In polar coordinates, the radial equation of motion for a particle
moving under a central force is
\[
m\left(\frac{d^2r}{dt^2}
-r\left(\frac{d\phi}{dt}\right)^2\right)=F(r).
\tag{1}
\]
Since the force is central, the angular momentum of the particle
about the centre of force remains constant. If \(J\) denotes the
angular momentum, then
\[
J=mr^2\frac{d\phi}{dt}.
\]
Therefore,
\[
\frac{d\phi}{dt}
=\frac{J}{mr^2}.
\tag{2}
\]
and
\[\because
\frac{dr}{dt}
=
\frac{dr}{d\phi}\frac{d\phi}{dt}.
\]
Using equation (2),
\[
\frac{dr}{dt}
=
\frac{J}{mr^2}\frac{dr}{d\phi}.
\tag{3}
\]
Since
\[ \boxed{
\frac{d}{dt}
=
\frac{J}{mr^2}\frac{d}{d\phi}}
\]
Hence,
\[
\frac{d^2r}{dt^2}
=
\frac{d}{dt}
\left(
\frac{dr}{dt}
\right)
\]
\[
\frac{d^2r}{dt^2}
=
\frac{J}{mr^2}
\frac{d}{d\phi}
\left(
\frac{J}{mr^2}
\frac{dr}{d\phi}
\right).
\tag{4}
\]
Since \(J\) and \(m\) are constants,
\[
\frac{d^2r}{dt^2}
=
\frac{J^2}{m^2r^2}
\frac{d}{d\phi}
\left(
\frac{1}{r^2}
\frac{dr}{d\phi}
\right).
\]
On differentiating,
\[
\frac{d^2r}{dt^2}
=
\frac{J^2}{m^2r^2}
\left[
\frac{1}{r^2}
\frac{d^2r}{d\phi^2}
-
\frac{2}{r^3}
\left(
\frac{dr}{d\phi}
\right)^2
\right].
\]
Therefore,
\[
\boxed{
\frac{d^2r}{dt^2}
=
\frac{J^2}{m^2}
\left[
\frac{1}{r^4}\frac{d^2r}{d\phi^2}
-
\frac{2}{r^5}
\left(\frac{dr}{d\phi}\right)^2
\right]
}.
\tag{5}
\]
Substituting equations (2) and (5) in equation (1), we get
\[
m
\left[
\frac{J^2}{m^2}
\left\{
\frac{1}{r^4}\frac{d^2r}{d\phi^2}
-
\frac{2}{r^5}
\left(\frac{dr}{d\phi}\right)^2
\right\}
-
\frac{J^2}{m^2r^3}
\right]
=
F(r).
\]
Thus,
\[
\frac{J^2}{m}
\left[
\frac{1}{r^4}\frac{d^2r}{d\phi^2}
-
\frac{2}{r^5}
\left(\frac{dr}{d\phi}\right)^2
-
\frac{1}{r^3}
\right]
=
F(r).
\tag{6}
\]
To simplify the equation, put
\[
\boxed{u=\frac{1}{r}}.
\]
Therefore,
\[
r=\frac{1}{u}.
\tag{7}
\]
Differentiating with respect to \(\phi\),
\[
\frac{dr}{d\phi}
=
-\frac{1}{u^2}\frac{du}{d\phi}.
\tag{8}
\]
Differentiating once again,
\[
\frac{d^2r}{d\phi^2}
=
\frac{2}{u^3}
\left(\frac{du}{d\phi}\right)^2
-
\frac{1}{u^2}
\frac{d^2u}{d\phi^2}.
\tag{9}
\]
Now,
\[
\frac{1}{r^4}\frac{d^2r}{d\phi^2}
=
u^4
\left[
\frac{2}{u^3}
\left(\frac{du}{d\phi}\right)^2
-
\frac{1}{u^2}
\frac{d^2u}{d\phi^2}
\right].
\tag{10}
\]
Hence,
\[
\frac{1}{r^4}\frac{d^2r}{d\phi^2}
=
2u
\left(\frac{du}{d\phi}\right)^2
-
u^2\frac{d^2u}{d\phi^2}.
\tag{11}
\]
Also,
\[
\frac{2}{r^5}
\left(\frac{dr}{d\phi}\right)^2
=
2u^5
\left[
\frac{1}{u^4}
\left(\frac{du}{d\phi}\right)^2
\right]
=
2u
\left(\frac{du}{d\phi}\right)^2.
\tag{12}
\]
Therefore, the first two terms in equation (6) combine to give
\[
\frac{1}{r^4}\frac{d^2r}{d\phi^2}
-
\frac{2}{r^5}
\left(\frac{dr}{d\phi}\right)^2
=
-u^2\frac{d^2u}{d\phi^2}.
\tag{13}
\]
Since
\[
\frac{1}{r^3}=u^3,
\]
equation (17) becomes
\[
\frac{J^2}{m}
\left[
-u^2\frac{d^2u}{d\phi^2}
-u^3
\right]
=
F\left(\frac{1}{u}\right).
\tag{14}
\]
Dividing by \(-u^2\), we obtain
\[
\frac{J^2}{m}
\left[
\frac{d^2u}{d\phi^2}+u
\right]
=
-\frac{F(1/u)}{u^2}.
\tag{15}
\]
Hence, the differential equation determining the trajectory of a
particle moving under a central force is
\[
\left[
\frac{d^2u}{d\phi^2}+u
\right]
=
-\frac{F(1/u)}{u^2} \frac{m}{J^2}.
\tag{16}
\]
where
\[
\boxed{u=\frac{1}{r}}
\qquad\text{and}\qquad
\boxed{J=mr^2\frac{d\phi}{dt}}.
\]
Important:
Equation (16) is known as the Binet equation or the
orbit equation for motion under a central force. It contains no
explicit time variable \(t\), and therefore directly relates the
radial distance \(r\) to the angular coordinate \(\phi\).
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