- Source: Binet equation
The Binet equation, derived by Jacques Philippe Marie Binet, provides the form of a central force given the shape of the orbital motion in plane polar coordinates. The equation can also be used to derive the shape of the orbit for a given force law, but this usually involves the solution to a second order nonlinear, ordinary differential equation. A unique solution is impossible in the case of circular motion about the center of force.
Equation
The shape of an orbit is often conveniently described in terms of relative distance
r
{\displaystyle r}
as a function of angle
θ
{\displaystyle \theta }
. For the Binet equation, the orbital shape is instead more concisely described by the reciprocal
u
=
1
/
r
{\displaystyle u=1/r}
as a function of
θ
{\displaystyle \theta }
. Define the specific angular momentum as
h
=
L
/
m
{\displaystyle h=L/m}
where
L
{\displaystyle L}
is the angular momentum and
m
{\displaystyle m}
is the mass. The Binet equation, derived in the next section, gives the force in terms of the function
u
(
θ
)
{\displaystyle u(\theta )}
:
F
(
u
−
1
)
=
−
m
h
2
u
2
(
d
2
u
d
θ
2
+
u
)
.
{\displaystyle F(u^{-1})=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u\right).}
Derivation
Newton's Second Law for a purely central force is
F
(
r
)
=
m
(
r
¨
−
r
θ
˙
2
)
.
{\displaystyle F(r)=m\left({\ddot {r}}-r{\dot {\theta }}^{2}\right).}
The conservation of angular momentum requires that
r
2
θ
˙
=
h
=
constant
.
{\displaystyle r^{2}{\dot {\theta }}=h={\text{constant}}.}
Derivatives of
r
{\displaystyle r}
with respect to time may be rewritten as derivatives of
u
=
1
/
r
{\displaystyle u=1/r}
with respect to angle:
d
u
d
θ
=
d
d
t
(
1
r
)
d
t
d
θ
=
−
r
˙
r
2
θ
˙
=
−
r
˙
h
d
2
u
d
θ
2
=
−
1
h
d
r
˙
d
t
d
t
d
θ
=
−
r
¨
h
θ
˙
=
−
r
¨
h
2
u
2
{\displaystyle {\begin{aligned}&{\frac {\mathrm {d} u}{\mathrm {d} \theta }}={\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {1}{r}}\right){\frac {\mathrm {d} t}{\mathrm {d} \theta }}=-{\frac {\dot {r}}{r^{2}{\dot {\theta }}}}=-{\frac {\dot {r}}{h}}\\&{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}=-{\frac {1}{h}}{\frac {\mathrm {d} {\dot {r}}}{\mathrm {d} t}}{\frac {\mathrm {d} t}{\mathrm {d} \theta }}=-{\frac {\ddot {r}}{h{\dot {\theta }}}}=-{\frac {\ddot {r}}{h^{2}u^{2}}}\end{aligned}}}
Combining all of the above, we arrive at
F
=
m
(
r
¨
−
r
θ
˙
2
)
=
−
m
(
h
2
u
2
d
2
u
d
θ
2
+
h
2
u
3
)
=
−
m
h
2
u
2
(
d
2
u
d
θ
2
+
u
)
{\displaystyle F=m\left({\ddot {r}}-r{\dot {\theta }}^{2}\right)=-m\left(h^{2}u^{2}{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+h^{2}u^{3}\right)=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u\right)}
The general solution is
θ
=
∫
r
0
r
d
r
r
2
2
m
L
2
(
E
−
V
)
−
1
r
2
+
θ
0
{\displaystyle \theta =\int _{r_{0}}^{r}{\frac {\mathrm {d} r}{r^{2}{\sqrt {{\frac {2m}{L^{2}}}(E-V)-{\frac {1}{r^{2}}}}}}}+\theta _{0}}
where
(
r
0
,
θ
0
)
{\displaystyle (r_{0},\theta _{0})}
is the initial coordinate of the particle.
Examples
= Kepler problem
=Classical
The traditional Kepler problem of calculating the orbit of an inverse square law may be read off from the Binet equation as the solution to the differential equation
−
k
u
2
=
−
m
h
2
u
2
(
d
2
u
d
θ
2
+
u
)
{\displaystyle -ku^{2}=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u\right)}
d
2
u
d
θ
2
+
u
=
k
m
h
2
≡
constant
>
0.
{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {k}{mh^{2}}}\equiv {\text{constant}}>0.}
If the angle
θ
{\displaystyle \theta }
is measured from the periapsis, then the general solution for the orbit expressed in (reciprocal) polar coordinates is
l
u
=
1
+
ε
cos
θ
.
{\displaystyle lu=1+\varepsilon \cos \theta .}
The above polar equation describes conic sections, with
l
{\displaystyle l}
the semi-latus rectum (equal to
h
2
/
μ
=
h
2
m
/
k
{\displaystyle h^{2}/\mu =h^{2}m/k}
) and
ε
{\displaystyle \varepsilon }
the orbital eccentricity.
Relativistic
The relativistic equation derived for Schwarzschild coordinates is
d
2
u
d
θ
2
+
u
=
r
s
c
2
2
h
2
+
3
r
s
2
u
2
{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {r_{s}c^{2}}{2h^{2}}}+{\frac {3r_{s}}{2}}u^{2}}
where
c
{\displaystyle c}
is the speed of light and
r
s
{\displaystyle r_{s}}
is the Schwarzschild radius. And for Reissner–Nordström metric we will obtain
d
2
u
d
θ
2
+
u
=
r
s
c
2
2
h
2
+
3
r
s
2
u
2
−
G
Q
2
4
π
ε
0
c
4
(
c
2
h
2
u
+
2
u
3
)
{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {r_{s}c^{2}}{2h^{2}}}+{\frac {3r_{s}}{2}}u^{2}-{\frac {GQ^{2}}{4\pi \varepsilon _{0}c^{4}}}\left({\frac {c^{2}}{h^{2}}}u+2u^{3}\right)}
where
Q
{\displaystyle Q}
is the electric charge and
ε
0
{\displaystyle \varepsilon _{0}}
is the vacuum permittivity.
= Inverse Kepler problem
=Consider the inverse Kepler problem. What kind of force law produces a noncircular elliptical orbit (or more generally a noncircular conic section) around a focus of the ellipse?
Differentiating twice the above polar equation for an ellipse gives
l
d
2
u
d
θ
2
=
−
ε
cos
θ
.
{\displaystyle l\,{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}=-\varepsilon \cos \theta .}
The force law is therefore
F
=
−
m
h
2
u
2
(
−
ε
cos
θ
l
+
1
+
ε
cos
θ
l
)
=
−
m
h
2
u
2
l
=
−
m
h
2
l
r
2
,
{\displaystyle F=-mh^{2}u^{2}\left({\frac {-\varepsilon \cos \theta }{l}}+{\frac {1+\varepsilon \cos \theta }{l}}\right)=-{\frac {mh^{2}u^{2}}{l}}=-{\frac {mh^{2}}{lr^{2}}},}
which is the anticipated inverse square law. Matching the orbital
h
2
/
l
=
μ
{\displaystyle h^{2}/l=\mu }
to physical values like
G
M
{\displaystyle GM}
or
k
e
q
1
q
2
/
m
{\displaystyle k_{e}q_{1}q_{2}/m}
reproduces Newton's law of universal gravitation or Coulomb's law, respectively.
The effective force for Schwarzschild coordinates is
F
=
−
G
M
m
u
2
(
1
+
3
(
h
u
c
)
2
)
=
−
G
M
m
r
2
(
1
+
3
(
h
r
c
)
2
)
.
{\displaystyle F=-GMmu^{2}\left(1+3\left({\frac {hu}{c}}\right)^{2}\right)=-{\frac {GMm}{r^{2}}}\left(1+3\left({\frac {h}{rc}}\right)^{2}\right).}
where the second term is an inverse-quartic force corresponding to quadrupole effects such as the angular shift of periapsis (It can be also obtained via retarded potentials).
In the parameterized post-Newtonian formalism we will obtain
F
=
−
G
M
m
r
2
(
1
+
(
2
+
2
γ
−
β
)
(
h
r
c
)
2
)
.
{\displaystyle F=-{\frac {GMm}{r^{2}}}\left(1+(2+2\gamma -\beta )\left({\frac {h}{rc}}\right)^{2}\right).}
where
γ
=
β
=
1
{\displaystyle \gamma =\beta =1}
for the general relativity and
γ
=
β
=
0
{\displaystyle \gamma =\beta =0}
in the classical case.
= Cotes spirals
=An inverse cube force law has the form
F
(
r
)
=
−
k
r
3
.
{\displaystyle F(r)=-{\frac {k}{r^{3}}}.}
The shapes of the orbits of an inverse cube law are known as Cotes spirals. The Binet equation shows that the orbits must be solutions to the equation
d
2
u
d
θ
2
+
u
=
k
u
m
h
2
=
C
u
.
{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {ku}{mh^{2}}}=Cu.}
The differential equation has three kinds of solutions, in analogy to the different conic sections of the Kepler problem. When
C
<
1
{\displaystyle C<1}
, the solution is the epispiral, including the pathological case of a straight line when
C
=
0
{\displaystyle C=0}
. When
C
=
1
{\displaystyle C=1}
, the solution is the hyperbolic spiral. When
C
>
1
{\displaystyle C>1}
the solution is Poinsot's spiral.
= Off-axis circular motion
=Although the Binet equation fails to give a unique force law for circular motion about the center of force, the equation can provide a force law when the circle's center and the center of force do not coincide. Consider for example a circular orbit that passes directly through the center of force. A (reciprocal) polar equation for such a circular orbit of diameter
D
{\displaystyle D}
is
D
u
(
θ
)
=
sec
θ
.
{\displaystyle D\,u(\theta )=\sec \theta .}
Differentiating
u
{\displaystyle u}
twice and making use of the Pythagorean identity gives
D
d
2
u
d
θ
2
=
sec
θ
tan
2
θ
+
sec
3
θ
=
sec
θ
(
sec
2
θ
−
1
)
+
sec
3
θ
=
2
D
3
u
3
−
D
u
.
{\displaystyle D\,{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}=\sec \theta \tan ^{2}\theta +\sec ^{3}\theta =\sec \theta (\sec ^{2}\theta -1)+\sec ^{3}\theta =2D^{3}u^{3}-D\,u.}
The force law is thus
F
=
−
m
h
2
u
2
(
2
D
2
u
3
−
u
+
u
)
=
−
2
m
h
2
D
2
u
5
=
−
2
m
h
2
D
2
r
5
.
{\displaystyle F=-mh^{2}u^{2}\left(2D^{2}u^{3}-u+u\right)=-2mh^{2}D^{2}u^{5}=-{\frac {2mh^{2}D^{2}}{r^{5}}}.}
Note that solving the general inverse problem, i.e. constructing the orbits of an attractive
1
/
r
5
{\displaystyle 1/r^{5}}
force law, is a considerably more difficult problem because it is equivalent to solving
d
2
u
d
θ
2
+
u
=
C
u
3
{\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u=Cu^{3}}
which is a second order nonlinear differential equation.
See also
Bohr–Sommerfeld quantization § Relativistic orbit
Classical central-force problem
General relativity
Two-body problem in general relativity
Bertrand's theorem
References
Kata Kunci Pencarian:
- Relasi perulangan
- Binet equation
- Fibonacci sequence
- Classical central-force problem
- Jacques Philippe Marie Binet
- Alfred Binet
- Kepler problem
- Bohr–Sommerfeld model
- Recurrence relation
- Cauchy–Binet formula
- Bertrand's theorem
