Transcript gravitational potential energy

LAW OF UNIVERSAL GRAVITATION
FG
Gm1m2

2
r
FG gravitational force (in two directions)
G
r
m1
universal gravitation constant
6.67x10-11 Nm2kg-2
distance between the objects
mass of the larger object
near the earth’s surface . . .
Gm1m2
m2 g 
2
r
Gm1
g 
2
r
both of these equations could be
applied to the surface of any planet
Planet X has a radius that is 3.5 times the
radius of the earth and a mass that is 2.0
times the earth’s. Compare the acceleration
due to gravity at the surface of each planet.
Gm1
g 2
r
Gmx
gx  2
rx
2me
gx 
2
(3.5re )
gx

 0.163
ge
2 Gme
gx  ( 2 ) 2
3 .5
re
2
gx  ( 2 )g e
3 .5
What happens to the gravitational attraction
between two particles if one mass is
doubled, the other tripled and the distance
between them cut in half?
G 2m1 3m2
r 2
( )
F2
2

Gm1m2
F1
2
r
F2
 24
F1
read p. 139-142
p. 141 1-6 extra
p. 143 8-13
p. 144 1-6
SATELLITES
A satellite is an object or a body that revolves
around another object, which is usually larger in
mass.
Planets, moons, space shuttles, space
stations, comets, and “satellites” are
satellites.
Satellites remain in a constant orbit because
they are acted upon by a centripetal force
and display centripetal acceleration.
FG  Fc
Gm1m2
m2 v

2
r
r
Gm1
2
v 
r
Gm1
v
r
2
remember m1
is the larger
mass and the
central object
v
2
Gm1

r
Gm1
 2r 

 
r
 T 
2
Gm1T
2
 4 r
2
3
What is the period of rotation of the moon about
the earth?
Gm1T  4 r
2
2 3
4 r
T
Gm1
2 3
4 (3.84 10 m)
T
2
11 Nm
24
6.67 10
5
.
98

10
kg
2
kg
2
8
T  2.367 10 s
6
T  27.40d
3
read 145-146
p. 151 1, 3-6
extra p. 147 2-4, 6
p. 160 14-20
GRAVITATIONAL FIELDS
A force field exists in the space surrounding an
object in which a force is exerted on objects
(e.g. gravitational, electric, magnetic). The
strength of gravitational force fields is determined by the Law of Universal Gravitation. If
two or more gravitational fields are acting on an
object then the net field is the sum of all the
individual fields.
read 274-275
p.276 2-6
p.277 1-8
KEPLER’S LAWS
In 1543 Copernicus proposes the heliocentric
model of the solar system in which planets
revolve around the sun in circular orbits. Slight
irregularities show up over long periods of study.
Tycho Brahe takes painstaking observations for 20
years with large precision instruments but dies
(1600) before he can analyze them properly.
A young mathematician continues Brahe’s work.
From his analysis the kinematics of the planets
is fully understood.
Kepler’s First Law of Planetary Motion
Each planet moves around the Sun in an orbit that
is an ellipse, with the Sun at one focus of the
ellipse.
Kepler’s Second Law of Planetary Motion
The straight line joining a planet and the Sun
sweeps out equal areas in space in equal intervals
of time.
Planets move faster when they are closer to the
Sun (centripetal force is stronger).
orbits are elliptical but are
not very elongated
equal areas
equal times
Kepler’s Third Law of Planetary Motion
The cube of the average radius of a
planet is directly proportional to the
square of the period of the planet’s orbit.
We have already proved this a few slides
back. Recall
Gm1m2 m2v

2
r
r
2
2r 2
m2 (
)
Gm1m2
T

2
r
r
3
r
Gm1

2
2
T
4
constant
For our solar system m1 is the mass of the
sun.
Mars’ average distance from the sun is
2.28 x1011 m while its period of rotation is 5.94
x 107 s. What is Jupiter’s average distance from
the sun if its period of rotation is 3.75 x 108 s ?
3
rJ
2
TJ

this equation
holds for objects
orbiting the same
mass
3
rm
2
Tm
rJ  7.79 10 m
11
read 278-283
p. 283 10-12
p. 284 4-7, 9
GRAVITATIONAL POTENTIAL
ENERGY, AGAIN
Recall the Law of Universal Gravitation
FG
Gm1m2

2
r
for constant masses, a graph of force vs. radius
would be . . .
The graph above is a F vs. d graph which
means the shaded area is the work required
to move an object from r1 to r2.
The shaded area is not easy to calculate but
can be done with a geometric mean. In this
case the work done by the lifter is equal to
DEp.
Another method involves calculus and
integration over a range from r1 to r2.
W  F1F2 (r2  r1 )
geometric mean
of force
Gm1m2 Gm1m2
DE p 
(
r

r
)
2
1
2
2
r1
r2
Gm1m2
DE p 
(r2  r1 )
r1r2
Gm1m2 Gm1m2
DE p 

r1
r2
Gm1m2
Gm1m2
DE p  

r2
r1
Gm1m2
Ep  
r
Know these two equations, you are not required
to know the previous development.
Which preceding equation can be simplified to
mgDh, the potential energy change near the
earth’s surface?
Potential energy
is a negative
function!
It increases until
it is zero.
PE stops here
because the
objects come
into contact and
cannot get
closer.
Recall
2
mv1
2
or
2
mv2
2
 mgh1 

2
mv1
2
2
mv2
2
 mgh2
 (mgh2  mgh1 )
so . . .
2
1
2
2
mv Gm1m2 mv Gm1m2



2
r1
2
r2
or
2
2
2
1
mv mv
Gm1m2
Gm1m2

 ( 

)
2
2
r2
r1
read p. 285-287 p. 287 1-5
Escape from a Gravitational Field
To escape a gravitational field an object must
have at least a total mechanical energy of zero!!
EM  EK  EP
for escape
EM  0
Escape energy - the minimum EK needed to project a mass (m2) from the surface of another
mass (m1) to escape the gravitational force of m1
Escape speed - the minimum speed needed to
project a mass (m2) from the surface of another
mass (m1) to escape the gravitational force of m1
Binding energy - the additional EK needed by a
mass (m2) to escape the gravitational force of m1
(similar to escape energy but applies to objects
that possess Ek i.e. satellites).
To calculate the escape energy or the escape
speed of a mass (m2):
EK  EP  0
To calculate the binding energy of a mass (m2):
EK  EP  EB  0
binding
energy
Calculate the escape velocity of any object on
the Earth’s surface.
EK  EP  0
2
m2v
Gm1m2

0
2
re
2Gm1
v 
re
2
v
2(6.67 10
11
2
Nm kg )(5.98 10 kg)
6
6.38 10 m
2
24
m
v  11182
s
The escape velocity is the same for all objects
on the Earth’s surface while the escape energy
is different for different massed object.
What is Ek and EM of an orbiting body
(satellite)?
FG  Fc
this is always true
of satellites
2
Gm1m2
m2 v

2
r
r
Gm1m2
2
 m2 v
r
Gm1m2
2
 m2 v
r
2
Gm1m2
m2 v

2r
2
 Ep
EK 
2
EM  E K  E P
EM  
EM 
Ep
2
Ep
 Ep
for an orbiting
satellite !!
2
Note that the total energy is negative since
the satellite is “bound” to the central body.
read p.288-293 p. 293 6-11
#12 is interesting!
extra p. 294 1-8 p. 300 1-17
25,26 look fun
a) What is the speed of Earth in orbit about
the Sun?
b) What is the total energy of Earth?
c) What is the binding energy of Earth?
d) If Earth was launched from the surface
of the Sun to its present orbit then what
velocity must it be launched with (Ignore
the radius of Earth.)?
e) If Earth came to rest and fell to the Sun
then what velocity would it have when it hit
the Sun (Ignore radius of Earth.)?
me= 5.98x1024 kg
ms= 1.99x1030 kg
re= 1.49x1011 m (of orbit)
rs= 6.96x108 m
(of the body)
G= 6.67x10-11 Nm2kg-2
2r
v
T
7
T  3.156  10 s
4 m
v  2.966  10
s
or
a)
Gm1m2
2
 m2v
r
m
v  2.985  10
s
4
b)
EM 
Ep
2
Gms me
EM  
2re
EM  2.664 10 J
33
c) The binding energy is 2.664 x 1033 J
d)
2
1
2
2
mv Gm1m2 mv Gm1m2



2
r1
2
r2
2
1
mv Gm1m2
Gm1m2


2
r1
2r2
2
1
mv
Gm1m2 Gm1m2


2
2r2
r1
2
1
mv
36
 1.138 10 J
2
5 m
v1  6.169 10
s
e)
2
1
2
2
mv Gm1m2 mv Gm1m2



2
r1
2
r2
2
2
Gm1m2 Gm1m2 mv



r1
r2
2
2
2
mv
36
 1.181 10 J
2
5 m
v2  6.285 10
s