PC235 Winter 2013
Classical Mechanics
Assignment #7 Solutions
#1 (5 points) JRT Prob. 8.2
Two masses m1 and m2 move in a uniform gravitational field g and
interact via a potential energy U (r).
(a) Show that the Lagrangian can be decomposed as in Eq. (8.13) of the
text.
(b) Write down Lagrange’s equations for the three CM coordinates X, Y, Z,
and describe the motion of the CM. Write down the three Lagrange
equations for the relative coordinates and show clearly that the motion
of r is the same as that of a single particle of mass equal to the reduced
mass µ, with position r and potential energy U (r).
Solution
(a) The Lagrangian is L = T − U , or
1
1
L = m1 ṙ21 + m2 ṙ22 − [m1 gz1 + m2 gz2 + U (r)]
2
2
1
1 2
2
= M Ṙ − M gZ + µṙ − U (r) = Lcm + Lrel ,
2
2
(1)
where we have chosen rectangular coordinates with z measured vertically upward and Z is the z component of the CM position, Z =
(m1 z1 + m2 z2 )/M .
(b) The Lagrange equations for the three components of R are simply
M Ẍ = 0,
M Ÿ = 0,
M Z̈ = −M g,
(2)
so the CM moves like a projectile of mass M . The Lagrange equations
for the relative coordinates are
µr̈ = −∇U (r).
(3)
This is Newton’s second law for a particle of mass µ, position r, and
potential energy U (r).
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#2 (10 points) JRT Prob. 8.10
Two particles of equal mass m move on a frictionless horizontal surface
in the vicinity of a fixed force center, with potential energies U1 = 12 kr12 and
U2 = 21 kr22 . In addition, the particles interact with each other via a potential
energy U12 = 12 αkr2 , where r is the distance between them and α and k are
positive constants.
(a) Find the Lagrangian in terms of the CM position R and the relative
position r = r1 − r2 .
(b) Write down and solve the Lagrange equations for the CM and relative
coordinates X, Y, x, and y. Describe the motion.
Solution
2
(a) The KE in terms of R and r is given by eq. (8.10): T = 12 M Ṙ + 21 µṙ2 ,
and the PE is just U1 +U2 +U12 . Using eq. (8.9) with M = 2m, µ = m/2
to rewrite the PE in terms of R and r, we have
1 2 2 1
1
1 2
2
2
U = U1 +U2 +U12 = k(r1 +r2 )+ αkr = kR + k α +
r . (4)
2
2
2
2
Therefore,
1
1 2
1
1 2
2
2
L = M Ṙ + µṙ − kR − k α +
r,
2
2
2
2
(5)
where r2 = x2 + y 2 and R2 = X 2 + Y 2 .
(b) There are four Lagrange equations. That for the CM coordinate X
reads
∂L
d ∂L
=
=⇒ −2kX = M Ẍ,
(6)
∂X
dt ∂ Ẋ
with an identical equation for Y . The equation for x reads
d ∂L
1
∂L
=
=⇒ −k α +
x = µẍ,
(7)
∂x
dt ∂ ẋ
2
with an identical equation for y. Rearranging the equations of motion
for X and Y , we see that Ẍ = −(2k/M )X and Ÿ = −(2k/M )Y . This
2
p
equation
represents
simple
harmonic
motion
with
frequency
2k/M =
p
k/m. In general, the CM moves in an elliptical path (which may
be a circle or a line, depending on the initial conditions). Meanwhile, rearranging the equations of motion for x and y, we see that
ẍ = −(k/µ)(α + 1/2)x and ÿ = −(k/µ)(α + 1/2)y. Therefore, both
components
of the relative position r oscillate with the same frequency
p
k(α + 1/2)/µ, and r also moves around an ellipse.
#3 (10 points) JRT Prob. 8.16
We have proved in (8.49) that any Kepler orbit can be written in the
form r(φ) = c/(1 + ǫ cos φ), where c > 0 and ǫ ≥ 0. For the case that
0 ≤ ǫ < 1, rewrite this equation in Cartesian coordinates (x, y) and prove
that the equation can be cast in the form (8.51), which is the equation of an
ellipse. Verify the values of the constants given in (8.52).
Solution
We will make use of the relations x = r cos φ and y = r sin φ. Multiplying
both sides of the r(φ) equation by (1+ǫ cos φ) and rearranging gives r = c−ǫx.
Squaring both sides, and writing r2 = x2 +y 2 , we find (1−ǫ2 )x2 +2cǫx+y 2 =
c2 . If we divide both sides by (1 − ǫ2 ) and define d = cǫ/(1 − ǫ2 ), this gives
(x2 + 2dx) +
y2
c2
=
.
1 − ǫ2
1 − ǫ2
(8)
Next, we can add d2 to both sides to “complete the square” on the left,
resulting in
ǫ2
c2
c2
c2
y2
2
2
1+
=
=
+d =
= a2 , (9)
(x+d) +
2
2
2
2
2
2
1−ǫ
1−ǫ
1−ǫ
1−ǫ
(1 − ǫ )
where we have defined a = c/(1 − ǫ2 ). Finally, dividing through by a2 , we
arrive at
y2
(x + d)2
(x + d)2 y 2
+ 2
=
+ 2 = 1,
(10)
a2
a (1 − ǫ2 )
a2
b
√
where we have introduced the notation b = a 1 − ǫ2 . Collecting our definitions of a, b, and d, we see that
c
c
a=
(11)
, d = aǫ,
, b= √
2
1−ǫ
1 − ǫ2
3
as required.
#4 (5 points) JRT Prob. 8.18
An earth satellite is observed at perigee to be 250 km above the earth’s
surface and traveling at about 8500 m/s. Find the eccentricity of its orbit
and its height above the earth at apogee.
Solution
We are given the satellite’s height hmin = 250 km; therefore the distance
from the earth’s center is rmin = Re + hmin =6650 km. For any known satellite, we can approximate µ as m, since the satellite’s mass is a tiny fraction
of the earth’s. Therefore, the angular momentum is ℓ = mvmax rmin , and the
parameter c is
ℓ2
(vmax rmin )2
c=
=
(12)
γµ
GMe
(where we have used the fact that γ = GMe m). With the given numbers, we
get c = 7960 km. Then, since rmin = c/(1 + ǫ), we have
ǫ = (c − rmin )/rmin = 0.197.
(13)
Similarly, rmax = c/(1 − ǫ)=9910 km, so hmax = rmax − Re = 3510 km.
#5 (10 points) JRT Prob. 8.30
The general Kepler orbit is given in polar coordinates by r(φ) =
c/(1 + ǫ cos φ). Rewrite this in Cartesian coordinates for the cases that ǫ = 1
and ǫ > 1. Show that if ǫ = 1, you get the parabola of eq. (8.60), and
if ǫ > 1, you get the hyperbola of eq. (8.61). For the latter, identify the
constants α, β, and δ in terms of c and ǫ.
Solution
If we multiply both sides of the r(φ) equation by (1 + ǫ cos φ), replace r cos φ
by x, and rearrange, we find that r = c − ǫx. Squaring both sides gives
x2 + y 2 = c2 − 2cǫx + ǫ2 x2 . We now have two cases to consider. (a) If ǫ = 1,
the terms in x2 cancel and we’re left with y 2 = c2 −2cx, which is the equation
of a parabola. (b) If ǫ > 1, we find (ǫ2 − 1)x2 − 2cǫx − y 2 = −c2 . Completing
4
the square for x gives
(ǫ2 − 1)(x − δ)2 − y 2 = −c2 +
ǫ2 c 2
c2
=
.
ǫ2 − 1
ǫ2 − 1
(14)
Finally, multiplying both sides by (ǫ2 − 1)/c2 , we get
y2
(x − δ)2
−
=1
α2
β2
(15)
where
c
c
, β=√
,
2
−1
ǫ −1
which is the equation of a hyperbola.
α=
ǫ2
and δ =
ǫ2
ǫc
,
−1
(16)
#6 (15 points)
A spaceship of mass m is drifting through space because its engines have
failed. It is moving in a straight line with speed v. It is approaching the
planet Newtonia, which has a mass MN ≫ m. With the present course,
it will miss Newtonia by a distance d. Newtonia is known for its strange
gravity - the force it exerts on any mass m has an inverse-cube relation with
distance:
γm
F = − 3 r̂.
(17)
r
The astronauts on board the spaceship take some readings and determine
the value of the constant γ to be
8d2 v 2
.
(18)
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Your goal here is to prove that the astronauts are about to get a free tour of
Newtonia. That is, they will orbit it exactly once, before continuing on their
current trajectory. This is shown in the figure below.
γ=
(a) Write down a differential equation for u(φ) = 1/r(φ). It will be a secondorder DE. (3 points)
(b) Solve the DE to find u(φ). There will be two unknowns. (2 points)
(c) From the figure, we see that one initial condition is that r = ∞ when φ = 0.
The second initial condition is a bit more subtle, so I’ll do it for you. In the
du
textbook, between Eqns (8.39) and (8.40), we see that ṙ = − µℓ dφ
. Therefore,
5
µṙ
µ(−v)
1
du
=−
=−
=
(19)
dφ
ℓ
µdv
d
(the radial velocity ṙ is negative because the spaceship is initially moving in
the −r̂ direction. The angular momentum ℓ is positive because the spaceship
is orbiting counterclockwise about Newtonia). Using these initial conditions,
determine the values of the unknowns from part (b). (3 points)
(d) Derive the equation of the orbit, r(φ). (2 points)
(e) In a couple of sentences, explain how your r(φ) represents the path shown
in the figure. (2 points)
(f) On this path, what is the closest distance between the spaceship and Newtonia, as a function of d and v? Assume that the radius of Newtonia is
negligible. (1 point)
(g) What is the fastest speed of the spaceship during this round-trip of Newtonia? (2 points)
initial approach
speed v
final trajectory
speed v
d
Newtonia
φ=0
Fig. 1: Question #6. Trajectory of the spaceship in the gravitational field of Newtonia.
Solution
(a) Eq. (8.41) from the class notes indicates that
u′′ = −u −
µF (1/u)
γmµu3
=
−u
+
2 2
ℓ 2 u2
γmµℓ u = u
−1 ,
ℓ2
(20)
(21)
where u = u(φ) and the primes denote differentiation with respect to
φ. This can be tidied up quite a bit by noticing that, from the given
initial conditions, ℓ = dmv (this must be constant). Therefore,
γmµ
8d2 v 2 m2 MN
′′
u =u 2 2 2 −1 = u
−1
(22)
dmv
9d2 m2 v 2 (m + MN )
8MN
= u
−1 ,
(23)
9(m + MN )
6
where we have used µ = mMN /(m + MN ) (MN is the mass of Newtonia). If we make the assumption that the spacecraft has a considerably
smaller mass than the planet, then the masses cancel out, and we are
left simply with
u
u′′ + = 0.
(24)
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Note carefully that this answer is tidy only because of the particular
form of γ.
(b) We’ve seen equations like this all throughout
p PC235. The solutions is
SHM in φ with an angular frequency of 1/9. That is,
u = A cos(φ/3) + B sin(φ/3).
(25)
(c) Since r = ∞ when φ = 0, we have u(0) = 0, and therefore A = 0.
From the hint in the problem statement, we see that B3 cos(0) = d1 , or
B = d3 .
(d) Therefore, the required solution is u =
r(φ) =
d
3 sin
3
d
sin(φ/3), or
φ
3
.
(26)
(e) Our answer for r(φ) indicates that r = ∞ when sin(φ/3)=0. That is,
when φ = 0 and 3π. The angle of the spacecraft’s position is therefore
bound between 0 and 3π, as the figure indicates.
(f) The distance of closest approach is formally found by setting dr/dφ =
0. Realistically though, we notice that - since d is constant - we can
minimize r simply by maximizing its denominator. Since sin(φ/3) has
a maximum value of 1, r has a minimum value of d/3. As expected, it
occurs when φ = 3π/2; halfway through the round trip.
(g) By conservation of angular momentum, the fastest speed occurs at the
point of closest approach. Since the speed is v when the perpendicular
distance is d, a perpendicular distance of d/3 means that the speed
must be 3v.
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#7 (5 BONUS points) JRT Prob. 8.31
Consider the motion of two particles subject to a repulsive inversesquare force (for example, two positive charges). Show that this system has
no states with E < 0 (as measured in the CM frame), and that in all states
with E > 0, the relative motion follows a hyperbola. Sketch a typical orbit.
Solution
If we write Fr = γ/r2 (with γ positive), then E = T + U = T + γ/r, which is
never negative. The transformed radial equation then reads u′′ = −u − γµℓ2 .
This has the solution
1
c
r= =
,
(27)
u
ǫ cos φ − 1
where c = ℓ2 /γµ > 0 and ǫ is a positive constant. Since E ≥ 0, it follows
that ǫ ≥ 1. If ǫ = 1, it is easy to check that the preceding equation defines
a parabola. If ǫ > 1, then multiplying the preceding equation by ǫ cos φ − 1
gives the equation ǫx−r = c, which can be rearranged as in previous problems
on this assignment to read
(x − d)2 y 2
− 2 = 1,
(28)
a2
b
√
where a = c/(ǫ2 − 1), b = a ǫ2 − 1, and d = aǫ. This is the equation of
a hyperbola, as in the accompanying sketch, which shows the path of the
relative position (i.e. the position of particle 1 as seen from particle 2).
Fig. 2: Question #7 - motion of two particles subject to a repulsive inverse-square force (in the rest
frame of one of the particles).
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