Download Gravitation and Other Central Forces - RIT

Gravitation and Other Central Forces
Vern Lindberg
July 14, 2010
1
Introduction, the Universal Law of Gravity, and Coulomb’s
Law
Central forces, forces between two point objects that point along the line joining the objects,
are of widespread interest in physics. We have already discussed the isotropic harmonic
oscillator in 2 and 3 dimensions. The general law of gravity, the force that holds the
universe together, is treated in the non-relativistic sense as a central force, and planetary
motion was one of the first applications of mechanics (and remains an active field, http:
//en.wikipedia.org/wiki/Asteroid_impact_avoidance). The Coulomb force between
charges is also a central force, involved in atomic physics (recall the Bohr model of the
atom) and in Rutherford scattering.
In this chapter we will discuss the gravitational problem in detail, but also point out
expressions that are true for all central forces, attractive or repulsive.
Historically, the motion of the stars and planets was of paramount philosophical, religious,
and physical study. Fixed stars were arrayed into constellations with great myths around
them. The planets, asteres planetai or wandering stars, Mercury, Venus, Mars, Jupiter,
Saturn, the Sun and the Moon were studied carefully in order to predict the fates of
mankind. Johannes Kepler, working with careful records of Tycho de Brahe determined
empirical laws of motion, and set the sun at the center of the solar system.
Galileo Galilei improved on the Dutch spyglasses and created a telescope (8x magnification),
looked at the planets and got in big trouble. Isaac Newton founded much of mechanics, as
well as calculus, and could derive Kepler’s Laws. We will not discuss General Relativity at
all.
Newton’s Law of Universal Gravitation begins with two point masses—objects with diameters much smaller than their separation.
1
Figure 1: Point masses, separation, and forces. The notation F~i→j means gravitational
force acting on j due to i.
The force is
mi mj
F~i→j = −G 2 r̂i→j
rij
(1)
Here r̂i→j is a unit vector at the location of mj pointing away from mi , that is along the
direction from mi toward mj . Note that the force is attractive, proportional to the masses,
and inversely proportional to the square of separation distance.
My form of the equation differs a bit from that of Fowles and Cassiday, Eq. 6.1.1, but says
the same thing in what I think is clearer notation.
Measuring the value of the gravitational constant G is very hard, and it is the least precisely
measured1 of the fundamental physical constants. The most recent value from NIST (more
recent than the text) is
G = (6.67428 ± 0.00067) × 10−11 N · m2 /kg2
2
(2)
Gravitational Force between a Spherical Shell and a Point
Object
By 1669 Newton had the universal law figured out, but did not publish for another 7
years. What he needed to figure out was the gravitation force between objects that are not
point-like. To do this he needed to develop calculus.
Consider a uniform spherical shell of mass M and radius R and a point object of mass
m. The separation of the point from the center of the shell is r > R. The text sets up
the integral needed to find the net gravitational force. The result is the same as the force
between point objects.
The fractional uncertainty in G is 1×10−4 , compared to the next most uncertain quantity, Boltzmann’s
constant, known to a fractional uncertainty of 2 × 10−6 .
1
2
By extension, any spherically symmetric object obeys this law: spherical symmetry means
that the density depends at most on the distance from the center of the spherical object,
but not on the angular location.
If the point object is inside the spherical shell, there is no gravitational force between
them. This is analogous to the electric case, and is easily derived using the Gauss’s Law
for Gravity. Define the gravitational field at the location of a test object of mass m as
~g = F~ /m, then for a Gaussian surface enclosing mass M ,
I
~ = 4πGM
~g · dA
(3)
The integral is evaluated over the closed surface A.
3
Kepler’s Laws
Working with data collected by Tycho de Brahe, Johannes Kepler deduced the three laws
named for him. Primarily he looked at the data for Mars, visible throughout the year and
having an eccentricity of 0.09, a value the text describes as “highly elliptical”. Ha! The
semi-major axis is 0.44% larger than the semi-minor axis. Not “highly” in my book.
I. Law of Ellipses 1609 The orbit of each planet is an ellipse with the sun located at
one focus.
II Law of Equal Areas 1609 A line drawn from the sun to a planet sweeps out equal
area in equal times as the planet orbits the sun.
III Harmonic Law 1618 The square of the sidereal period of a planet is directly proportional to the cube of the semi-major axis of the planet. Sidereal period is the
period measured relative to the “fixed” stars.
The text has an interesting historical description of how Robert Hooke, Edmond Halley, and
Christopher Wren interacted with Newton, and how Newton proved these laws starting with
the Law of Universal Gravitation. More is at http://en.wikipedia.org/wiki/Newton’
s\_law\_of\_universal\_gravitation.
4
Kepler’s Second Law: Law of Equal Areas
Newton determined that this law is simply a statement of Conservation of Angular Momentum.
3
~ (relative to an origin) of a point particle moving with linear
The angular momentum, L,
momentum p~ located at ~r from the origin is defined as
~ = ~r × p~
L
(4)
The time rate of change of this angular momentum is easily shown to be the net torque,
~.
N
~
dL
d(~r × p~)
d~
p
~
=
= ~v × p~ + ~r ×
= 0 + ~r × F~ ≡ N
(5)
dt
dt
dt
If only central forces act the cross product, ~r × F~ = 0, the torque is zero, and
the angular momentum is a constant.
This has several consequences. First, the direction of the angular momentum is fixed. This
means that the object must move in a two-dimensional plane. We will describe the motion
using cylindrical polar coordinates.
The position is ~r = rêr and the linear momentum is p~ = m(ṙêr + rθ̇êθ ). Then
~ = (rêr ) × m(ṙêr + rθ̇êθ ) = mr2 θ̇k̂
L
(6)
Now we need the area swept out. Figure 2 shows the motion of the object in a small time
dt. The area of the shaded triangle is
1
dA = r(r dθ)
(7)
2
Figure 2: Object moves a small distance in a time dt
and
Ȧ =
dA
1
L
= r2 θ̇ =
= constant
dt
2
2m
(8)
Thus for all central forces, the areal velocity, Ȧ is constant, as is the angular
momentum.
Example 6.4.1 An object is subject to a central force, and all circular orbits have the
same area velocity. Find an expression for the force.
4
5
Review of Conic Sections
Recall that one definition for a conic section uses a focus and a directrix separated by a
distance q, with the type of conic section determined by the eccentricity, .
If we place the origin of our coordinate system at a focus (rightmost focus if there is more
than one) and deal with the on-axis case, a general conic section can be written
r=
q
α
=
1 + cos θ
1 + cos θ
1
1 + cos θ
=
r
r0 (1 + )
(9)
where α = q is the latus rectum, the distance to the conic section at θ = π/2. The closest
approach of the mass to the focus, r0 , is
r0 =
q
(1 + )
In Cartesian coordinates with the focus at the origin,
(1 − 2 )x2 + 22 qx + y 2 = 2 q 2
5.1
Ellipse: < 1
We can characterize an ellipse by semi-major and semi-minor axes, a and b,with
a2 − b2 = c2
a=
q
(1 − 2 )
b= √
q
1 − 2
r0 = (1 − )a
=
c
a
α=
b2
= (1 − 2 )a
a
r1 = (1 + )a
If we use the center of the ellipse as an origin, the equation for the on-axis ellipse is
x2 y 2
+ 2 =1
a2
b
(10)
with two foci located at x = ±c.
5.2
Parabola: = 1
For the parabola opening to the right, define r0 ≡ c as the distance from the focus to the
apex. Then q = α = 2c.
5
With the focus as the origin, the equation of the parabola is
y 2 = ±4c(c − x)
If the apex is used as the origin, then the equation for the parabola is
y 2 = 4cx
5.3
Hyperbola: > 1
Relative to the center of the branches of the hyperbolae, the foci are located at ±c and the
apexes are located at ±a. The equation of the hyperbola is
x2 y 2
− 2 =1
a2
b
where a2 + b2 = c2 . The eccentricity is = c/a. The branches are asymptotic to lines
y = ±bx/a.
6
6.1
Kepler’s First Law: The Law of Ellipses
For General Central Force
We start with a general central force,
F~ = f (r)êr
(11)
We have already established that motion is in a plane with constant angular momentum
~ Define angular momentum per unit mass
L.
~
~` = L = ~r × ~v
m
` = |~r × ~v | = r2 θ̇
(12)
Radial acceleration in polar coordinates is ar = r̈ − rθ̇2 so the radial equation of motion
is
m`2
mr̈ − 3 = f (r)
(13)
r
and the tangential equation of motion, using aθ = 2ṙθ̇ + rθ̈ = d(r2 θ̇)/dt, is
d 2
(r θ̇) = 0
dt
6
(14)
The tangential equation can be solved to give what we already know,
r2 θ̇ = L/m = ` = constant.
The radial equation becomes easier if we change variables to u = 1/r, so
1 du dθ
du
ṙ = − 2
= −`
u dθ dt
dθ
dθ d du
d2 u
d du
= −`
= −`2 u2 2
r̈ = −`
dt dθ
dt dθ dθ
dθ
and the radial equation becomes
d2 u
f (1/u)
+u=−
2
dθ
mu2 `2
The energy equation will be written later.
6.2
(15)
(16)
(17)
Universal Law of Gravity
Let’s write the Universal Law of Gravity for a small planet of mass m orbiting a massive star
of mass M assuming that the massive star does not move (we will relax this requirement
in a later chapter.) The text uses k = GM m, I will also use K = k/m = GM .
f (r) = −GM mu2 = −ku2 = −Kmu2
(18)
The equation of motion is then
k
K
d2 u
+u=
= 2
2
2
dθ
m`
`
(19)
with solution
K
(20)
`2
For simplicity, orient the axes so that θ0 = 0 and so θ = 0 is the point of closest approach.
Expressing the solution for r,
u = A cos(θ − θ0 ) +
r=
`2 /K
1 + (A`2 /K) cos θ
(21)
Comparing Equations 9 and 21 we recognize that the motion is a conic section with
A`2
`2
α=
(22)
k
K
Consider elliptical motion. The aphelion and perihelion (farthest and closest) distances are
found for θ = 0, π
α
α
r0 =
r1 =
(23)
1+
1−
=
7
E.g. The sun has a mass of 1.98892 × 1030 kg. The perihelion and aphelion for the earth
are 147.3 Gm and 152.1 Gm. Show that the eccentricity is 0.0160 and find the latus
rectum, the value of A, the semi-major axis, and the value of `.
Example 6.5.3 in Text Find the speed of a satellite in circular orbit in terms of g at the
surface of the earth of radius Re and the orbital radius rc .
Example 6.5.4 in Text The Hohmann Transfer. The Hohmann transfer is the most fuel
efficient way to take a satellite from a circular orbit of radius rc to an elliptical orbit
with an apogee of r1 2 . The transfer is an engine impulse that increases the speed of
the satellite from vc to v0 .
p
Show that the required increase in the speed is v0 /vc = 2r1 /(rc + r1 ) and evaluate
for transfer from a low earth circular orbit to an ellipse with apogee at the moon.
7
Kepler’s Third Law: The Harmonic Law
This relates the period of a satellite, τ , to its semi-major axis,a.
We know from Kepler’s Second Law that Ȧ = `/2 = constant. Integrating over a period
we get A = `τ /2 and hence
2A
τ=
(24)
`
It is straightforward to show that the area of an ellipse3 is A = πab, and from our knowledge
of conic sections we can write
√
2πab
2πa2 1 − 2
τ=
=
(25)
`
`
Now we know α = a(1 − 2 ) and that α = `2 /K, so with a little algebra we get
τ2 =
7.1
4π 2 3
4π 2 3
a =
a
K
GM
(26)
Units
We could use SI units of meters, kilograms, and seconds, but it is easier to use units natural
to the problem. If we are looking at planetary motion around the sun, and measure the
period in years, and the semi-major axis in A.U. (astronomical units, 1 A.U. = semi-major
axis of the earth), then K = GM = 4π 2 and τ 2 = a3 .
2
3
Another Hohmann transfer would then put it into circular orbit at the larger radius.
Start with the equation for an ellipse relative to its center, Eq. 10
8
For satellites around the earth, measure the period in lunar months and distance in terms
of L.U. = semi-major axis of the lunar orbit, and the same result occurs.
For satellites of the sun velocities would have units of A.U./year and the angular momentum
per unit mass would have units (A.U.2 /year.)
E.g. A gps satellite has a period of 12 hours. What is its altitude?
Express the period in lunar sidereal months τ = 0.5/27.322 = 0.018300 months.
Then the semi-major axis from τ 2 = a3 is a = 0.069445 L.U. and since 1 L.U. =
384748 km, the radius of the gps satellite is 26718 km and the altitude above the
surface of the earth is 20340 km.
E.g. 39P Oterma Comet Oterma has a semi-major axis of 7.228 A.U. and an eccentricity of 0.2434. Find the aphelion and perihlion distances, the period, the angular
momentum per unit mass, and the speed at perihelion.
7.2
Is Universal Gravitation Universal? Dark Matter
For 150 years from the time of Newton’s statement of the Universal Law of Gravity it
ruled supreme—not withstanding the extremes of General Relativity. The Law was used
to discover of Neptune and Pluto, and to predict the motion of objects in the solar system
and beyond.
In 1934 Fritz Zwicky noticed a problem. The text deals with it in some detail, but the
idea is this. Many galaxies have a central core and surrounding arms of low density. If
we use Gauss’s Law for gravity we can determine the gravitational field inside and outside
the core. Then using Newton’s second law we can determine the speed of stars at different
distances from the center of the core.
Inside the core the speed should vary as v ∼ r while outside the speed should vary as
v ∼ r−1/2 . Figure 6.6.1 shows the measured galactic rotation curve that varies dramatically
from the Keplerian/Newtonian curve. To account for the variation Zwicky posited dark
matter filling up much of the space around the galactic core.
Dark matter is now assumed to be 80% of the mass of the universe. What it is, and what
dark energy is is an active research area in Astrophysics.
8
Potential Energy, Potential, Field
It is easy to show that gravity is a conservative force, therefore we can define a potential
energy, V , for it. Consider motion from point A to point B as shown in Figure 3. For
9
conservative forces, work is independent of path, so rather than the general curve, evaluate
work along the dashed line, ACB.
Along the arc, AC, no work is done since the force is perpendicular to the path (êr · êθ = 0)
Along the radial portion, CB, the work is
Figure 3: Motion in a gravitational filed. Since the force is conservative, the work done
by gravity while the object moves from A to B does not depend on path. Rather than the
solid arbitrary path, use the dashed path to evaluate the work.
Z
r2
W = −(V2 − V1 ) =
r1
−GM m
êr ) · (drêr ) = −GM m
(
r2
Z
r2
r1
dr
= GM m
r2
1
1
−
r2 r1
(27)
NOTE: The text computes the work done by a force required to overcome
gravity, and so has the opposite sign.
For convenience we choose V = 0 at r = ∞, leaving us with an expression for the potential
energy,
GM m
V (r) = −
(28)
r
We have already mentioned gravitational field, ~g defined so that if a mass m is at a
particular location where the field is ~g , then the gravitational force on the mass is
F~ (r) = m~g (r)
(29)
GM
êr . If there are several point
For a single mass M the gravitational field is ~g (r) = −
r
masses acting on m, we must do a vector sum and the unit vectors êr are all different.
We can also define a gravitational potential, Φ, so that for a mass m the gravitational
potential energy is
V (r) = mΦ(r)
(30)
For a point mass, Φ(r) = −
GM
r
10
Finding the potential form several point masses is simply a scalar sum of the individual potentials, an easier task than adding vectors. Once we have Φ(~r) we can get the gravitational
field and force using a gradiant,
F~ (~r) = −∇V (~r)
~g (~r) = −∇Φ(~r)
(31)
Example 6.7.1 Find the gravitational potential for a uniform spherical shell (i) outside
and (ii) inside the shell. Show that this gives the expected fields at these locations.
Example 6.7.2 Find the gravitational potential and field for a uniform ring of radius R
in the plane of the ring and (i) outside and (ii) inside the ring. Evaluate for the far
field (r R), approximating and keeping two non-zero terms. Evaluate near the
center (r ≈ 0) and show that a mass near the center is repelled from the center.
9
Energy Equation for Central Forces
Earlier we determined the equation of motion (Newton’s Second Law) for central forces:
mr̈ −
m`2
= f (r)
r3
d2 u
f (1/u)
+u=−
dθ2
mu2 `2
(32)
Since the gravitational force is conservative, T + V = E = constant. We compute the
kinetic energy in polar coordinates,
1
1
T = m~v · ~v = m(ṙ2 + r2 θ̇2 )
2
2
(33)
Hence we can write the energy equations of motion for a general central force as
1
m(ṙ2 + r2 θ̇2 ) + V (r) = E = constant
2
(34)
or in terms of u = 1/r,
1 2
m`
2
9.1
"
du
dθ
2
#
+ u2 + V (1/u) = E
(35)
Energy Equation of Motion for Gravity
Using k = GM m and K = GM , we can write the potential energy as
V =−
k
= −ku
r
Φ=−
11
K
= −Ku
r
(36)
The text shows how the energy equation of motion can be solved to show that the motion
is a conic section. Since we have already shown this, I will instead determine an expression
for the eccentricity in terms of the energy.
We know that
r=
α
1 + cos θ
At
θ = 0,
Then
r = r0 =
α=
`2
K
α
`2
=
1+
K(1 + )
` = v0 r0
1 `2
Km
1
Km
= m 2−
E = mv02 −
2
r0
2 r0
r0
(37)
Putting in the expression for r0 and rearranging we get
2E`2
= (1 + )2 − 2(1 + ) = −1 + 2
mK 2
and
r
=
1+
2E`2
mK 2
(38)
(39)
Using α = (1 − 2 )a we can write the total energy, kinetic plus potential, as
E=−
Km
2a
(40)
We can use the energy to determine the type of conic section.
E < 0, < 1 ellipse
a>0
E = 0, = 1 parabola a = 0
E > 0, > 1 hyperbola a < 0
E.g. 1 A projectile is launched from a location a distance r0 from a planet of mass M .
Find the required initial speed so that the projectile escapes the planet.
In order to escape, the orbit must either be a parabola or an hyperbola, E ≥ 0.
Hence
1
Km
mv02 −
≥0
(41)
2
r0
which is easily solved to give a result that is independent of the projectile mass
r
2K
v0 ≥
(42)
r0
12
In real planetary systems perturbations from other planets will modify this: a projectile launched with a speed less than escape velocity and interact with other planets in
a “gravitational slingshot” process and then escape. This is commonly used by NASA
in planetary probe missions. http://en.wikipedia.org/wiki/Gravity_assist
9.2
Units again
Earlier we discussed measuring planetary motion about the sun by measuring time in
years, and distance in A.U. Doing this is defining time and distance in dimensionless units,
T = t/1 year, R = r/1 A.U. Then from Eq. 26, K = GM = 4π 2 , and τ 2 = a3 .
In these units the speed of the earth (assumed circular orbit) is
2πae
= 2π A.U./yr
τe
(43)
K = GM = ae ve2 = 4π 2
(44)
ve =
We also recognize that
Now define a dimensionless quantities for velocity, V , and distance, R.
V =
v
ve
R=
r
ae
(45)
This lets us simplify the expression for eccentricity
Suppose that a planet/comet is located at a radius R from the sun with velocity V making
an angle φ with the radius (see Figure 6.10.1), then
` = rv sin φ = (RV sin φ)ae ve = 2πRV sin φ
(46)
The energy per unit mass is E/m = (1/2)v 2 − K/r. We can write the eccentricity
2
2E`2
Kae
RV ae re sin φ 2
2 2
= 1+
= 1 + V ve −
mK 2
R
K
2
(RV sin φ)2
= 1+ V2−
R
(47)
(48)
Likewise we can express the energy, Eq. 40, in terms of the dimensionless units. Since
E = mv 2 /2 − mK/r,
a=
Km
ae ve2
ae
1
=
=
=
2
2
2
−2E
(2ae ve /r) − v
(2ae /r) − (v/ve )
(2/R) − V 2
13
(49)
E.g. 2 You observe comet 103P Hartley (see Figure 4) at a distance 2.07 A.U. from the
sun, with a speed 0.816 of the earth’s speed and the angle between the radius vector
and the velocity vector is 52.1◦ . Find
(a) The eccentricity.
(b) The semi major axis, semi-minor axis, the perihelion and aphelion distances.
(c) The angular momentum per unit mass of the comet.
(d) The speeds of the comet at perihelion and aphelion.
(e) The period of the comet.
Figure 4: The orbit of Comet 103P Hartley, expected perhelion Oct 28, 2010. The square
is the sun, the circle is the orbit of the earth. Shown with axes at the center of the ellipse,
units of A.U.
(a) R = 2.07
V = 0.816 so using Equation 48, the eccentricity is = 0.683
√
(b) From Eq. 49, a = 3.33 A.U. We know b = 1 − 2 a = 2.43 A.U., α = (1−2 )a =
14
1.777 A.U., and then we can use r = α/(1 + cos θ) to get r0 = 1.06, r1 = 5.61
A.U.
(c) Use ` = 2πRV sin φ = 8.37 A.U.2 /yr
(d) At aphelion and perihelion, sin φ = 1, so from ` = 2πRV we get VP = 1.26 and
VA = 0.237.
(e) Finally, using τ 2 = a3 we get τ = 6.08 yr.
Hohmann Transfer Using Energy Arguments, E.g. 6.10.2 A spacecraft is in a circular low-earth orbit. Some energy is added to take it to an elliptical orbit with
apogee at the orbit of gps satellites. Finally additional energy is added to take it into
a circular gps orbit with period of 12 hours. What is the required velocity boost at
each of the two burns, and what are the corresponding energy per unit mass boosts?
10
Effective Potential: Limits of the Radial Motion
The general energy equation for a central potential energy V (r) is
1
m(ṙ2 + r2 θ̇2 ) + V (r) = E = constant
2
(50)
but we also know that the angular momentum is a constant of the motion, ` = r2 θ̇ =
constant so the energy equation can we written
m 2 `2
ṙ + 2 + V (r) = E = constant
(51)
2
r
It is convenient and common to write an effective potential energy (often called just the
effective potential not to be confused with Φ) that is the sum of a centrifugal potential
energy, m`2 /(2r2 ) and the real potential energy V ,
U (r) =
m`2
+ V (r)
2r2
(52)
The energy equation of motion then looks like a one-dimensional differential equation,
m 2
ṙ + U (r) = E
2
(53)
An energy plot, U (r) versus r can be drawn, and the turning points are the intersection of
this curve with a horizontal line at energy E, Figure 5 for the gravitational case.
15
Applying this to the inverse square gravitational field, V (r) = −k/r, the turning points
occur for ṙ = 0, U (r) = E, and this can be written
−2Er2 − 2kr + m`2 = 0
(54)
with solutions, the turning points, at
√
k 2 + 2Em`2
(55)
−2E
Note that if E < 0, there are two positive real turning points, as expected for an ellipse.
r1,0 =
k±
Figure 5: Effective Potential for Gravitational Force, units are arbitrary. The gravitational
and centrifugal components are shown, and add to the efective potential. Horizontal lines
for two negative energies are shown: one has two turning points, the other is the lowest
energy circular orbit.
The circular orbit occurs when k 2 + 2Em`2 = 0. For E = 0 there is one positive turning
point and the other root is ∞. This is the parabola. Finally, if E > 0 there is one positve
real root, and one negative, meaningless, root, and we have an hyperbola.
You will see effective potentials again in quantum mechanics of the hydrogen atom.
11
Stabilty Analysis of Nearly Circular Orbits
A circular orbit is possible for any attractive central force, V (r) < 0. A circle implies ṙ = 0,
so the energy equation becomes m`2 /(2r2 ) + V (r) = E Given r = a we can then find the
required angular momentum per unit mass and speed.
16
For attractive central forces the vector component f (r) < 0. At a circular orbit with r̈ = 0
(no radial acceleration) the radial equation of motion is
0=
m`2
+ f (a)
a3
(56)
So circular orbits of radius a are an equilibrium state, but is that equilibrium stable? That
is, if we disturb the satellite by a small amount x so that the radius is r = a + x, will the
satellite feel a restoring force towards the circular orbit, or a repulsive force?
The equation of motion mr̈ = m`2 /r3 + f (r) becomes
mẍ = m`2 [a + x]−3 + f (a + x)
m`2 h
x i−3
=
1
+
+ f (a + x)
a3
a
(57)
(58)
Expanding the two expressions in a Taylor series about x = a, keeping first order terms,
m`2
a3
m`2
=
3
ha
mẍ =
=
i
x
1 − 3 + · · · + [f (a) + f 0 (a)x + · · · ]
a
h xi
−3 + [f 0 (a)x]
a
i
x
f (a)3 + f 0 (a)x
a
h
(59)
(60)
(61)
We can write this as
mẍ + Ax = 0
(62)
where
3f (a)
− f 0 (a)
(63)
a
If A > 0, the solution to the differential equation is oscillatory, and the circular orbit is
stable if small disturbances are made. If A < 0 the solution is a combination of exponentials
and the circular motion is unstable. If A = 0 we must use higher order terms in the Taylor
expansion to determine the stability.
A=−
So circular orbits are stable providing
a
f (a) + f 0 (a) < 0
3
(64)
E.g. Power law central force Suppose that f (r) = −crn . For what values of n are
circular orbits stable?
We have f 0 (r) = −ncrn−1 , so the criterion for stability is
a
−can − cnan−1 < 0
3
17
(65)
and this reduces to
n > −3
(66)
So inverse square (n = −2) and the isotropic oscillator (n = +1) both allow stable
circular orbits. If n = −4 the circular orbit is unstable, and if n = −3 more analysis
must be done (it turns out to be unstable.)
12
Apses, Apsidal Angles, and Orbital Precession
The points where the radius reaches an extreme value (maximum or minimum) is called an
apsis or apse. Perihelion and aphelion points are examples of apses (the plural of apsis).
The angle between two successive apses is called the apsidal angle, and we will determine
this angle for the nearly circular orbit discussed in the last section.
The period of the oscillatory motion discussed in the last section is
s
r
m
m
τr = 2π
= 2π
3f
(a)
A
− a − f 0 (a)
(67)
The time to move from the minimum to the maximum radius about the circle r = a is
τr /2, and we want to know by how much the polar angle θ increases in this time. Since
our orbit is nearly circular we can approximate
r
`
`
f (a)
θ̇ = 2 ≈ 2 = −
(68)
r
a
ma
Hence the apsidal angle in this case is
1
π
Ψ = τr θ̇ = p
2
3 + af 0 (a)/f (a)
(69)
E.g. Apsidal Angle for power law force Using f (r) = −arn we find that the apsidal
angle is
π
Ψ= √
(70)
3+n
This is independent of the radius of the orbit. For an inverse square law, Ψ = π and
for an isotropic oscillator Ψ = π/2, and both of these are termed reentrant since the
orbit just repeats itself without any precession.
Other powers lead to non-reentrant orbits. If n = −1.9, Ψ = 0.953π = 172◦ , so the
apses precess in a direction opposite the orbit.
18
E.g. 6.13.1 Precession of Mercury Recall that near the center of a ring of mass, F~ =
rêr , a repulsive force. (The text uses > 0 but it is NOT the eccentricity.) As a
first approximation let us describe the net gravitational force on Mercury as a linear
combination of the attractive force of the sun and the small repulsive force due to
the exterior planets, treated as rings of mass. Thus
f (r) = −
k
+ r
r2
(71)
It is easy to then get the apsidal angle
−1/2
2ka−3 + Ψ = π 3+a
−ka−2 + a
1/2
1 − k −1 a3
= π
1 − 4k −1 a3
(72)
(73)
Now expand the numerator and denominator separately as Taylor series, multiply
them out and keep terms to first order in a3 /k to get
3 3
Ψ≈π 1+ a
(74)
2k
With > 0 this means that the apses precess in the same direction as the orbit. In
fact Urbain Leverrier in 1877 had done a much more detailed and correct analysis and
found that the known planets would produce an effect leading to the precession of
the perihelion of Mercury being 52700 of arc per century while the observed precession
was 56500 of arc per century4 .
Two solutions were proposed: An undiscovered planet named Vulcan orbited at about
half the radius of Mercury (Vulcan never found), or the inverse square law was not
quite accurate and should be (2 + 1.612 × 10−7 ). A third alternative, an oblate
(football-shape) sun, could also cause this precession, however no oblateness has
been seen.
We now know that the discrepancy is due to the onset of general relativistic effects,
as shown by Einstein, and alternate explanations have been abandoned.
13
Rutherford Scattering
Between 1909 and 1914 Ernest Rutherford and his graduate students Hans Geiger and
Ernst Marsten did a series of experiments to determine the distribution of charge in an
4
By 1900 the numbers had been improved to a predicted 53400 and measured 57500 of arc per century.
19
atom. The electron had been discovered in 1899, a “particle” having little mass and a
negative charge. Atoms are neutral, so there must be positive charge in the atom, and
associated with the positive charge is most of the mass of the atom.
At the time J. J. Thompson’s model of the nucleus was a “plum-pudding” model of a homogeneous sphere of uniformly distributed mass and positive charge in which the negative
electrons were embedded, like raisins in a plum pudding5 .
Rutherford and his students set out to probe the distribution of mass and charge by sending
massive alpha particles at thin foils of metal. A foil was used to reduce the number of
multiple collisions, one alpha with several nuclei.
They expected small deviations, based on Thompson’s model but instead, in Rutherford’s
words, “It was quite the most incredible event that has ever happened to me in my life.
It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it
came back and hit you.”
I will sketch the basic set-up on the board. We send N particles at the foil that has n
atoms per unit area in the foil. We can measure the number of particles, dN , scattered
between angles θs and θs + dθs and write this in terms of a differential scattering cross
section σ(θs ) and an infinitesimal solid angle dΩ,
dN
= nσ(θs )dΩ
N
(75)
The differential cross-section has units of area, with the area of 10−28 m2 being called
a barn (b). The area of a uranium nucleus is approximately 1 b. A shed is 10−24 b =
10−52 m2 .
Solid angle, Ω, is defined in an analogous manner as an angle in radians.
θ(radians) = s/R
Ω(steradians) = Area/R2
(76)
(77)
The solid angle for a complete sphere is thus 4π, and this is the solid angle for any closed
surface. The differential solid angle, dΩ, that the particles are scattered into is the area of
an annulus (radius R sin θs and width R dθs ) divided by R2 , or
dΩ = 2π sin θs dθs
(78)
5
From Wikipedia’s Plum Pudding article, “Christmas pudding is a steamed pudding, heavy with dried
fruit and nuts, and usually made with suet. It is very dark in appearance — effectively black — as a
result of the dark sugars and black treacle in most recipes, and its long cooking time. The mixture can be
moistened with the juice of citrus fruits, brandy and other alcohol (some recipes call for dark beers such as
mild, stout or porter).”
20
so
dN
= nσ(θs )2π sin θs dθs
(79)
N
The differential cross section is the fraction of incident particle scattered into a particular
solid angle per scattering site.
Rutherford and his students measured the cross section. We will derive the theoretical
cross section assuming a nuclear model of the atom with the mass concentrated in a small
region at the center of the atom. We shall ignore the light negative electrons and just
consider scattering a positive particle (an α-particle) from a very heavy positive nucleus,
so heavy that we will consider it to be fixed.
Begin with the repulsive Coulomb force,
f (r) =
kqQ
r2
(80)
The text uses cgs units (charge in esu, distance in cm, and force in dynes) and k =1. We
can use SI units (coulombs, m, N) with k = 9 × 109 N m2 /C2 .
The energy is
kqQ
1
>0
E = mv 2 +
2
r
so the trajectories are hyperbolas. The solutions to the differential equation 17
(81)
d2 u
f
+u=− 2 2
dθ2
m` u
then becomes (compare to Equation 21)
r=
−kQq/m`2
1
m`2 /(kQq)
=
2
+ A cos(θ − θ0 )
−1 + m` A/(kQq) cos(θ − θ0 )
(82)
or for the energy solution (compare to text Eq. 6.10.7c)
r=
m`2 /kQq
q
2
−1 +
1 + k2Em`
cos(θ − θ0 )
2 Q2 q 2
(83)
A and θ0 , or E and θ0 are determined from initial conditions. The incoming particle is
closest to the scattering atom at θ0 .
Unlike the gravitational case we do not make θ0 = 0. Instead we imagine the incoming
particles are moving along the x-axis as shown in Figure 6. On that diagram we define
incident and scattering angles θ0 and θs and it is easily seen that
θs = π − 2θ0
21
(84)
Figure 6: Rutherford Scattering parameters.
Incoming particles start at r = ∞ and θ = 0. This means that the denominator of Equation
83 must be zero. Square and rearrange to get
2Em`2
(85)
1 + 2 2 2 cos2 θ0 = 1
k Q q
This can be rearranged to give
√
`
kQq
(86)
θs √
`
= 2Em
2
kQq
(87)
tan θ0 =
2Em
From Equation 84 we rewrite this
cot
Next we introduce the impact parameter b. Incoming particles that enter through an
annulus of radii b and b + db scatter between θs and θs + dθs (logically dθs < 0) . Using
` = bv0 and the total energy E = mv02 /2,
cot
θs
2bE
=
2
kQq
(88)
b=
kQq cot θ2s
2E
(89)
or
Take the derivative to get
kQq
db
=−
dθ
4E sin2 θs /2
22
(90)
Logically we can look at the incoming side and see
dN
= (2πbdb)n = nσ(θs )2π sin θs dθs
N
where the last expression is from Equation 79. Rearrange to get
b db σ(θs ) =
sin θs dθs (91)
(92)
Now use Equations 89 and 90 in 92, and the trig identity sin θs = 2 sin θ2s cos θ2s to
yield
1
Q2 q 2
(93)
σ(θs ) =
4
2
16E sin θs /2
This is the formula that Rutherford derived and was able to fit to the data (I’ll show you
an example).
In addition to showing that a nuclear model of the atom is the appropriate one, Rutherford
scattering is used an analysis tool. A beam of alpha particles is sent into a sample, and
scattering at an angle very close to 180◦ , “backscattering”, is measured. In University
Physics you discussed one-dimensional elastic scattering of a particle from a stationary
target and found the rebounding velocity in terms of the incident velocity,
v10 =
m2 − m1
v1
m2 + m1
This leads to a relation between the scattered alpha energy (mass 4) and the incident
energy,
m2 − m1 2
M −4 2
0
Eα =
Eα =
Eα
(94)
m2 + m1
M +4
Consider a sample that consists of a silicon substrate with a thin film of TaSi. We would
like to know how thick the film is, and what the elemental composition of the film is.
Sample data are shown in Figure 7.
Thus for incident energy of 2.2 MeV, alphas scattered from Tantalum should have energy
2.01 MeV, and those scattered from silicon should have energy 1.26 MeV. The edges of the
two features in the figure are at these energies.
As the alphas penetrate the sample, they will lose energy primarily in interactions with
electrons, so the data in Fig 7 shows alphas with less energy than that predicted by elastic
collisions.
23
Figure 7: Rutherford Backscattering (RBS) from two different samples of a TaSi alloy film
on top of a Si substrate. Incident energy is about 2.2 MeV. Composition as a function of
depth can be extracted from this data.
The Ta peaks are simpler to understand. Sample 1 shows a tall narrow feature for Ta,
indicating that the Ta is spread over a narrower depth in Sample 1 than in Sample 2.
The maximum count is higher for Sample 1, so there must be more Ta at the surface.
Looking at the Si features confirm this, the Si signal near the surface is smaller for Sample
1 meaning less Si.
Near the surface the alphas can scatter either from the Ta or the Si. Once through the
film, the alphas only scatter from the Si, so once the thin film layer is passed, the Si signal
becomes larger, and is the same for both samples.
Full analysis programs allow us to extract the composition as a function of depth, and
to get the depth in nm. RBS can ample to a depth of about 1 µm. For more, http:
//www.eaglabs.com/training/tutorials/rbs_theory_tutorial/
24