Download Gravitational Force is

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By Mrs Lim
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What do you know about
gravitation?
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• It caused an apple (lucky not durian!) to
fall on Newton’s head about 400 years ago.
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 4 Jan 1643 – 31 March 1727
 In Manor House at Woolsthorpe, Lincolnshire
 Educated at Grantham Grammar School,
Lincolnshire
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cool isn’t it?
A one pound note with Newton’s portrait!
(Imagine one with your portrait)
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Gravitational Force is ….
 attractive in nature.
 exerted by stars and planets. (As well as
the apple)
 Gravity is stronger on Earth than on
Moon.
 Gravitational effects diminishes with
distance….
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1.2 Newton’s Law of Gravitation
The Newton’s law of gravitation states that every
particle in the universe attracts every other
particle with a force which is directly proportional
to the product of their masses and inversely
proportional to the square of their distance apart.
Gm1m2
F
r2
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F
F
m1
m2
r
1. - equal and opposite attractive force on each other
2. - directed along the line joining the two particles.
3. - for a sphere of uniform density, the force that it
exerts on other objects can be obtained by treating
the sphere as a point mass at the centre of the
sphere.
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F
F
m1
m2
r
Qn: What should we do if not spherical or not point
mass?
Ans:
The formula is still applicable if the objects are placed
sufficiently far apart such that their sizes become
negligible compared to the distance separating them.
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Quick Check 1
Earth and the Moon are gravitationally
attracted to each other. Does the more
massive Earth attract the moon with
(a) a greater force
(b) the same force
(c) or less force
than the moon attracts Earth?
Ans: (b)
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Quick Check 2
The weight of an object on peak of
Himalayas ( about 8848 m above sea level)
is
(a) the same as
(b) slightly smaller than
(c) much smaller than
the weight of that object at sea level.
Ans: (b)
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Quick Check 3

A)
B)
C)
D)
On the ground the gravitational force on
an object is W. What is the gravitational
force at a height R, where R is the radius
of the earth?
0.25W
0.5W
W
2W
Ans: [ A ]
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Explanation:
Gm1m2
W 
;
2
R
Gm1m2 W
W' 

 0.25W
2
(2R)
4
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Worked Example 1 (in tutorial worksheet)
Assuming that they can be treated as point masses,
find the gravitational force between
(a) two persons of masses 60 kg and 1.0 m apart,
(b) a person of mass 60 kg at the surface of the
Earth, and the Earth of mass 6.0  1024 kg and
radius 6400 km.
(a)Force =
G m1 m 2
r2
(6.67  10 11 ) (60 ) (60 )
=
(1) 2
= 2.4  10–7 N
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Example 1
Assuming that they can be treated as point masses, find
the gravitational force between
(a) two persons of masses 60 kg and 1.0 m apart,
(b) a person of mass 60 kg at the surface of the Earth,
and the Earth of mass 6.0  1024 kg and radius 6400 km.
11
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(
6
.
67

10
)
(
6
.
0

10
) (60)
G m1 m 2
(b) Force =
=
6 2
2
(
6
.
4

10
)
r
=
590 N
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TWO NEW CONCEPTS
• Gravitational Field
• Gravitational Field Strength
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Gravitational field is a region in which a mass
experiences a gravitational force.
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2.2 Gravitational Field Strength
The gravitational field strength at a point is
defined as the force per unit mass acting at that
point.
Field Strength at a point g =
FG /m
Unit: N kg-1
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Field Strength of the gravitational field
of a point mass or spherical mass
M
r
P
Gravitational force experienced by a small mass ms
placed at point P is
G M ms
F
r2
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Field Strength of the gravitational field
of a point mass or spherical mass
Field strength at point P is g
=
F
ms
 G M ms 
 2 

=  r
ms
GM
=
r
2
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Field Strength of the gravitational field of a point
mass or spherical mass
Gm1m2
F 
r2
g
GM
r2
• Minus sign indicates attractive nature of the force
(rather than repulsive).
• NEED NOT be included during calculations
• Usually included in graphical representation of F and g.
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Gm1m2
Force, F 
2
r
Field Strength,
g
GM
r
2
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Quick Check 4
(a)
(b)
(c)
The gravitational field generated by (or
due to) the more massive Earth will be
a greater
the same
or less
than the gravitational field generated by
the moon?
Ans: (a)
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Quick Check 5
The gravitational field experienced by an
object on the peak of Himalayas ( about
8848 m above sea level) due to the Earth is
(a)
(b)
(c)
the same as
slightly smaller than
much smaller than
the gravitational field experienced by the
object at sea level.
Ans: (b)
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Quick Check 6
Karen’s mass is 50 kg and Kenny’s mass is
75 kg. They are both sitting in the TPJC
Auditorium.
(Considering
only
their
interaction with Earth) They will experience
the same
A. Gravitational force
B. Gravitational field strength
Ans: (B)
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Worked Example 2 (in tutorial)
The mass and radius of the Earth is 6.0  1024 kg and
6.4  106 m respectively.
(a) Find the gravitational field strength at
(i) the surface of the Earth,
(ii) 3000 km above the Earth’s surface.
(b) What is the assumption made in the calculations?
(a)
(i) At Earth’s surface, distance from centre of Earth,
r = radius = 6.4  106 m
G M (6.67 1011 ) (6.0 1024 )
g 2 
 9.8 N kg–1
6 2
r
(6.4 10 )
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(a) (ii) At a height 3000 km above Earth’s surface, distance
from centre of Earth,
r
= 6.4  106 + 3000  103
= 9.4  106 m
g
GM
r2

(6.67  10 11 ) (6.0  10 24 )
(9.4  10 6 )2

4.5 N kg–1
(b) In our calculations, we have assumed Earth as a
uniform sphere.
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Quick Check 7
If a third object sits between two massive
objects like (say when the moon orbits round
the earth, it will experience the Sun’s
gravitational field as well. ) What will be the
total gravitational field? How should they be
added?
(A) as a scalar
(B) as a Vector
ANS : (B)
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 look at worked example 3 in tutorial
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2.5 Graphical representation of Earth’s
field strength
Earth
M
field strength g
RE
distance r
g

GM
RE
The formula
r2
Why?
(see appendix 6)
2
g
GM
GM
r2
is only valid for r > RE
For r < RE the relation is linear?
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2.6 Acceleration of free fall
Field Strength
g
=
F/m
Acceleration of a free falling object a
So
a
=
=
F/m
g
Gravitational field strength of the Earth is 9.81 N kg-1 (or
9.81 m s-2) near its surface and is approximately
constant due to the large size of the Earth.
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SUMMARY
Gm1m2
Force, F 
r2
Field Strength at a point g =
Field Strength,
g
GM
r2
FG /m
(Definition)
(M being a point mass or
spherical mass)
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A closer look at
gravitational potential
energy . . .
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What does gravitational potential energy depends on?
• It depends on the mass of the object
• It depends on the position
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3.2 Formal Definition of Gravitational
Potential Energy
The gravitational potential energy of a mass at a
point is defined as the work done by an external
agent in bringing the mass from infinity to that
point.
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displacement
M
P
gravitational
force
external
force
m
at infinity
• The work is done by an external force acting in a
direction opposite to the gravitational force.
• The direction of external force is opposite to the
displacement of the mass. Hence the work done by
the external force is negative.
• There is no change in velocity or kinetic energy.
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It can be shown that for two point or spherical masses
attracting each other with gravitational force (look at
lecture notes)
Potential Energy U

GMm

r
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Potential Energy 
GMm

r
or Potential Energy = mgh ?
Oops so which equation is correct?
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Lets Explore:
A 3-kg mass is projected to a height h above the Earth’s
surface. Use U = mgh and to find the change in
gravitational potential energy if
(i)
h = 1000 m
(ii)
h = 1000 km
Radius of the Earth is 6400 km; Mass of Earth = 6 x 1024 kg.
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(i)
Using U = mgh,
U
Using U
U
= 3 x 9.8 x 1000


=
2.94 x 104 J
GMm
r

1
1 
  GMm 


R

1000
R
E 
 E
1
1


  6.672  10 11  6  10 24  3 

3
6400  103 
 6401  10
 2.93  10 4 J
So there is not much difference between the
two formulae for h = 1000 m.
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(ii) Using U = mgh
U
Using U
U
=
3 x 9.8 x 1000 000


=
2.94 x 107 J
GMm
r

1
1 
  GMm 


3
R

1000

10
R
E
 E
1
1


  6.672  1011  6  1024  3 

3
3
 7400  10 6400  10 
 2.54  106 J
This time there is a significant difference.
U = mgh is inaccurate in this case.
Why?
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This is because the acceleration due to free
fall (in other words “g” on the object) is
not constant, hence the equation mgh will
not be accurate.
(see “A” level Physics is DIFFERENT from
“O” level)
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3.3 PROJECTION OF A BODY FROM
SURFACE OF A PLANET
low projection
velocity
high projection
velocity
The maximum height reached is obtained by equating
PEgained
= ½mv2 - 0
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Graphical Representation
Potential
Energy
EK
RE
r
0
U
ET = Ep + Ek

GMm

r
dist. from
centre of
Earth
EK
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Condition for Escape to Infinity
v
to infinity
rocket
Earth
k.e.  0
k.e. = ½mv2
p.e. = 0
p.e. = -GMm/r
Note that at infinity, total energy

0
By conservation of energy, the condition for escape is . . .
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K.E. + P.E.
0
i.e.
½mv2 + (-GMm/r)

0

v

2G M
r
So the escape speed is
v 
2GM
r
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SUMMARY
• Increase in GPE is work done against gravity.
• Formal definition of GPE (must remember).
For two point or spherical masses attracting each other
with gravitational force,
Potential Energy U

GMm

r
• This formula gives the absolute potential energy with
infinity set as the reference point (i.e. GPE = 0)
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4 GRAVITATIONAL POTENTIAL 
The gravitational potential at a point is defined as the
work done per unit mass by an external agent in bringing a
mass from infinity to that point.
Based on this definition, we get  
U
m
and hence for point mass or spherical mass,

GM

r
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For example, we use
a 1 kg mass….
W.D. = - 6 J
U = - 6 J
 = -8 J kg-1
Infinity
W.D. = - 2 J
U = - 2 J
OR
=0
OR
 = - 2 J kg-1
How will the workdone
change if the mass of the
object is 2 kg?
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Worked Example 5
X and Y are two points at respective distances d and 3d from
the centre of the Earth, where d is greater than the radius of
the Earth. The gravitational potential at X is –900 kJ kg–1.
What is the work done on a 2 kg mass when it is taken from
X to Y?
Solution
Y
rX

Since potential   -1/r, so
X
rY

Y
d

 900 3 d
 Y  -300 kJ kg–1
 WD = m (Y - X) = 2 [ (–300) – (–900) ] = 1200 kJ
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5 RELATION BETWEEN FIELD
STRENGTH AND POTENTIAL
Verify, for point mass, that
g
d
dr
gravitational
potential 
gradient
gives field
strength
distance r


GM

r
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Similarly
dU
F 
dr
These two equations are ALWAYS correct.
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Worked Example 6
The following table gives the values for the gravitational
potential due to the Earth at the various heights above the
Earth’s surface.
Height / km
580
600
620
Gravitational potential / MJ kg–1
-57.51
-57.34
-57.17
Deduce a value for the Earth’s gravitational field strength at a
height of 600 km.

580
600
620
r
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Solution
Field strength 
d
g 
dr
=
=
[  57.17  (57.51) ]  106

(620  580)  103
– 8.5 N kg–1
Magnitude of field strength is 8.5 N kg–1
The minus sign indicates that the direction of field
strength is towards the Earth’s centre.
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In the auditorium, an invisible friend is also
attending lecture and he is quite worried after
learning about the attractive gravitational force…..
If Gravitation
force (due to
the Earth) acts
on the moon;
then the Moon
should collapse
onto the Earth….
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THE SKY IS FALLING!!!!!
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Do You agree?
Since there is attractive gravitation
force exerted by the Earth on the Moon
and since the moon is less massive, it
should move and collide with the earth
…Since the Sun has the same effect
on the Earth, then Earth should move
and hit the Sun…. Then what is left?
…….
Is the sky falling ? Is it the end of the
world ?
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6 ORBITS
• There are many naturally occurring orbital motions.
• Gravitational forces between heavenly bodies are
keeping them in their curved orbits.
• Most orbits are elliptical, but we will only consider
circular ones.
• In circular orbital motion, the gravitational force
acts as the centripetal force.
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Johannes Kepler
63
r
Gravitational force
GMm
r2
=

F
centripetal force
mv 2
r
or
mr2
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Worked Example 7
(a) Assume that the planets move in circular orbits about
the Sun, show that the squares of their periods of
revolution are proportional to the cubes of the radii of
their orbits.
(b) The Earth is 1.50  1011 m from the centre of the Sun
and takes exactly one year to complete one orbit. The
planet Mars takes 1.88 years to complete an orbit of
the Sun. Calculate the radius of Mars’ orbit.
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Solution
(a)
gravitational force
=
GMm
r2
T2
 T2

centripetal force
mr2
 2 
 mr 

T


4 2 3

r
GM

2
r3
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(b)
Since T 2  r 3,
then


TM2
rM3
 3
TE2
rE
(1.88)2
rM3

2
(1)
(1.50 1011 )3
rM = 2.28  1011 m
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6.2 What about energies associated with a
satellite?
Imagine a satellite of mass m moves in a circular orbit
about the Earth of mass M. The radius of the orbit is r.
We’ll derive an expression for
(a)
the kinetic energy T of the satellite,
(b)
the gravitational potential energy V of the satellite,
(c)
the total energy E of the satellite.
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(a)
Gravitational force
= centripetal force
GMm
r2
 kinetic energy, T =

1
mv 2
2
mv 2
r
GMm

2r
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(b)
Gravitational potential energy, V = 
(c)
Total energy, E = T + V
GM m
r
GMm
 GMm 

 

2r
r


GMm

2r
Note: A negative total energy indicates that the satellite
does not have enough energy to escape from
Earth’s gravitational field to infinity.
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6.3
The GEOSTATIONARY (Geosynchronous) Orbit
A Geostationary orbit is one in which a satellite orbiting in
it will always appear as stationary when viewed from any
point from the Earth.
It has the following characteristics:
•
Period = one day ( 24 hours).
•
The whole orbit is in the equatorial plane.
•
The direction of revolution is the same as the direction
of Earth's rotation.
•
Since T2  r3, it also has a particular radius.
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Application…

If you own a communication empire and
your company will be launching a new
geostationary communication satellite.
What would be the radius of the orbit?
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Find the radius of the geostationary orbit.
gravitational force
=
centripetal force
GMm
r2

mr 2

 4 2 
mr  2 
 T 
r

 GM 2 
T 

2
 4

1
3
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For geostationary orbit, we substitute T = 24 hours =
86400 s, M = 6.0 x 1024 kg, and obtain
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r
 (6.67  10 11 ) (6.0  10 24 )
2
 
(86400) 
2
4


 4.23  107 m
The radius of a geostationary orbit is 4.23  107 m.
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Summary
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Orbital Motion
(a)
gravitational force
GMm
r2
T2
=
centripetal force

mr2
 2 
 mr 

T 
4 2 3

r
GM
2
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