Selected Topics in Physics
a lecture course for 1st year students
by W.B. von Schlippe
Spring Semester 2007
Lecture 4
Kepler Problem (continued)
1
Kepler Problem (continued)
The Kepler problem is the problem of the motion of two masses
in gravitational interaction.
For two point masses, m1 and m2,
located at r1 and r2, respectively
the gravitational force exerted by
m2 on m1 is
m1
y
F12
F21=-F12
r1
m2
r2
x
G
mm
F12 = G G 1 G2 2
r1 − r2
G G
r2 − r1
G G
r1 − r2
and the force exerted by m1 on m2 is
by Newton’s third law
G
G
F21 = − F12
2
The equations of motion for the two bodies are:
m1m2 G G
G
m1r1 = G G G 3 ( r2 − r1 )
r1 − r2
(4.1)
m1m2 G G
G
m2 r2 = −G G G 3 ( r2 − r1 )
r1 − r2
(4.2)
There follows the conservation of total momentum:
G
G
G
G
P = MR = m1r1 + m2 r2 = constant; M = m1 + m2
G
G
G
R = ( m1r1 + m2 r2 ) ( m1 + m2 )
where
of the centre of mass.
is the radius vector
Equivalent with the conservation of total momentum is the following
equation of motion:
G
G
P = MR = 0
(4.3)
3
Then we have found a second eqn of motion in terms of the
G G G
relative coordinate r = r1 − r2
G
r =−
μ γ
r
2
rˆ with γ = G μ M
(4.4)
which is the equation of motion of a particle of mass μ in the
gravitational field of a mass M that is at rest at the origin of the
coordinate system.
μ=
m1m2
m1 + m2
is the reduced mass.
Then we have found two more conservation laws:
conservation of energy:
conservation of angular momentum
G G G G
n̂ = r × r r × r
1 G 2 γ
μr − = E
2
r
G
G G
L = μr × r
= μ r ϕ nˆ = constant
2
(4.5)
(4.6)
4
hence
1 2
L2
γ
− =E
μr +
2
r
2
2μ r
(4.7)
L2
γ
Veff ( r ) =
−
2μ r 2 r
effective P.E.:
10
which has a minimum at
5
r0 = L2 μγ
0.2
-5
-10
0.4
0.6
0.8
(4.8)
1
Veff ( r0 ) = −
μγ 2
2
2L
(4.9)
5
hence
E≥−
μγ 2
2 L2
The equal sign corresponds to circular motion with radius
r = − γ 2E
(4.10)
For Emin<E<0 the motion lies between a minimum and a maximum
value of
r, which we get from the condition
T = 0 hence E = Veff ( r )
γ
L2
E=
−
2
2μ r
r
i.e.
hence
rmin,max
γ
L
⎛ γ ⎞
=−
± ⎜
+
⎟
2E
⎝ 2 E ⎠ 2μ E
2
2
(4.11)
6
Then we eliminate the time from Eq. (4.7) using (4.6) and end
up with two Deqs.:
ϕ =
L
μr 2
(4.6a)
2
⎛ du ⎞
2μγ
2μ E
1
2
u
u
,
u
+
−
=
=
⎜
⎟
2
2
d
L
L
r
ϕ
⎝
⎠
(4.12)
Eq. (4.12) has the solution
or
1 μγ
u = = 2 + A cos (ϕ + α )
r L
(4.13)
L2 μγ
r=
1 + A ( L2 μγ ) cos (ϕ + α )
(4.14)
This is the eqn of the trajectory; the time dependence can be found
by substituting
r
into (4.6a) and carrying out one final integration.
7
The constants A and α are integration constants of the 2nd order
DEq. (3.28), but the solution we have found must satisfy the 1st order
DEq. (3.27), which has only one integration constant.
Substituting the solution (4.13) into (4.12) we get:
μγ
⎛ μγ
⎞⎛
A2 sin 2 (ϕ + α ) + ⎜ 2 + A cos (ϕ + α ) ⎟⎜ A cos (ϕ + α ) − 2
L
⎝L
⎠⎝
hence
And since
2 μ E ⎛ μγ ⎞
2
A = 2 +⎜ 2 ⎟
L
⎝L ⎠
⎞ 2μ E
⎟= 2
L
⎠
2
E ≥ − μγ 2 2 L2
A2 ≥ 0
8
Thus there are solutions of four classes:
(i )
(i )
A = 0;
(ii )
0 < A < γμ L2 ;
(iii )
A = γμ L2 ;
(iv)
A > γμ L2 .
A=0
E = Emin = − μγ 2 2 L2 ,
Emin < E < 0,
E = 0,
E > 0.
is the circular motion of radius
r = L2 μγ = − γ 2 E
Tracing our transformations back to r1 and r2 in the CMS, we have
G
G
m2 r2 ( t ) = − m1r1 ( t ) ,
r1 ( t ) r2 ( t ) = m2 m1
Conclusion: the two bodies move in concentric circles of radii,
inversely proportional to their masses, and are always in opposition.
9
(ii )
0 < A < γμ L2
In this case
r
remains finite for all values of φ
and we note that
r (ϕ + 2π ) = r (ϕ )
i.e. the orbit is closed.
From the theory of conic sections we know that it is an ellipse.
If we choose α = 0, then the major axis of the ellipse corresponds to
φ = 0, and we have
rmax + rmin = 2a = − γ E
and hence
a = − γ 2E
10
The distance r at φ=π/2 is the semi-latus rectum:
A=
L2
(4.15)
γμ
and the equation of the ellipse becomes
r=
with the eccentricity
A
1 + ε cos ϕ
ε = 1− A a
(4.16)
and
A > 0 was assumed.
Had we chosen A < 0 (but of the same magnitude), then we would have
got instead of (4.16) (Exercise!):
r = A (1 − ε cos ϕ )
which is the same ellipse but referred to the other focus.
11
Now recall once more the equation of angular momentum conservation:
L = μ r 2ϕ
If we divide by 2μ and multiply by dt, then we get
L
1
dt = r 2 dϕ = dF
2μ
2
which is the area swept by the radius vector in the time interval dt, and
since by angular momentum conservation L/2μ is constant, we can
integrate to get the area swept by the radius vector in a finite interval of time:
ΔF = ( L 2μ ) Δt
and if we choose that time to be the time
F
T
of one revolution, then
is the area of the ellipse,
F = ( L 2 μ ) T = π ab
where b is the semi-minor axis:
b = aA
12
Using (4.15) to eliminate the angular momentum
L
we get
3
1 γ
T =πa 2
2 μ
and with
γ = Gμ M
after a little calculation:
a 3 GM
=
2
T
4π 2
(4.17)
Now, by Kepler’s third law, the ratio a3/T2 is the same for all planets known
by his time. But we see that in Eq. (4.17) M is the sum of the masses of
the sun and the planet. We conclude that the mass of the sun is so much
greater than the masses of the planets that the discrepancy is not observable.
Therefore, applying (4.17) to the solar system,
within observational accuracy.
M
is the mass of the sun
13
Then, since
G
is measured in terrestrial experiments (e.g. using a
torsion balance) and a and T are found by astronomical observation,
Eq. (3.32) gives us the solar mass:
3
a
M = 4π 2
GT 2
and hence with values appropriate for the earth:
a = 1 a.u. = 1.4960 ×1011 m, T = 1 yr = 3.1557 × 107 s,
G = 6.6726 ×10−11 m3 kg −1s −2
M : = 1.9893 ×1030 kg
14
(iii )
A = γμ L2
r=
A
1 + cos ϕ
This is the equation of a parabola in polar coordinates and we have
r → ∞ for ϕ → ±π
(iv)
A > γμ L2 .
r=
A
, ε >1
1 + ε cos ϕ
This is the equation of a hyperbola in polar coordinates.
The requirement of positivity of
r
implies
cos ϕ > − 1 ε
15
In astronomical terms hyperbolic orbits are the orbits of non-recurring
comets.
Hyperbolic orbits also arise in the case of repulsive forces, such as
the Coulomb force between a gold nucleus and an alpha particle
(Rutherford experiment!); in this case the equation of motion is
G zZe 2
μ r = 2 rˆ
r
Ze the charge of
the target, which is supposed to be stationary; e is the elementary
where
ze
is the charge of the projectile and
charge (= charge of a proton).
In the case of Rutherford scattering we have z = 2 and
Z = 79.
Of interest is to find the distance of closest approach, i.e. the shortest
distance between the alpha particle and the gold nucleus – Exercise!
We shall come back to this question in our lectures on nuclear
physics.
16
perihelion: the vertex
nearest the sun.
Kepler’s 1st Law: a planet moves
on an ellipse with the sun standing
in one of the foci of the ellipse.
aphelion: the vertex
furthest from the sun.
Kepler’s 2nd Law: the radius
vector of the planet sweeps
equal areas in equal intervals
of time.
17
Tides
In the discussion of tides it will be convenient to use the concept of a potential.
Thus consider the potential energy of a particle of mass m in the gravitational
field of a mass
M
located at the origin:
V (r ) = −
GmM
= m Φ (r )
r
This defines the gravitational potential
Φ of M
Φ (r ) = −
at
r:
GM
r
18
M
at r
If
is at a point with radius vector
a, then the gravitational potential
is
GM
G
Φ (r ) = − G G
r −a
If
r/a < 1, then we can expand the potential in powers of r/a:
2
3
⎡ r
⎛
1⎛ r ⎞
r
⎛ ⎞ ⎞⎤
2
⎢1 + μ + ⎜ ⎟ ( 3μ − 1) + O ⎜⎜ ⎜ ⎟ ⎟⎟ ⎥
2⎝a⎠
⎢⎣ a
⎝ ⎝ a ⎠ ⎠ ⎥⎦
G
G
where μ = cos α , and α is the angle between a and r
GM
G
Φ (r ) = −
a
(4.18)
Thus we can find the gravitational potential on the surface of the earth
due to the moon.
Let the centre of the moon lie on the
centre of the earth, and let vector
r
z axis at a distance a
from the
be the radius vector of a point …
19
… on the surface of the earth: see figure
earth
r
moon, mass M
α
z
a
The gravitational field at
r
is defined by
G G
G
g ( r ) = −∇Φ ( r )
20
2
3
⎛
r
1⎛ r ⎞
G G
G GM ⎡ r
⎛ ⎞ ⎞⎤
2
g ( r ) = −∇ Φ ( r ) =
∇ ⎢1 + μ + ⎜ ⎟ ( 3μ − 1) + O ⎜ ⎜ ⎟ ⎟ ⎥
⎜ ⎝ a ⎠ ⎟⎥
a
2⎝a⎠
⎢⎣ a
⎝
⎠⎦
2
3
⎛
⎤
GM ⎧⎪ ⎛ r ⎞ 1 ⎡⎛ r ⎞
r
⎛ ⎞ ⎞ ⎫⎪
2
=
⎨∇ ⎜ μ ⎟ + ∇ ⎢⎜ ⎟ ( 3μ − 1) ⎥ + O ⎜⎜ ∇ ⎜ ⎟ ⎟⎟ ⎬
a ⎪⎩ ⎝ a ⎠ 2 ⎣⎢⎝ a ⎠
⎦⎥
⎝ ⎝ a ⎠ ⎠ ⎪⎭
Now
r μ = z , hence ∇ ( r μ ) = ( 0, 0,1) = zˆ
and that means that the first term contributes a constant force in z direction
which is the same for all parts of the earth: this is the overall gravitational
attraction of the earth by the moon.
21
In the second term we have
2
1
⎛r⎞
2
2 2
2
3
1
3
r
r
μ
=
μ
−
−
(
)
(
)
⎜ ⎟
2
a
a
⎝ ⎠
1
1 2
2
2
2
2
⎡
⎤
= 2 ⎣3 z − ( x + y + z ) ⎦ = − 2 ( x + y 2 − 2 z 2 )
a
a
and hence the tidal field:
2
⎤
GM
G G GM ⎡⎛ r ⎞
2
gt ( r ) =
∇ ⎢⎜ ⎟ ( 3μ − 1) ⎥ = − 3 ∇ ( x 2 + y 2 − 2 z 2 )
2a ⎢⎣⎝ a ⎠
2a
⎥⎦
GM
= − 3 ( x , y , −2 z )
a
22
Here, for reference, again the final result:
GM
G G
g t ( r ) = − 3 ( x , y , −2 z )
a
Discussion (see also the figure on the next slide):
x = y = 0, then there is a tidal force towards the moon for z > 0
and away from the moon at z < 0;
1.) If
2.) If
z = 0, then the tidal force is directed towards the centre of the earth;
3.) [Exercise!] find the polar angle θ at which the radial component of the
tidal force is equal to zero.
At this angle the tidal force is tangential to the earth’s surface, giving
rise to tidal currents. These tidal currents lead to large differences in the
levels of the sea surface near the coast, especially in river mouths.
23
GM
G G
g t ( r ) = − 3 ( x , y , −2 z )
a
moon
Earth
In the figure, tidal forces are shown on the earth’s surface caused by
the gradient of the lunar gravitational potential.
There are also tidal forces caused by the field of the sun: these are
by magnitude about half of the tidal forces caused by the moon.
They enhance the lunar tidal forces when earth, sun and moon are aligned
and diminish them when the moon is in the first or third quarter.
24
Compare the tidal forces on the earth’s surface
GM
G G
g t ( r ) = − 3 ( x , y , −2 z )
a
with the gravitational attraction of the earth:
GM E
g ( R) =
R2
The tidal force has its maximum when x = y = 0 and z = R [Exercise!]:
g
max
t
2GMR
=
a3
The ratio of the (maximum!) tidal force to the earth’s attraction is
25
g tmax 2 M ⎛ R ⎞
=
⎜ ⎟
g
ME ⎝ a ⎠
3
Now,
M M E 1 80
and
R a 1 60
hence
gtmax g 1.2 × 10−7
Similarly we get for the solar tidal force
−7
gtmax
g
0.6
×
10
Sun
and we see that the tidal force of the sun is half that of the moon.
26