Equation giving the form of a central force
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 [disambiguation needed ] ordinary differential equation . A unique solution is impossible in the case of circular motion about the center of force.
Equation [ edit ] 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 [ edit ] 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 [1]
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 [ edit ] Kepler problem [ edit ] Classical [ edit ] 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 [ edit ] The relativistic equation derived for Schwarzschild coordinates is[2]
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 [ edit ] 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[3]
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
[4] ).
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 [ edit ] 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 [ edit ] 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 [ edit ] References [ edit ]