Download Gravitational Potential Energy

11/08/2010
Gravitational Potential Energy
To explore how much energy a spacecraft needs to escape from Earth’s gravity,
we must expand on the topic of gravitational potential energy.
SPH4UI
To calculate the change in gravitational potential energy that a mass
undergoes when it undergoes a vertical displacement near the Earth’s surface,
we used:
UG  mg  h 
Gravitational Potential energy
Mr. Burns
Where U G is the change in gravitational potential energy, m is the mass, g
is the magnitude of the gravitational field constant, and h is the vertical
displacement, where when we choose hi to be the ground, we simply obtain
U=mgh. This equation assumes that g remains reasonably constant during
the change in vertical displacement.
We define the work done on a system by a specific conservative force as the
negative of the change in a potential energy function associated with the force.
Changes are always taken as final minus initial values
Wconserv   U f  U i 
 U
Gravitational Potential Energy
As we recall from the Kinetic energy Theorem:
 change in the Kinetic   net work done on 



 energy of a particle   the particle

EK  EKf  EKi  W
Gravitational Potential Energy
W   F  x  dx
f
Since the work done a we move a mass between two points is
independent upon the path, the force is conservative
Wconserv  U  EK
Calculus definition of work
i
U    F  x  dx
f
A force does positive work when it has a vector component in the same
direction as the displacement. If the work done on a particle is positive, then
the particles kinetic energy increases.
Change in potential energy is equal to the negative
of the work done by the gravitational force on a
mass
U  W
i
U  
hf
hi
U  mgh
This gives us a Calculus definition of
change in potential energy
 mg  dh  mgh f
 mghi
Equation for change in
potential energy
If we set, hi to be zero (the ground), we obtain
an simple formula to determine the potential
energy
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11/08/2010
Gravitational Potential Energy
Gravitational Potential Energy
Work done by the gravitational force FG during the
vertical motion of a mass from initial height y1 to
final height y2 as the body rises. The gravitational
potential energy U increases (negative work done
by gravity, Kinetic Energy decreases).
But, how do we handle Gravitational Potential Energy, when g is not a
constant (such as when we are not on Earth)?
Work done by the gravitational force FG during the
vertical motion of a mass from initial height y1 to
final height y2 as the body rises. The gravitational
potential energy U decreases (positive work done
by gravity, Kinetic Energy increases)
Gravitational Potential Energy
Gravitational Potential Energy
Recall that the law of universal gravitation is given by:
GMm
FG  2
r
As the result of the work being done to
increase the separation from r1 to r2,
the gravitational potential energy of the
system increases.
Where FG is the magnitude of the force of gravitational attraction between
any two objects, M is the mass of one object, m is the mass of the other
object, and r is the distance between the centres of the two
To increase the separation of the two masses requires work to be done. That
is, we must do work to overcome the gravitational attraction between the two
masses when moving mass m from distance r1 to distance r2
Work is done to overcome this distance
r1
M
m
r2
The work done to change the
separation from r1 to r2 is
equal is equal to the change in
gravitational energy from r1 to
r2. However, recall that work
done by a varying force is
equal to the area under the
force-displacement graph for
that interval.
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Gravitational Potential Energy
Gravitational Potential Energy
To find an expression for this work, we consider a body of mass m outside the
Earth and compare the work Wgrav done by the gravitational force when a body
moves directly away from or towards the centre of the Earth (from r=r1 to r=r2).
r2
r2
r1
r1
Wgrav   Fdr   
GmE m
GmE m GmE m
dr 

r2
r2
r1
Because
Force
The path does not have
to be a the
straight
line, it could also be a
directly upon the initial and
curve, but the work done points
only depends
toward
the
centre
of
final values of r, not the path taken (this is definition of a
the Earth, F is
conservative force).
negative
We now define the corresponding potential energy U so that
Wgrav=U1-U2. Therefore we have a final definition of gravitational
potential energy
Gm1m2
U 
r
Gravitational Potential Energy
When gravity
field is constant
(near Earth)
U 
Gm1m2
r
In using our equation, we have
chosen U to be zero when the body
of mass is infinitely far away from
the Earth  r    .
As the body moves towards the Earth,
gravitational potential energy
decreases and becomes negative.
Gravitational Potential Energy
These formula seen initially confusing when first encountered.
U  mgh
You may be troubled by the equation for
gravitational potential energy because it
states that it is always negative. But we
have encountered this before, recall when
we used U=mgh, we found that U could be
negative whenever the body of mass was
below the arbitrary height we chose h to be
zero at.
U 
Gm1m2
r
Universal Potential energy,
Equals zero when r
approaches infinity
What is the change in gravitational potential energy of a 72.1 kg astronaut, lifted
from Earth’s surface into a circular orbit at an altitude of 4.35 x 10 2 km?
rE  6.38 106 m
M E  5.98 1024 kg
U  U f  U i
 GMm   GMm 
 



rf  
ri 

2
2
 
  

11 N  m 
24
11 N  m 
24
  6.67 10
  5.98 10 kg   72.1kg     6.67 10
  5.98 10 kg   72.1kg  
kg 2 
kg 2 
 

  
6
5
6




 6.38 10 m  4.35 10 m 
 6.38 10 m 

 


 

But, they provide us with the same
meaning. The change in gravitational
energy at the Earth's Surface is just a
special case of the general solution. This
is the gravity well scientists talk about.
  4.2198 109 J    4.5076 109 J 
 2.88 108 J
Let’s approximate the answer using U=mgh
U  mgh
Note: h is
distance above
the ground

N
  72.1kg   9.80   4.35 105 m 
kg 

 3.07 108 J
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Gravitational Potential Energy
Gravitational Potential Energy
We can use the fact that if no other external forces besides gravity affect a
mass, then the total energy is conserved (the gravitational force does work and
thus the mechanical energy is conserved). The Kinetic Energy gained by the
system is matched with the Potential Energy lost, and thus our mass will have
the following statement of energy conservation..
EK1  U1  EK 2  U 2
Suppose you want to place a 1000 kg weather satellite into a circular orbit 300
km above Earth’s surface.
a) What speed, period, and radial acceleration must it have?
b) How much work has to be done to place the satellite in orbit?
c) How much additional work must be done to make this satellite escape
Earth’s gravity
Now by inserting the formulas for the Kinetic and Potential Energy and
interpreting the scenario where the object reaches the r location where both EK2
and U2 have a zero value (where the object comes to a stop).
E K 1  U1  E K 2  U 2
E K 1  U1  0
1 2  GMm 
mvi   
0
2
r 

1 2 GMm
mv 
2
r
2GM
2
v 
r
2GM
v
r
This gives us the initial
speed v1, needed for a
body to escape from the
surface of a spherical
mass M with radius r
First we need the radius of the satellites orbit:
r  6380km  300km  6680km  6.68 106 m
Gravitational Potential Energy
Suppose you want to place a 1000 kg weather satellite into a circular orbit 300
km above Earth’s surface.
a) What speed, period, and radial acceleration must it have?
b) How much work has to be done to place the satellite in orbit?
c) How much additional work must be done to make this satellite escape
Earth’s gravity
r  6.68 106 m
Suppose you want to place a 1000 kg weather satellite into a circular orbit 300
km above Earth’s surface.
a) What speed, period, and radial acceleration must it have?
b) How much work has to be done to place the satellite in orbit?
c) How much additional work must be done to make this satellite escape
Earth’s gravity
For a stable orbit: FG=mac
 4 2 r 
GM E m
r2
 v2 
GM E m
 m 
r2
 r 
v
GM E
r
RE=6380 km
This is the escape
velocity of the mass
Gravitational Potential Energy
For a stable orbit: FG=mac
ME=5.97x1024 kg
2

11 N  m 
24
 6.67 10
  5.97 10 kg 
kg 2 

v
6
6.68 10 m
m
 7720
s
 m 2 
 T 
T
4 2 r 2 r
GM E
r  6.68 106 m
v  7720
m
s
2 r
v
2  6.68 106 m 

m
7720
s
 5440 s

 90.6 min
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11/08/2010
Gravitational Potential Energy
Gravitational Potential Energy
Suppose you want to place a 1000 kg weather satellite into a circular orbit 300
km above Earth’s surface.
a) What speed, period, and radial acceleration must it have?
b) How much work has to be done to place the satellite in orbit?
c) How much additional work must be done to make this satellite escape
Earth’s gravity
b) How much work has to be done to place the satellite in orbit?
The work required is the difference between the total mechanical energy when the
satellite is in orbit (Ef) and when the satellite was on the launch pad (Ei).
ac 
r  6.68 10 m
6
v2
r
2
m

 7720 
s


6.68 106 m
m
 8.92 2
s
m
v  7720
s
Gravitational Potential Energy
c) How much additional work must be done to make this satellite escape
Earth’s gravity
• We recall that for a satellite to escape to infinity, the total mechanical energy
must be zero.
• The total mechanical energy in orbit Ef was -2.99 x 1010 J
• To increase this to zero, an amount of work equal to 2.99 x 1010 J must be
done.
W  E f  Ei
RE=6380 km
R =6680 km
ME=5.97x1024 kg
  EKf  U f    EKi  U i 
E f  EKf  U f
1
 GM E m     GM E m  
  mv 2   

 0 
r    
rE  

2

1 2  GM E m 
mv   

2
r 

2


N m
 m2 E m
6.67 1011
5.97 10 24 kg  1000 kg  
5.97 1024 kg  1000kg 12 m 6.67GMr 10E 11 NkgGM
2
1
2 
m
kg 2

r





  1000kg   7720  


6
s
6.68 106 m
6.38

10
m
2
 GM



m


  E
 circular orbit 

  2r

2
 2.99 1010 J   6.25 1010 J 
 3.26 1010 J
Gravitational Potential Energy
A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous
orbit (one orbit is 24 hours) around Earth .
M E  5.98 1024 kg
rE  6.38 106 m
a) What is the radius of the satellite’s orbit?
b) What is the velocity of the satellites orbit?
c) What is the gravitational potential energy (universal) of the satellite when it is
at rest on Earth’s surface?
d) What is the total energy of the satellite when it is in geosynchronous orbit?
e) How much work the launch rocket do on the satellite to place it into orbit?
f) Once in orbit, how much additional energy would the satellite require to
escape from Earth’s potential well?
g) What should the launch velocity be it the satellite is required to escape from
Earth’s potential well?
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11/08/2010
Gravitational Potential Energy
Gravitational Potential Energy
A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous
orbit (one orbit is 24 hours) around Earth .
M E  5.98 10 kg
24
rE  6.38 10 m
6
a) What is the radius of the satellite’s orbit?
Fc  FG
Since we are given the period of the
orbit is 24 hours. We will use the
Period version of centripetal
acceleration.
GM E mS
r2
GM E
ac  2
r
mS ac 
2

2
11 N  m 
24
 6.67 10
  5.98 10 kg   24h  3600s 
3
kg 2 
 
2
4
 4.23 107 m
4 r GM E
 2
T2
r
2
r3 
r
GM ET 2
4 2
3
Gravitational Potential Energy
rE  6.38 10 m
6
c) What is the gravitational potential energy (universal) of the satellite when it is
at rest on Earth’s surface?
UG  
rE  6.38 106 m
b) What is the velocity of the satellites orbit?
v
GM E
r
2

11 N  m 
24
 6.67 10
  5.98 10 kg 
kg 2 
 
7
4.23 10 m
 3070
m
s
km
h
Gravitational Potential Energy
A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous
orbit (one orbit is 24 hours) around Earth .
M E  5.98 10 kg
M E  5.98 1024 kg
 1.11104
GM ET 2
4 2
24
A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous
orbit (one orbit is 24 hours) around Earth .
GM E ms
rE
A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous
orbit (one orbit is 24 hours) around Earth .
M E  5.98 1024 kg
d) What is the total energy of the satellite when it is in geosynchronous orbit?
ET  K  U


11 N  m 
24
2
 6.67 10
  5.98 10 kg  5.00 10 kg 
kg 2 

 6.38 106 m 
rE  6.38 106 m
GM E ms
1
ms v 2 
2
r
2
 3.13 1010 J

1 GM E ms
2
r
 2.36 109 J
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11/08/2010
Gravitational Potential Energy
5.00x102
A
kg weather satellite is to be placed into a circular geosynchronous
orbit (one orbit is 24 hours) around Earth .
M E  5.98 1024 kg
rE  6.38 106 m
e) How much work the launch rocket do on the satellite to place it into orbit?
Gravitational Potential Energy
A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous
orbit (one orbit is 24 hours) around Earth .
M E  5.98 1024 kg
rE  6.38 106 m
f) Once in orbit, how much additional energy would the satellite require to
escape from Earth’s potential well?
W  E
The negative of
Total Energy to
get the satellite
there.
 E f  Ei
 2.36 109 J   3.13 1010 J 
 2.89 1010 J
Gravitational Potential Energy
A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous
orbit (one orbit is 24 hours) around Earth .
M E  5.98 1024 kg
rE  6.38 106 m
2.36 109 J
Gravitational Potential Energy
Suppose your hand moves up 0.50m while you are throwing a ball ( 0.145kg),
which leaves your hand with an upward velocity of 20.0 m/s. Assuming your
hand exerts a constant upward force on the ball, determine the magnitude of
this force. Determine the speed of the ball 15.0 m above the point where it
leaves your hand.
g) What should the launch velocity be it the satellite is required to escape from
Earth’s potential well?
vE 
2GM E
rE


N  m2 
24
  5.98 10 kg 
kg 2 
6.38 106 m
We will look at the
energy diagrams
for y1, y2, and y3 to
guide us in solving
this problem
 2   6.67 1011

 1.12 104
m
s
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11/08/2010
Gravitational Potential Energy
Gravitational Potential Energy
Suppose your hand moves up 0.50m while you are throwing a ball (0.145kg), which leaves your hand
with an upward velocity of 20.0 m/s. Assuming your hand exerts a constant upward force on the ball,
determine the magnitude of this force. Determine the speed of the ball 15.0 m above the point where
it leaves your hand.
The ball’s motion has
two stages: while it is
in contact with your
hand and after it
leaves your hand
To keep track of these
stages, we let point 1
be where your hand
first starts to move,
and point 2 where the
ball leaves your hand,
and point 3 where the
ball is 15.0m above
point 2.
3
2
1
Gravitational Potential Energy
Whand  F  y2  y1 
F
3
The kinetic energy increases by
29.0J and the potential energy
increases by 0.71J, the change in
total mechanical energy 29,7J is
due to the work by your hand.
3
 EK 2  U 2

1 2
mv2  mgy2
2
2

1
m
m
 0.145kg   20.0    0.145kg   9.8 2   0.0m 
2
s
s 


2
  EK 2  EK 1   U 2  U1 
  29.0 J  0 J1   0   0.71J  
 29.7 J
Suppose your hand moves up 0.50m while you are throwing a ball (0.145kg), which leaves your hand
with an upward velocity of 20.0 m/s. Assuming your hand exerts a constant upward force on the ball,
determine the magnitude of this force. Determine the speed of the ball 15.0 m above the point where
it leaves your hand.
EK 2  U 2To
 Efind
U 3 speed at point
K 3 the
EK 33, we
EK 2 note
 U 2 that
U 3 the total
3
energy
1 2mechanical
1
mv3between
 mv22 points
mgy2  mgy
3
2 and
3
2
2
are conserved, Therefore
we can use:
2
EK2+U
m
 2=EmK3+U3 
v3  v22  2 g  y2  y3 
2
Whand
y2  y1
29.7 J
0.50m
 59 N
The non gravitational force
of your hand acts only
between points 1 and 2.
using a coordinate system
with point 2 being at 0m
K 1  U1  Whand
and point 3 at 15.0m E
and
point 1 at -0.50m,1 provides
2
mv1  mgy1  Whand
us with a easy application
2
of conservative energies
to
m

0 J solve
kg problem.
 0.145this
 9.80 2   0.50m   Whand
s 

Whand
Gravitational Potential Energy
Suppose your hand moves up 0.50m while you are throwing a ball (0.145kg), which leaves your hand
with an upward velocity of 20.0 m/s. Assuming your hand exerts a constant upward force on the ball,
determine the magnitude of this force. Determine the speed of the ball 15.0 m above the point where
it leaves your hand.
Since the force done by your
hand was a constant force,,
the W hand done by this force
is equal to the force
multiplied by the upward
displacement (y2-y1)
Suppose your hand moves up 0.50m while you are throwing a ball (0.145kg), which leaves your hand
with an upward velocity of 20.0 m/s. Assuming your hand exerts a constant upward force on the ball,
determine the magnitude of this force. Determine the speed of the ball 15.0 m above the point where
it leaves your hand.

1
  20.0   2  9.8 2   0m  15m 
s
s 


m
 10.3
s
2
1
8