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Sect. 13.4: The Gravitational Field
Gravitational Field
• A Gravitational Field  g, exists at all points in space.
• If a particle of mass m is placed at a point where the
gravitational field is g , it experiences a force:
Fg = mg
Fg
• The field exerts a force on the particle.
g
m
• The gravitational field g is defined as
• Gravitational field = Gravitational Force experienced by a “test”
particle placed at that point divided by the mass of the test particle.
– The presence of the test particle is not necessary for the field to exist
• The source particle creates the field
• The gravitational field g
vectors point in the direction
of the acceleration a particle
would experience if it were
placed in that field. Figure
• The magnitude is that of the
freefall acceleration, g, at that
location.
• The gravitational field g
describes the “effect” that any
object has on the empty space
around itself in terms of the
force that would be present if
a second object were
somewhere in that space
Fg
GM
g
  2 rˆ
m
r
Sect. 13.5: Gravitational Potential Energy
• The gravitational force is conservative
• Recall Ch. 8 discussion of conservative forces: Only for
conservative forces can a potential energy U be defined &
Total Mechanical Energy is Conserved for
Conservative Forces ONLY.  Just as in Ch. 8, define
the change in Gravitational Potential Energy associated
with a displacement of a mass m is as the negative of the work
done by the gravitational force on m during the displacement.
r
That is:
f
U  U f  U i    F  r  dr
ri
F(r) is the Gravitational Force. For a mass m in the Earth’s
gravitational field,
• Doing this integral:
This gives
As always, the reference point where the potential energy is zero is
arbitrary. Usually choose it at ri  , so (1/ri)  0 and Ui  0.
This gives
GME m
U (r )  
r
• For example, as a particle moves from
point A to B, as in the figure

its gravitational potential energy
changes by
 1 1
U = UÄ  UÄ  GME m   
 rÄ
ri 
Notes on Gravitational Potential Energy
in the Earth’s Gravity
• We’ve chosen the zero for the gravitational
potential energy at ri   where the gravitational
force is also zero.
This means that Ui = 0 when ri   or that
GME m
U (r )  
r
This is valid only for r  RE & NOT for r < RE
– That is, it is valid outside the Earth’s surface but NOT
inside it!
– U is negative because of the choice of Ui
Gravitational Potential Energy in Earth’s Gravity
• The figure is a graph of the
gravitational potential energy U
versus r for an object above the
Earth’s surface.
• Note that the potential energy
goes to zero as r approaches
infinity
Gravitational Potential Energy:
General Discussion
• For any two particles, masses m1 & m2, the
gravitational potential energy function is
Gm1m2
U
r
• The gravitational potential energy between any two
particles varies as 1/r. (Recall that the force varies as 1/r2)
• The potential energy is negative because the force is attractive &
we’ve chosen the potential energy to be zero at infinite separation.
• Some external energy must be supplied to do the positive
work needed to increase the separation between 2 objects
– The external work done produces an increase in
gravitational potential energy as the particles are separated
& U becomes less negative
Binding Energy
• The absolute value of the potential energy for
mass m can be thought of as the binding
energy of m. (The energy of binding of m to the
object which is attracting it gravitationally).
• Consider two masses m1 & m2 attracting each
other gravitationally. If an external force is
applied to m1 & m2 giving them an energy
larger than the binding energy, the excess
energy will be in the form of kinetic energy of
m1 & m2 when they are at infinite separation.
Systems with Three or More Particles
• For systems with more than two masses, the total gravitational
potential energy of the system is the sum of the gravitational
potential energy over all pairs of particles.
• Because of this, gravitational potential energy is said to obey
the superposition principle. Each pair of particles in the system
contributes a term to Utotal.
• Example; assume 3 particles as in
the figure.
The result is shown in the equation
• The absolute value of Utotal
represents the work needed to
separate the particles by an
infinite distance
U total  U12  U13  U23
 m1m2 m1m3 m2m3 
 G 



r
r
r
13
23
 12

Sect. 13.6: Energy Considerations in Satellite Motion
Energy and Satellite Motion
• Consider an object of mass m moving with a speed v
in the vicinity of a large mass M
– Assume that M >> m as is the case for a small
object orbiting a large one.
• The total mechanical energy is the sum of the
system’s kinetic and potential energies.
• Total mechanical energy: E = K +U
1
Mm
2
E  mv  G
2
r
• In a system in which m is bound in an orbit around
M, can show that E must be less than 0
Energy in a Circular Orbit
• Consider an object of mass m
moving in a circular orbit about a
large mass M, as in the figure.
• The gravitational force supplies the
centripetal force:
Fg = G(Mm/r2) = ma = m(v2/r)
Multiply both sides by r & divide by 2:
G[(Mm)/(2r)] = (½)mv2
Putting this into the total mechanical energy
& doing some algebra gives:
1
Mm
2
E  mv  G
2
r
=
GMm
E
2r
• The total mechanical energy is
negative in for a circular orbit.
• The kinetic energy is positive and is
equal to half the absolute value of the
potential energy
• The absolute value of E is equal to the
binding energy of the system
Energy in an Elliptical Orbit
• It can be shown that, for an elliptical orbit, the
radius of the circular orbit is replaced by the
semimajor axis, a, of the ellipse. This gives:
GMm
E
2a
• The total mechanical energy is negative
• The total energy is conserved if the system is
isolated
Escape Speed from Earth
• Consider an object of mass m projected
upward from the Earth’s surface with an
initial speed, vi as in the figure.
• Use energy to find the minimum value of the
initial speed vi needed to allow the object to
move infinitely far away from Earth. E is
conserved (Ei = Ef), so, to get to a maximum
distance away (rmax) & then stop (v = 0):
(½)mvi2 - G[(MEm)/(RE)] = -G[(MEm)/(rmax)]
We want vi for rmax    the right side is 0 so
(½)mvi2 = G[(MEm)/(RE)]. Solve for vi = vescape
This is independent of the
2GME
v esc 
direction of vi & of the
RE
object mass m!
Escape Speed, General
• The result for Earth can be extended
to any planet
2GM
v esc 
R
• The table gives escape speeds from
various objects.
• Note: Complete escape from an
object is not really possible
– Gravitational force extends to infinity so
some force will always be felt no matter
how far away you get
• This explains why some planets have
atmospheres and others do not
– Lighter molecules have higher average
speeds and are more likely to reach
escape speeds