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Following points are the summary of these Lecture Slides : Universal Gravitation, Newtonian Gravitation, Fall acceleration, Gravitational Force, Gravitational Potential, Energy, Escape Speed, Objects, attracted, Universe
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April 28, 2009
Newtonian Gravitation
Free-fall Acceleration & the Gravitational Force
Gravitational Potential Energy
Escape Speed
Kepler 1
st
Law
Kepler 2
nd
Law
Kepler 3
rd
Law
April 28, 2009
April 28, 2009
G is the constant of universal gravitation
G = 6.673 x 10
N m² /kg²
This is an example of an
Determined experimentally
Henry Cavendish in 1798
2
2
April 28, 2009
The force that mass 1 exertson mass 2 is equal andopposite to the force mass 2exerts on mass 1
The forces form a Newton’sthird law action-reaction
The gravitational force exerted by a uniformsphere on a particle outside the sphere is thesame as the force exerted if the entire mass of thesphere were concentrated on its center
April 28, 2009
Consider an object of mass m near the Earth’ssurface
Acceleration a
g
due to gravity
Since
Near the Earth’s surface
km
1 .
6378
E
R
kg
10
9742 . 5
23
E
M
2
2
m/s
8 .
9
E
E
g
M R
G
a g
E
E
ma
R mM
G
F
2
2
2
2
1
E
E
R mM
G
r
m
m
G
F
April 28, 2009
Consider an object of mass m at a height h abovethe Earth’s surface
Acceleration a
g
due to gravity
a
g
2 )
(
h
R
M
G
a
E
E
g
g
E
E
ma
R mM
G
F
2
2
2
2
1
)
(
h
R
mM
G
r
m
m
G
F
E
E
April 28, 2009
The escape speed is the speed needed for anobject to soar off into space and not return
For the earth, v
esc
is about 11.2 km/s
Note, v is independent of the mass of theobject
E
E
esc
R GM
v
2
0
1 2
2
r
m
M
G
mv
PE
KE
E
April 28, 2009
All planets move inelliptical orbits with theSun at one of the focalpoints.
A line drawn from theSun to any planetsweeps out equal areasin equal time intervals.
The square of the orbitalperiod of any planet isproportional to cube ofthe average distancefrom the Sun to theplanet.
April 28, 2009
A line drawn fromthe Sun to anyplanet will sweepout equal areas inequal times
Area from A to Band C to D are thesame
April 28, 2009
The square of the orbital period of any planet isproportional to cube of the average distancefrom the Sun to the planet.
T is the period of the planet
a is the average distance from the Sun. Or a is thelength of the semi-major axis
For orbit around the Sun, K = K
S
= 2.97x
s
2
/m
3
K is independent of the mass of the planet
s
s
GM
K
2
4
3
2
April 28, 2009
From a telecommunications point of view, it’sadvantageous for satellites to remain at thesame location relative to a location on theEarth. This ca occur only if the satellite’sorbital period is the same as the Earth’s periodof rotation, 24 h. (a) At what distance from thecenter of the Earth can this geosynchronousorbit be found? (b) What’s the orbital speed ofthe satellite?