figure 8.2.9 : Variation of gravitational potential with distance from the
centre of the eatrh
19
Example 8 :
1. Show that the gravitational potential at the surface of the earth is –63 MJ/kg,
given the surface gravitational field strength is 9.8 N/kg and the mean radius of
the earth is 6400 km.
Solution :
V
U
U
= - 63 x 106 J/kg
m
 Gm E m
Gm
, g  2 E = 9.8 N/kg
r
r
GmE = gr2 = 9.8 r2
Therefore
U
m
 Gm E m 1
=
r
m
2
 gr
=
r
= - 9.8 r
= - 9.8(6.4 x 106)
= -63 x 106 J/kg.
V
20
Exercises 3 :
1. A diagram below represents two points X and Y at distance L and 2L from the
centre of the earth. The gravitational potential at X is unknown. When a 1 kg
mass is taken from X to Y the work done on the mass is +4kg.Calculate the
gravitational potential at X.(- 4 x 103 J/kg)
Earth
X
Y
L
2L
2. a) Difine gravitational potential.
b) Find the gravitational potential 100 km above the earth’s surface.(-5.4 x 107
J/kg)
3. Assuming that the earth is a uniform sphere of radius 6.4 x 106 m and mass 6.0 x
1024 kg, calculate
a) the gravitational potential at
I ) the earth’s surface(-6.3 x 107 J/kg)
ii) a point 6.0 x 105 m above it(-5.7 x 107 J/kg)
b) the work done in taking a 5.0 kg mass from the earth’s surface to a point
6.0 x 105 m above it (2.7 x 107 J/kg)
c) the work done in taking a 5.0 kg mass from the earth’s surface to a point
where the earth’s gravitational effect is negligible.(31 x 107 J/kg)
21
8.3 DERIVATION AND APPLICATION OF ESCAPE VELOCITY
Escape velocity
Suppose we want to project a mass upward with sufficient speed so that it will escape the
gravitational clutches of the earth altogether and travel out into the solar system, never
refering to earth.
This minimum speed that a mass must be have, called the escape velocity.
The escape velocity is obtained from the fact that the potential energy gained by the body
equals its loss of the kinetic energy, if air resistance is neglected.
The work done measures the energy transferred. Let m be the mass of the earth when it is
distance x from the centre of the earth is
F
GMm
x2
Therefore work done by gravity when the body moves a further short distance upwards
is
W  Fx
 GMmx

x2
(the –ve sign shows the force acts in the opposite direction to the displacement)
Therefore total work done while body escapes

W  
r
Mm
dx , (r = radius of earth)
x2

 1
W  GMm  
 x r

1
W  GMm  
 x r

GMm
r
22
If the body leaves the earth with speed v and just escapes from its gravitational field,
1
GMm
mV e 2 
2
r
Ve
But,

2GM
r
GM
g 2
r
Ve  2gr
=
Subtituting, r = 6.37 x 106m and g= 9.81 ms-1; we get v= 11.2 km/s
An object must have a speed of fast over 11.2 km/s to escape from the earth into outer
space. (neglect air friction)
Example 9 :
Calculate the escape velocity from the earth for a 5000kg spacecraft, and determine the
kinetic energy it must have at the earth’s surface in order to escape the earth’s
gravitational field.
Solution :
2GM E
, with
RE
Using equation, Ve 
M E  5.98x10 24 kg
R E  6.37 x10 6 m
Ve 
2(6.67 x1011)(5.98x1024 )
6.37 x106
 1.12x104 ms 1
The kinetic energy of the spacecraft is,
K
1
1
mVe2  (5.00x103 )(1.12x104 ) 2
2
2
 3.14x1011J
23
Example 10 :
The moon’s radius is 1600km and its surface gravity is 1.6 N/kg. Calculate the escape
velocity from the lunar space.(2.3 x 103 m/s).
Solution :
R= 1.6x106m, g = 1.6Nkg-1
To calculate the escape velocity,
We use, Ve = 2gr
V e  2 x1.6 x1.6 x106
 2.3x103 ms 1
Exercises 4 :
1. The moon has mass 7.7 x 1022 kg and radius 1.7 x 106 m. Calculate the work
needed to completely remove a 1.5 x 103 kg space craft from its surface into outer
space. Neglect the air friction.(4.5 x 109 J).
2. Calculate the minimum speed which a body must have to escape from the moon’s
gravitational field, given that the moon has 7.7 x 1022 kg and radius 1.7 x
106m.(2.5 x 103 m/s)
3. A planet has radius 5.0 x 105 m and mean density 3.0 x 103 kg/m-3. Calculate the
escape speed of bodies on its surface.(6.47 x 102 m/s)
24
8.4 SATELLITE MOTION IN A CIRCULAR ORBIT
Satellite objects which are in orbit around a larger mass as a results of gravitational
attraction.
For example :
1. Moon is a satellite of the earth.
2. Our mother earth are satellites of the sun
earth
r
moon,m(satellite)
Figure 8.4.1
Gravitational attraction provides the centripetal acceleration and force.
See Figure 8.4.1, a satellite of mass,m in circular orbit of radius, r of an object(planet) of
mass,M.
The gravitational force that act :
Fg 
GMm
;
r2
Fg - gravitational force
G – Universal gravitational constant
M – mass of the planet
m- mass of the satellite
r – radius from the centre of the planet to the satellite
------ 8.4.1
the gravitational force provides the centripetal force that required by the satellite
mv 2
Fc 
;
------ 8.4.2
r
m- mass of the satellite
v – orbital velocity
r – radius from the centre of the planet to the satellite
25
Thus, in circular motion, we can get the period of the roration, T by :
2r
T
------- 8.4.3
v
These three equations are useful for us to solve problems on satellite in orbit.
First steps:
Equating 8.4.1 and 8.4.2
Fg=Fc
mv 2
r
r2
we will get the orbital speed, v;
GM
v
r
GMm

------- 8.4.4
The orbital speed is an only speed that a satellite can have if the satellite is to remain in
an orbit with fixed radius.
Consequently, for a given orbit, a satellite with large mass has exactly the same orbital
speed as a satellite with a small mass.
By substituting v 
2r
from equation 8.4.3,
T
2r
GM

T
r
T
3
2
2r
T2 
GM
4 2 r 3
GM
------- 8.4.5
Since G and M are constant then T2  r3 –this is Kepler’s Third Law and can be applied
to any satellite in orbit around a massive body.
26
DO YOU KNOW…..
1. Once a satellite in orbit at a correct speed, it continues in uniform circular motion
forever, assuming that effects such as friction due to residual atmosphere do not
reduce the speed.
2. One of the applications of satellites is a network of satellites called Global
Positioning System (GPS), which is use to determine the position of an object on
the earth’s surface within 15m or less.
3. An important application of equation 8.4.5 occurs in the field of communications,
where “synchronous satellite” are put into a circular orbit that I sin the plane of
the equator.
Receiver
Satellite
Station
Figure 8.4.2
The period of the satellite is chosen to be one day, which is also the time it takes for the
earth to turn once about it axis.
Therefore, these satellitemove around their orbits in a way that is synchronized with the
rotation of the earth.
27
Example 11 :
Determine the speed of the Hubble space telescope orbiting at a height of 596km above
the earth’s surface.
(RE = 6.38106m, ME = 5.981024kg)
Solution :
GM
r
v
6.67  10 5.98  10 
11
v
24
6.98  106
v  7.56  103 ms 1
Example 12 :
Satellites which orbit the earth with a time period of 24 hours are used for
communication purposes since they appear stationary above a given point on earth.
Calculate the height of such a satellite above the earth’s surface.
(RE = 6.4106m, ME = 6.01024kg, G = 6.6710-11Nm2kg-2)
Solution :
T = 246060 = 86400s
From,
4 2 r 3
T2 
GM
2
T GM
r3 
4 2
r
3
864002 6.67  1011 6.0  1024 
4 2
 r  4.23  10 7 m
since the earth has a radius of 6.4106m
then ;
the height of the satellite above the earth surface :
r’ = 4.23107 - 6.4106m
= 3.59107m
28
Example 13 :
Use the following data to calculate the time in earth years for Mars to orbit the sun.
Average radius of Earth’s orbit, R = 151010m
Average radius of Mars’s orbit, r = 231010m
Solution :
From,
T2 
4 2 r 3
GM
T2  r3
Let t = time in earth years.
T12 r13
 ;
T22 r23
T1 is the orbit time of Earth.
T2 is the orbit time of Mars
r1 is the radius of Earth’s orbit
r2 is the radius of Mars’s orbit.
T22
r13
 3  T12
r2
23  10 

15  10 
10 3
T2
10 3
1
 T2  1.9 of earth years
29
Exercise 5 :
A spacecraft travels in a circular orbit around the moon at a speed of 1650ms-1. What is
its mean distance above the surface of the moon?
(mass of the moon, mm = 7.341022kg, radius of the moon, rm = 1.74106m, G = 6.671011
Nm2kg-2)
(58km)
Exercise 6 :
One of Jupiter’s moon is called Ganymede. The radius of Ganymede’s orbit around
Jupiter is 2.8 times that of the moon around the Earth. The mass of Jupiter is
approximately 300 times of the Earth.
Determine the ratio of the velocity of Ganymede to the velocity of the moon.
(10)
30
8.5 PLANETARY MOTIONS
In early history, scientists regarded the Earth as the centre of the universe.
This hyphothesis is known as geocentric model.
In 1543, Nicolaus Copernicus (1473 – 1543 ), showed that the Earth and the other
planets revolve in circular orbits about the sun, known as heliocentric hyphothesis.
Johannes Kepler ( 1571- 1630 ), the German astronomer is best known for developing
the laws of planetary motions based on the careful observations of Tycho Brahe ( 15461601 ).
Kepler generalized the mathematical analysis made by Tycho to include the motions of
all planets in three statements known as Kepler’s Laws.
8.5.1 Planetary Motions – Kepler’s Laws
Kepler’s Laws describe the orbital motion of the planets round the sun.
Newton showed that the same laws apply to the motion of any satellite.
Kepler’s First Law : Each planet moves in an elliptical orbit with one of the two focal
points of the ellipse. Some of the planets move in orbits which are almost circular, with
the sun at the centre.
Kepler’s Second Law : The line from the sun to each planet sweeps out equal areas in
equal time.
S2
S1
S1 = S2
Figure 8.5.1
31
Kepler’s
Third Law : The ratio of
r3
is the same (constantly) for all the
T2
planets, where
r = average radius of planets from the sun.
T = time taken for one complete orbit.
The proof of Kepler’s Third Law
Velocity, v
Sun,
(mass, M)
rrrrrr
rrrr
Fc
Planet of mass,m
r
Figure 8.5.2
Since the planet follows the circular orbit of radius, r at steady speed, v the
centripetal acceleration of the planet,
v2
a=
.
r
v2
The centripetal force, Fc = ma = m
.
r
The centripetal force is provided by the force of gravitation,
v2
GMm
Fc :
= m
.
r
r2
Fg = Fc
v2
GMm
=
m
r
r2
32
 v2 =
GM
r
(1)
For steady speed, the time ,T to make one complete orbit is given by the equation,
v=
2r
.
r
 2r  GM
=
r
 r 
Therefore, from (1) : 
r 3 GM
= constant

T 2 4 2
or
4 2 r 3
T 
GM
r3
T  2
GM
2
Example 8.5.1
1.
The moon has an orbital time period of 27.3 days. Calculate its average
distance from earth. Use g E = 9.81 Nkg-1 and RE = 6400 km from the earth’ s
surface gravity and radius respectively.
Solution :
Let r = the average distance from earth to the moon.
4 2 r 3
Use Kepler s Third Law : T 
GM
’
2

3
r =
GM 2
T ,
4 2
gT 2
 r =
4 2
33
g
GM
 gr 2 = GM
2
r
9.81x 27.3x 24 x 60 x 60 
=
4 2
2
8
= 3.8x10 m.
Example 8.5.2
Use the following data to calculate the time in earth years for Mars to orbit the sun.
Average radius of Earth’s orbit, R = 151010m
Average radius of Mars’s orbit, r = 231010m
From,
T2 
4 2 r 3
GM
T2  r3
Let t =time in earth years.
T12 r13
 ;
T22 r23
T1 is the orbit time of Earth.
T2 is the orbit time of Mars
r1 is the radius of Earth’s orbit
r2 is the radius of Mars’s orbit.
T22 
r13
 T12
3
r2
23  10 

15  10 
10 3
T2
10 3
1
 T2  1.9 of earth years
34
Additional Exercise :
1.
The first artificial satellite to orbit the earth was Sputnik I which had
time period of 96 minutes. Calculate its average height above the earth. Use the same
values for g and R as in the previous example.
( use g = 9.81 Nkg-1 , R = 6.4 x 106 m )
( h = 500 km)
2.
Communications satellites orbit the earth with a time period of 24 hours, in orbits
which make them appear stationary when seen from the earth surface. Their orbits are
synchronous orbits because they move around at the same rate as the earth. Use Kepler’s
Third Law to calculate of a communications satellite.
( h = 35700 km )
3.
Satellite dishes do not have to change directions in order to stay focused on a
signal from a satellite. This means that the satellite always have to be found at the same
location with respect to the surface of the earth. In order for this to occur, the satellite
must be at the height such that its revolution period is the same as that of the earth, 24
hours. At what height must a satellite be to achieve this ?
( RE = 6.4 x 106 m, ME = 6.0 x 1024 kg )
( h = 3.59 x 107 m )
Bibliography/References:
1.
“ Understanding Physics for Advanced Level “ 4 th. Edition (2000) Jim
Breithaupt.
2.
“ College Physics “ (1999) Raymond A Serway and Jerry S. Faughn. 5 th.
Edition. College Publishing.
3.
“ University Physics “. Sears and Zemansky’s.
4.
Raymond A. Serway & Jerry S. Fauglin (1999), “College Physics”,
5th.Edition, College Publishing.
5.
Physics For A-Level
6.
Calculation For A Level Physics.
35