Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
9.2.1 Define gravitational potential and
gravitational potential energy. Understand
that the work done in a gravitational field
is independent of path.
9.2.2 State and apply the expression for
gravitational potential due to a point mass.
9.2.3 State and apply the formula relating
gravitational field strength to gravitational
potential gradient.
9.2.4 Determine the potential due to one or more
point masses.
9.2.5 Describe and sketch the pattern of
equipotential surfaces due to one and two
point masses.
9.2.6 State the relation between equipotential
surfaces and gravitational field lines.
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
Define gravitational potential and gravitational
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
where G = 6.67×10−11 N m2 kg−2 of gravitation
If we multiply the above force by a distance r we
get
EP = -Gm1m2/r
gravitational
potential
where G = 6.67×10−11 N m2 kg−2
energy
FYI
The actual proof is beyond the scope of this
course.
Note, in particular, the minus sign.
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
Define gravitational potential and gravitational
potential energy.
EP = -Gm1m2/r
gravitational
potential
where G = 6.67×10−11 N m2 kg−2
energy
EXAMPLE: 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 9: Motion in fields
9.2 Gravitational field,potential,energy
Define gravitational potential and gravitational
potential energy.
The previous formula is for large-scale
gravitational fields (say, some distance from a
planet).
Recall the “local” formula for gravitational
potential energy:
∆EP = mg∆y where g = 9.8 m s-2.
local ∆EP
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 9: Motion in fields
9.2 Gravitational field,potential,energy
Understand that the work done in a gravitational
field is independent of path.
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.
d
2
d1
The total work done by gravity is
thus Wg = -mgd1.
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
Understand that the work done in a gravitational
field is independent of path.
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!
d2
s4
d1
s1
s2 s 3
s6
s5
FYI
The work done by gravity
is independent of the path
followed.
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
Understand that the work done in a gravitational
field is independent of path.
We call any force which does work independent of
path a conservative force.
Thus gravity is a conservative force.
Path 2
FYI
Only conservative forces have associated
potential energy functions.
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
Path 1. The work is different. A
Thus its work is not path-independent.
Thus it is not conservative.
B
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
Understand that the work done in a gravitational
field is independent of path.
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 9: Motion in fields
9.2 Gravitational field,potential,energy
Understand that the work done in a gravitational
field is independent of path.
If work is done in a conservative force field
then there is an associated potential energy
function.
∆EP = -W = -Fd cos 
potential energy
function
where F is a conservative force
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 9: Motion in fields
9.2 Gravitational field,potential,energy
Define gravitational potential and gravitational
potential energy.
EP = -Gm1m2/r
Gravitational
potential
where G = 6.67×10−11 N m2 kg−2
energy
We now define gravitational potential as
gravitational potential energy per unit mass just
as we did in Topic 6:
∆V = ∆EP/m
Gravitational
potential
V = -Gm/r
where G = 6.67×10−11 N m2 kg−2
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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.
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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
concentric spheres:
surfaces
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.
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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.
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
9.2.7 Explain the concept of escape speed from a
planet.
9.2.8 Derive an expression for the escape speed
of an object from the surface of a planet.
9.2.9 Solve problems involving gravitational
potential energy and gravitational potential.
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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
escape speed
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!)
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
Solve problems involving gravitational potential
energy and gravitational potential.
To escape we need vesc2 = 2GM/Re.
The kinetic energy alone must then be
E = (1/2)mvesc2 = (1/2)m(2GM/Re) = GMm/Re.
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 9: Motion in fields
9.2 Gravitational field,potential,energy
Solve problems involving gravitational potential
energy and gravitational potential.
Recall that Ep = -GMm/r.
Thus ∆EP = -GMm(1/R – 1/Re).
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
Solve problems involving gravitational potential
energy and gravitational potential.
Just know it!
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
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 9: Motion in fields
9.2 Gravitational field,potential,energy
Solve problems involving gravitational potential
energy and gravitational potential.
V = -GM/r so that V0 = -GM/R0.
But –g0R0 = -(GM/R02)R0 = -GM/R0 = V0.
Thus V0 = -g0R0.
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
Solve problems involving gravitational potential
energy and gravitational potential.
At R0 = 0.5107
V0 = -3.8107.
From previous
problem
g = -V0/R0
0.5107 = 5.0106 = R0.
g = - -3.8107 / 0.5107 = 7.6 m s-2.
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
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
v2 = 2(3.0107)
v = 7700
ms-1.
This solution assumes
probe not in orbit but
merely reaches altitude
(and returns).
∆V = (-0.8 - -3.8)107
∆V = 3.0107
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
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.
This
Thus mv2/R = GMm/R2 or mv2 = GMm/R.
solution
2
Then EK = (1/2)mv = GMm/(2R).
assumes
From energy ∆EK = -∆EP or EK – EK0 = -∆EP. the
satellite
Then EK0 = ∆EP + EK = ∆EP + GMm/(2R).
is in
From g0 = GM/R02 we have GM = g0R02.
orbit.
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.
Topic 9: Motion in fields
9.2 Gravitational field,potential,energy
Solve problems involving gravitational potential
energy and gravitational potential.
Just know it!