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Fields
Essential idea: Electric charges and masses each
influence the space around them and that influence
can be represented through the concept of fields.
Nature of sciences: Paradigm shift: The move from
direct, observable actions being responsible for
influence on an object to acceptance of a field’s
“action at a distance” required a paradigm shift in
the world of science.
Understandings:
• Gravitational fields
• Electrostatic fields
• Electric potential and gravitational
potential
• Field lines
• Equipotential surfaces
Applications and skills:
• Representing sources of mass and charge,
lines of electric and gravitational force, and
field patterns using an appropriate symbolism
• Mapping fields using potential
• Describing the connection between
equipotential surfaces and field lines
Guidence:
• Electrostatic fields are restricted to the radial fields
around point or spherical charges, the field between
two point charges and the uniform fields between
charged parallel plates
• Gravitational fields are restricted to the radial fields
around point or spherical masses and the (assumed)
uniform field close to the surface of massive celestial
bodies and planetary bodies
• Students should recognize that no work is done in
moving charge or mass on an equipotential surface
Data booklet reference:
• W = q ΔVe
• W = m ΔVg
Theory of knowledge
• Although gravitational and electrostatic forces decrease with
the square of distance and will only become zero at infinite
separation, from a practical standpoint they become
negligible at much smaller distances. How do scientists decide
when an effect is so small that it can be ignored?
Utilization:
• Knowledge of vector analysis is useful for this
sub-topic (see Physics sub-topic 1.3)
Aims:
• Aim 9: models developed for electric and
gravitational fields using lines of forces allow
predictions to be made but have limitations in
terms of the finite width of a line.
Topic 10: Fields
10.1 Describing fields
Gravitational fields
The Earth has a radial field of gravity, which
means that the gravitational field is circular and
acts from the centre point.
You can see on the diagram that near the
Earth's surface the lines are closer
together than higher up. The closeness of
the lines represent the relative strength of
the field, so from the diagram, you can tell
that the strength of the field decreases
with altitude. Further apart lines represent
points where the field is weaker.
Topic 10: Fields
10.1 Describing fields
Gravitational fields
• A uniform field, however, has the lines perfectly parallel. The
Earth's gravitational field can be considered to be uniform on
the scale of small things such as cars, balls, and planes.
• For small heights at this scale (a few dozen kilometres), the
strength of the field doesn't change enough to be noticeable.
Again, the arrows point towards the centre of the Earth, since
that is the way objects fall..
The Earth's gravitational field is represented by
parallel lines on small scales.
Topic 10: Fields
10.1 Describing fields
Gravitational fields
The
normal
force,
the
force,
tension
drag
FYI: School
A must
assume
thatfriction
the force "signal"
travels
infinitely and
fast, to
are
all
classified
as contact
forces, occurring where
react
to relative
motion between
the masses.
two objects contact each other.
FYI: gravitational
School B assumes that
the forceand
"signal"
already in place
in the space
The
force,
theis electric
force,
surroundingdo
thenot
source
of theobjects
force. Thus
force
does NOT
have
to
however,
need
tothebe
insignal
contact
(or
even
keep traveling
to the receiving
mass.
close
proximity).
These
two forces are sometimes
called
action
a distance
FYI: Relativity
has at
determined
that the forces.
absolute fastest ANY signal can
There
are
schools
ofspeed
thought
8 m/s, the
propogate
is ctwo
= 3.010
of light.on action at a
distance.
School A: The masses know where each other are at
all times, and the force is instantaneously felt by
both masses at all times.
School B: The masses deform space itself, and the
force is simply a reaction to the local space,
rather than the distant mass.
Topic 10: Fields
10.1 Describing fields
Gravitational fields
Thus School B is currently the "correct" view.
If,
for example, the gravitational forces were truly
"action at a distance," the following would occur:
SUN
c
c
The planet and the sun could not "communicate" quickly
enough to keep the planet in a stable orbit.
Topic 10: Fields
10.1 Describing fields
Gravitational fields
Instead, look at the School B view of the space
surrounding the sun:
FYI: You have probably seen such a model at the museum. A coin is allowed to
roll, and it appears to be in orbit.
FYI: Of course, if the slope isn't just right to match the tangential speed, the coin
will spiral into the central maw.
The planet "knows" which
way
the
the
the
to "roll" because of
local curvature of
space surrounding
sun.
Long distance
communication is not
needed.
FYI: The view of School B is called
the FIELD VIEW.
Topic 10: Fields
10.1 Describing fields
Gravitational fields
Of course, if the mass isn't moving, it will roll
downhill if placed on the grid:
m0
m0
Depending on where we
place the "test" mass m0,
it will roll differently.
Question: Why is the arrow at the location of the first test mass smaller than
that at the second one?
Topic 10: Fields
10.1 Describing fields
Gravitational fields
We can assign an arrow to each position surrounding
the sun, representing the direction the test mass
will go, and how big a force the test mass feels.
We represent fields with
vectors of scale length.
In the case of the
gravitational field, the
field vectors all point
toward the center:
Topic 10: Fields
10.1 Describing fields
Electrostatic fields
If we view our
gravitational field
arrows from above, we
get a picture that
looks like this:
What the field arrows tell
us is the magnitude and the
direction of the force on a
particle placed anywhere in
space:
The blue particle will
feel a "downward" force.
The red particle will feel
SUN
FYI: We don't even have to
draw the object that is
creating the field.
a "leftward" force whose
magnitude is LESS than that
of the blue particle.
Topic 10: Fields
10.1 Describing fields
Electrostatic fields
FYI: Inward-pointing fields are called SINKS. Think of the field arrows as
Consider, now, a
water flowing into a hole.
negative charge:
have is an attractive
force.
FYI: Test charges are by convention POSITIVE. Therefore, field vectors around
a charge show the direction a POSITIVE charge would want to go if placed in
the field.
For masses, all we
SUN
-
All of the field
If we place a very small
arrows point inward.
POSITIVE test charge in
the vicinity of our
negative charge, it will
be attracted to the
center:
We can map out the
field vectors just as
we did for the sun.
Topic 10: Fields
10.1
Describing
fields
Outward-pointing
fields
called SOURCES.
thetwo
field
arrows as
FYI: We
call
such a system
anare
ELECTRIC
DIPOLE. A Think
dipoleofhas
poles
water flowing
a fountain.
(charges)
whichout
areofopposite
in sign.
Electrostatic fields
Question:
T or F: Gravitational
fields
alwaysand
sinks.
FYI:
The ELECTRIC
DIPOLE consists
ofare
a source
a sink.
We can place a negative
Question: is
T orpositive.
F: Electric fields are always
sinks. here.
charge
charge
Now suppose our
+
+
-
If we place a very small
The fields of the two
POSITIVE test charge in
the vicinity of our
positive charge, it will
be repelled from the
center:
We can map out the field
vectors just as we did
for the negative charge.
charges will interact:
Keep in mind that the
field lines show the
direction a POSITIVE
charge would like to
travel if placed in the
field.
A negative charge will do
the opposite:
Topic 10: Fields
10.1 Describing fields
Electrostatic fields
You may have noticed
that in the last field
diagram some of our
arrows were unbroken.
We can draw solid
field arrows as long as
we note that THE CLOSER
THE FIELD LINES ARE TO
ONE ANOTHER, THE
STRONGER THE FIELD.
+
-
FYI: There are some simple rules for drawing these solid "electric lines of
force."
Rule 1: The closer together the electric lines of force are, the stronger the
electric field.
strong
Rule 2: Electric lines of force originate on positive charges and end on
negative charges.
weak
Rule 3: The number of lines of force entering or leaving a charge is
proportional to the magnitude of that charge.
Topic 10: Fields
10.1 Describing fields
Field lines
A convenient way of visualizing electric field patterns is to draw
lines that follow the same direction as the electric field vector at
any point. These lines, called electric field lines, are related to
the electric field in any region of space in the following manner:
– The electric field vector E is tangent to the electric field line at each
point.
– The number of lines per unit area through a surface perpendicular to
the lines is proportional to the magnitude of the electric field in that
region. Thus, E is great when the field lines are close together and
small when they are far apart.
FYI: The same thing pays for field lines of gravitational field.
Question: Which charge is the strongest negative one?
Topic
10: Fields
Question: Which charge is the weakest positive one?
10.1 Describing fields
Question: Which charge is the strongest positive one?
Electrostatic
fields
Question: Which
charge is the weakest negative one?
Example 1.
D
E
A
Determine the sign and the magnitude of the charges
by looking at the electric lines of force. C
A
B
C
D
E
F
Topic 10: Fields
10.2 Fields at work
Potential and potential energy
You can think of potential energy as the capacity
to do work. And work is a force times a distance.
Recall the gravitational force (from Newton):
F = Gm1m2/r2
universal law of
gravitation
where G = 6.67×10−11 N m2 kg−2
If we multiply the above force by a distance r we
get
EP = -Gm1m2/r
gravitational
where G = 6.67×10−11 N m2 kg−2 potential energy
FYI
The actual proof is beyond the scope of this
course.
Note, in particular, the minus sign.
Topic 10: Fields
10.2 Fields at work
Gravitational fields
EP = -Gm1m2/r
gravitational
where G = 6.67×10−11 N m2 kg−2 potential energy
EXAMPLE 1. Find the gravitational potential
energy stored in the Earth-Moon system.
M = 5.981024 kg
m = 7.361022 kg
d = 3.82108 m
SOLUTION: Use EP = -Gm1m2/r.
EP = -Gm1m2/r
= -(6.67×10−11)(7.36×1022)(5.98×1024)/3.82×108
= -7.68×1028 J.
Topic 10: Fields
10.2 Fields at work
Gravitational fields
The previous formula is for large-scale
gravitational fields (say, some distance from a
planet).
Recall the “local” formula for gravitational
potential energy:
local ∆EP
∆EP = mg∆y where g = 9.8 m s-2.
The local formula treats y0 as the arbitrary
“zero value” of potential energy. The general
formula treats r =  as the “zero value”.
FYI
The local formula works only for g = CONST,
which is true as long as ∆y is relatively small
(say, sea level to the top of Mt. Everest). For
larger distances use ∆EP = -Gm1m2(1/rf – 1/r0).
Topic 10: Fields
10.2 Fields at work
Gravitational fields
Recall the definition of work:
work
W = Fd cos  where  is the angle
between the force F and the displacement d.
Consider the movement of a ball from the ground
to the table via displacements d1 and d2:
Along the displacement d1, gravity does work
given by
W1 = mgd1 cos 180° = -mgd1.
Along the displacement d2, gravity does work
given by
W2 = mgd2 cos 90° = 0.
d2
d1
The total work done by gravity is
thus Wg = -mgd1.
Topic 10: Fields
10.2 Fields at work
Consider the movement of a ball from the ground
to the table via the 6 new displacements s:
Along s1, s3 and s5 gravity does work given by
Wg = -mgs1 + -mgs3 + -mgs5 = -mg(s1 + s2 + s3).
Along s2, s4 and s6 gravity does no work. (Why?)
Superimposing d1 and d2 from the previous path we
see that Wg = -mg(s1 + s2 + s3) = -mgd1.
This is exactly the same as we got before!
s6
d2
s4
d1
s1
s2 s3
s5
FYI
The work done by gravity
is independent of the path
followed.
Topic 10: Fields
10.2 Fields at work
We call any force which does work independent of
path a conservative force.
Thus gravity is a conservative force.
FYI
Only conservative forces have associated
potential energy functions.
Path 2
PRACTICE: Show that friction is not a conservative force.
SOLUTION: Suppose you slide a crate across the floor along paths 1 and 2:
Clearly the distance along
Path 2 is greater than on
Path 1. The work is different.
Thus its work is not path-independent.
Thus it is not conservative.
A
Path 1
B
Topic 10: Fields
10.2 Fields at work
If work is done in a conservative force field
then there is an associated potential energy
function.
EXAMPLE: Show that for a conservative force
∆EP = -W = -Fd cos 
potential energy
function
where F is a conservative force
SOLUTION:
From conservation of mechanical energy we have
∆EP + ∆EK = 0 so that ∆EP = -∆EK.
From the work-kinetic energy theorem we have
W = ∆EK.
Thus
∆EP = -W = -Fd cos .
Topic 10: Fields
10.2 Fields at work
If work is done in a conservative force field
then there is an associated potential energy
function.
∆EP = -W = -Fd cos 
where F is a conservative force
potential energy
function
EXAMPLE:
yf
Show that for the gravitational force F = -mg
that the potential energy function is ∆EP = mg∆y.
SOLUTION:
d
Consider lifting a weight. The work done by
gravity on the weight is independent of the path,
so let us lift it straight up for simplicity.
Observe:  = 180º, d = ∆y, F = -mg. Thus
yo mg
∆EP = -W = -Fd cos  = -mg∆y cos 180º = mg∆y.
Topic 10: Fields
10.2 Fields at work
Define gravitational potential and gravitational
potential energy.
EP = -Gm1m2/r
Gravitational
where G = 6.67×10−11 N m2 kg−2 potential energy
We now define gravitational potential as
gravitational potential energy per unit mass:
∆V = ∆EP/m
V = -Gm/r
where G = 6.67×10−11 N m2 kg−2
Gravitational
potential
FYI
The units of V are J kg-1.
Gravitational potential is the work done per
unit mass done in moving a small mass from
infinity to r. (Note that V = 0 at r = .)
Topic 10: Fields
10.2 Fields at work
State and apply the expression for gravitational
potential due to a point mass.
∆V = ∆EP/m
Gravitational
potential
V = -Gm/r
where G = 6.67×10−11 N m2 kg−2
EXAMPLE: Find the change in gravitational potential in
moving from Earth’s surface to 5 Earth radii (from
Earth’s center).
SOLUTION: Use ∆V = -Gm/r2-(-Gm/r1), m = 5.98×1024 kg.
At Earth’s surface r1 = 6.37×106 m.
But then r2 = 5(6.37×106) = 3.19×107 m. Thus
∆V = -Gm(1/r2 - 1/r1)
= -Gm(1/3.19×107 - 1/6.37×106) = -Gm(-1.26×10-7)
= -(6.67×10−11)(5.98×1024)(-1.26×10-7)
= +5.01×107 J kg-1.
Addition of potential
• Potential is a scalar quantity, so adding the
magnitudes. If we tak the example:
Potential at P:
GM A  GM B 

V 
  
rA
rB 

Topic 10: Fields
10.2 Fields at work
Define gravitational potential and gravitational
potential energy.
FYI
A few words clarifying the gravitational
potential energy and gravitational potential
formulas are in order.
EP = -Gm1m2/r
gravitational potential energy
V = -Gm/r
gravitational potential
Be aware of the difference in name. Both have
“gravitational potential” in them and can be
confused during problem solving.
Be aware of the minus sign on both formulas.
The minus sign is there so that as you separate
two masses, or move farther out in space, their
values increase (as in the last example).
Both formulas become zero when r becomes
infinitely large.
Topic 10: Fields
10.2 Fields at work
State and apply the formula relating
gravitational field strength to gravitational
potential gradient.
The gravitational potential gradient is the
change in gravitational potential per unit
distance. Thus the GPG = ∆V/∆r.
EXAMPLE: Find the GPG in moving from Earth’s
surface to 5 radii from Earth’s center.
SOLUTION: In a previous problem (slide 11) we
found the value for the change in gravitational
potential to be ∆V = + 5.01×107 J kg-1.
5RE: r2 = 3.19×107 m. Earth: r1 = 6.37×106 m.
∆r = r2 – r1 = 3.19×107 - 6.37×106 = 2.55×107 m.
Thus the
GPG = ∆V/∆r
= 5.01×107 / 2.55×107 = 1.96 J kg-1 m-1.
Topic 10: Fields
10.2 Fields at work
State and apply the formula relating
gravitational field strength to gravitational
potential gradient.
The gravitational potential gradient is the
change in gravitational potential per unit
distance. Thus the GPG = ∆V/∆r.
PRACTICE: Show that the units for the gravitational potential gradient are the
units for acceleration.
SOLUTION:
The units for ∆V are J kg-1.
The units for work are J, but since work is force times distance we have 1 J = 1 N
m = 1 kg m s-2 m.
Therefore the units of ∆V are (kg m s-2 m)kg-1 or
[∆V] = m2 s-2.
Then the units of the GPG are
[GPG] = [∆V/∆r] = m2 s-2/m = m/s2.
Topic 10: Fields
10.2 Fields at work
State and apply the formula relating
gravitational field strength to gravitational
potential gradient.
EXAMPLE: Show that for the gravitational field
near the surface of Earth
g = ∆V/∆y
local potential gradient
SOLUTION: Recall that ∆V = ∆EP/m and ∆EP = mg∆y.
Thus
∆V = ∆EP/m
∆V = mg∆y/m
∆V = g∆y
g = ∆V/∆y.
FYI
From the last example we already know that the
units work out. The above formula only works
where g is constant (for small ∆y’s).
Topic 10: Fields
10.2 Fields at work
State and apply the formula relating
gravitational field strength to gravitational
potential gradient.
The following general potential gradient (which
we will not prove) works over greater range:
g = -∆V/∆r
general potential gradient
EXAMPLE: The gravitational potential in the
vicinity of a planet changes from -6.16×107 J kg-1
to -6.12×107 J kg-1 in moving from r = 1.80×108 m to
r = 2.85×108 m. What is the gravitation field
strength in that region?
SOLUTION:
g = -∆V/∆r
g = -(-6.12×107 - -6.16×107)/(2.85×108 - 1.80×108)
g = -4000000/105000000 = -0.0381 m s-2.
Topic 10: Fields
10.2 Fields at work
Determine the potential due to one or more point
masses.
Gravitational potential is derived from
gravitational potential energy and is thus a
scalar. There is no need to worry about vectors.
EXAMPLE: Find the gravitational potential
r
at the midpoint of the circle of masses
shown. Each mass is 125 kg and the
radius of the circle is 2750 m.
SOLUTION: Because potential is a scalar,
it doesn’t matter how the masses are arranged on
the circle. Only the distance matters.
For each mass r = 2750 m. Each mass contributes V
= -Gm/r so that
V = -(6.6710-11)(125)/2750 = -3.0310-12 J kg-1.
Thus Vtot = 4(-3.0310-12) = -1.2110-11 J kg-1.
Topic 10: Fields
10.2 Fields at work
Determine the potential due to one or more point
masses.
Gravitational potential is derived from
gravitational potential energy and is thus a
scalar. There is no need to worry about vectors.
EXAMPLE: If a 365-kg mass is brought in
r
from  to the center of the circle of
masses, how much potential energy will
it have lost?
SOLUTION: Use ∆V = ∆EP/m and the fact that
gravitational potential is zero at infinity.
∆EP = m∆V
FYI
0
Since ∆EP = -W we see
= m(V – V0)
that the work done in
= mV
bringing the mass in
-11
= 365(-1.2110 )
from infinity is
-9
= -4.4210 J.
+4.4210-9 J.
Topic 10: Fields
10.1 Describing fields
Describe and sketch the pattern of equipotential
surfaces due to one and two point masses.
Equipotential surfaces are imaginary surfaces at
which the potential is the same.
Since the gravitational potential
for a point mass is given by
m
V = -Gm/r it is clear that the
equipotential surfaces are at
fixed radii and hence are
equipotential surfaces
concentric spheres:
FYI
Generally equipotential surfaces are drawn so
that the ∆Vs for consecutive surfaces are equal.
Because V is inversely proportional to r the
consecutive rings get farther apart as we get
farther from the mass.
Worked example
Reffering to the figure.
1. What is the potential at A?
2. If a body is moved from A do B what is
the change in potential?
3. How much work is done moving a 2 kg
mass from A do B?
1.
VA = gh so potential at A = 10 x 3 = 30 J kg-1
2.
VA = 30 J kg-1
VB = 80 J kg-1
Change in potential = 80 – 30 = 50 J kg-1
3.
The work done moving from A to B is equal
to the change in potential x mass = 50 x 2 =
100 J.
Topic 10: Fields
10.1 Describing fields
Describe and sketch the pattern of equipotential
surfaces due to one and two point masses.
EXAMPLE: Sketch the equipotential surfaces due to
two point masses.
SOLUTION: Here is the sketch:
m
m
Topic 10: Fields
10.1 Describing fields
State the relation between equipotential surfaces
and gravitational field lines.
We know that for a point mass the gravitational
field lines point inward.
Thus the gravitational field lines
are perpendicular to the
equipotential surfaces.
m
FYI
A 3D image of the same
picture looks like this:
Topic 10: Fields
10.1 Describing fields
State the relation between equipotential surfaces
and gravitational field lines.
EXAMPLE: Use the 3D view of the equipotential
surface to interpret the
gravitational potential
gradient g = -∆V/∆r.
SOLUTION: We can choose
any direction for our r
value, say the
∆y
y-direction:
Then g = -∆V/∆y.
∆V
This is just the
gradient (slope) of
the surface.
Thus g is the (-) gradient
of the equipotential surface.
Topic 10: Fields
10.1 Describing fields
State the relation between equipotential surfaces
and gravitational field lines.
EXAMPLE: Sketch the gravitational field lines
around two point masses.
SOLUTION: Remember
that the gravitational
field lines point
inward, and that
they are
perpendicular to
m
m
the equipotential
surfaces.
Topic 10: Fields
10.1 Describing fields
State the relation between equipotential surfaces
and gravitational field lines.
EXAMPLE: Use a 3D view of
the equipotential surface
of two point masses to
illustrate that the
gravitational potential
gradient is zero somewhere
in between the two masses.
SOLUTION:
Remember that the
gravitational potential
gradient g = -∆V/∆r is just
the slope of the surface.
The slope is zero on the
saddle point. Thus g is also
zero there.
Practice
The Moon has a mass of 7.4 x 1022 kg and the Earth has
mass of 6.0 x 1024 kg. The average distance between the
Earth and the Moon is 3.8 x 105 km. If you travel directly
between the Earth and the Moon in a rocket of mass
2000 kg
(a) Calculate the gravitational potential when you are 1.0
x 104 km from the Moon.
(b) Calculate the rocket‘s PE at the point in part (a).
(c) Draw a sketch graph showing the change in potential.
(d) Mark the point where the gravitational field strength
is zero.
Topic 10: Fields
10.2 Fields at work
Explain the concept of escape speed from a
planet.
We define the escape speed to be the minimum
speed an object needs to escape a planet’s
gravitational pull.
We can further define escape speed vesc to be that
minimum speed which will carry an object to
infinity and bring it to rest there.
Thus we see that as r then v0.
M
R
m
r=R
r=
Topic 10: Fields
10.2 Fields at work
Derive an expression for the escape speed of an
object from the surface of a planet.
From the conservation of mechanical energy we
have ∆EK + ∆EP = 0. Then
∆EK + ∆EP = 0
EK – EK0 + EP – EP0 = 0
0
0
(1/2)mv2 – (1/2)mu2 + -GMm/r - -GMm/r0 = 0
(1/2)mvesc2 = GMm/R
vesc = 2GM/R
PRACTICE: Find the escape speed from Earth.
SOLUTION:
M = 5.981024 kg. R = 6.37106 m. Thus
vesc2 = 2GM/R
= 2(6.6710-11)(5.981024)/6.37106
vesc = 11200 ms-1 (= 24900 mph!)
escape speed
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
The ship MUST slow down and reverse (v becomes -).
The force varies as 1/r2 so that a is NOT linear.
Recall that a is the slope of the v vs. t graph.
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
Escape speed is the minimum speed needed to travel from the
surface of a planet to infinity.
It has the formula vesc2 = 2GM/R.
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
To escape we need vesc2 = 2GM/Re.
The kinetic energy alone must then be
(1/2)m(2GM/Re) = GMm/Re.
E = (1/2)mvesc2 =
This is to say, to escape E = 4GMm/(4Re).
Since we only have E = 3GMm/(4Re) the probe will not make it into
deep space.
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
Recall that Ep = -GMm/r.
Thus ∆EP = -GMm(1/R – 1/Re).
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
The probe is in circular motion so Fc = mv2/R.
But FG = GMm/R2 = Fc.
Thus mv2/R = GMm/R2 or mv2 = GMm/R.
Finally EK = (1/2)mv2 = GMm/(2R).
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
The energy given to the probe is stored in potential and kinetic energy. Thus
∆EK + ∆EP = E
GMm/(2R) - GMm(1/R–1/Re) = 3GMm/(4Re)
1/(2R) - 1/R + 1/Re = 3/(4Re)
1/(4Re) = 1/(2R)
R = 2Re
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
Just know it!
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
From ∆V = ∆EP/m we have ∆EP = m∆V.
Thus ∆EP = (4)( -3k - -7k) = 16 kJ
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
It is the work done per unit mass in bringing a small mass from infinity to
that point.
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
V = -GM/r so that V0 = -GM/R0.
But –g0R0 = -(GM/R02)R0 = -GM/R0
Thus V0 = -g0R0.
= V0.
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
At R0 = 0.5107
3.8107.
V0 = -
From previous problem
g = -V0/R0
g = - -3.8107 / 0.5107 = 7.6 m s-2.
0.5107 = 5.0106 = R0.
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
0
∆EK = -∆EP
EK – EK0 = -∆EP
(1/2)mv2 = ∆EP
v2 = 2∆EP/m
v2 = 2∆V
This solution assumes probe not in
orbit but merely reaches altitude
(and returns).
v2 = 2(3.0107)
v = 7700
ms-1.
∆V = (-0.8 - -3.8)107
∆V = 3.0107
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
∆EP = 3.0107m.
The probe is in circular motion so Fc = mv2/R.
But FG = GMm/R2 = Fc.
Thus mv2/R = GMm/R2 or mv2 = GMm/R.
Then EK = (1/2)mv2 = GMm/(2R).
From energy ∆EK = -∆EP or EK – EK0 = -∆EP.
Then EK0 = ∆EP + EK = ∆EP + GMm/(2R).
From g0 = GM/R02 we have GM = g0R02.
Then EK0 = ∆EP + g0R02m/(2R) so that
EK0 = m[3107 + 7.6(0.5107)2/(22107)] = 3.5107m.
Then (1/2)mv2 = 3.3107m and v = 8100 ms-1.
This solution
assumes the
satellite is in
orbit.
Topic 10: Fields
10.2 Fields at work
Solve problems involving gravitational potential
energy and gravitational potential.
Just know it!