Theory, worked examples and 39 previous-year questions with full solutions — IIT JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards.
Theory
6.1 Central forces
A central force is a force that is always directed along the line joining the particle to a fixed point \(O\), the centre of force, and whose magnitude depends only on the distance \(r\) of the particle from that point. Taking \(O\) as origin, it has the form
\[\boxed{\mathbf F=F(r)\,\hat{\mathbf r}},\]
where \(\hat{\mathbf r}=\mathbf r/r\). The force is attractive when \(F(r)<0\) and repulsive when \(F(r)>0\). Newtonian gravitation, \(F=-GMm/r^2\), the Coulomb force between two charges and the isotropic Hooke force \(F=-kr\) are the standard examples. The central-force problem is to find the motion of a particle under such a force. Its solution for the inverse-square law, the Kepler problem, explains the motion of the planets and was the first great triumph of Newtonian mechanics.
(a) A central force is conservative
In spherical polar coordinates a central force has the single component \(F_r=F(r)\), with \(F_\theta=F_\phi=0\). The curl then reduces to
because \(F\) depends on \(r\) alone. A central force is therefore conservative (Chapter 4), and it can be derived from a potential energy that depends only on \(r\):
Hence the angular momentum \(\mathbf L=\mathbf r\times\mathbf p\) about \(O\) is constant in magnitude and direction. Since \(\mathbf r\cdot\mathbf L=\mathbf r\cdot(\mathbf r\times\mathbf p)=0\) at every instant, the position vector always lies in the fixed plane through \(O\) perpendicular to \(\mathbf L\). Motion under a central force is planar, and it is the conservation of angular momentum, not of energy or of linear momentum, that confines it to a plane. (If \(\mathbf L=0\) the particle moves along a straight line through \(O\).) We may therefore describe the motion by plane polar coordinates \((r,\theta)\) in the plane of the orbit, as in Figure 6.1.
Figure 6.1 A particle of mass \(m\) at polar coordinates \((r,\theta)\), acted on by the central force \(F(r)\,\hat{\mathbf r}\).
A particle moving in a plane has two degrees of freedom. Since the energy \(E\) and the angular momentum \(L=\lvert\mathbf L\rvert\) are both conserved, the problem can always be reduced to quadratures, as the following sections show.
6.2 Equations of motion in plane polar coordinates
The components of acceleration in plane polar coordinates were derived in Chapter 3: the radial component is \(\ddot r-r\dot\theta^2\) and the transverse component is \(r\ddot\theta+2\dot r\dot\theta\). Since a central force has no transverse component, Newton’s second law gives
The transverse equation can be written \(\dfrac1r\dfrac{d}{dt}\left(mr^2\dot\theta\right)=0\), so that
\[\boxed{L=mr^2\dot\theta=\text{constant}},\]
which is the conservation of angular momentum found in Section 6.1(b), now expressed in polar coordinates. Using \(\dot\theta=L/mr^2\), the radial equation becomes an equation for \(r\) alone:
\[\boxed{m\ddot r-\frac{L^2}{mr^3}=F(r)}.\]
(a) The areal velocity
In a time \(dt\) the radius vector sweeps out a thin triangle of base \(r\) and height \(r\,d\theta\) (Figure 6.2), of area \(dA=\frac12r^2\,d\theta\). The areal velocity is therefore
The radius vector sweeps out equal areas in equal times. This is Kepler’s second law, and the argument shows that it holds for every central force, not only for gravitation: it is simply the conservation of angular momentum. For a closed orbit of area \(S\) and period \(T\), integration over one revolution gives
\[T=\frac{S}{L/2m}=\frac{2mS}{L}.\]
Figure 6.2 In time \(dt\) the radius vector sweeps out the area \(dA=\frac12r^2d\theta\).
(b) Circular orbits
A circular orbit of radius \(r_0\) has \(\dot r=\ddot r=0\), and the radial equation gives \(-mr_0\dot\theta^2=F(r_0)\). Writing \(\dot\theta=\omega_0\) for the constant angular velocity of the circular motion,
A circular orbit is possible only where the force is attractive, \(dV/dr>0\). Its angular momentum is \(L=mr_0^2\omega_0\), so the condition can also be written \(L^2/mr_0^3=V'(r_0)\).
6.3 Energy and the effective potential
The total energy is conserved, because the force is conservative and does not depend on time. With \(\mathbf v=\dot r\,\hat{\mathbf r}+r\dot\theta\,\hat{\boldsymbol\theta}\) and \(\dot\theta=L/mr^2\),
The term \(L^2/2mr^2\) is the kinetic energy of the transverse motion, expressed through the conserved angular momentum as a function of \(r\) alone. It is convenient to combine it with the true potential energy into the effective potential
The radial motion is thus exactly that of a particle of mass \(m\) moving in one dimension in the potential \(V_{\text{eff}}\), with \(r>0\); indeed the radial equation of Section 6.2 is \(m\ddot r=-dV_{\text{eff}}/dr\). The extra term \(L^2/2mr^2\) is called the centrifugal barrier: it corresponds to the outward “centrifugal force” \(L^2/mr^3=mr\dot\theta^2\), and for \(L\neq0\) it dominates any potential less singular than \(1/r^2\) at small \(r\), keeping the particle away from the centre. The methods of Chapter 5 for one-dimensional energy diagrams now apply directly. Since \(\frac12m\dot r^2\geq0\), the motion is confined to the region \(V_{\text{eff}}(r)\leq E\), and the points where \(E=V_{\text{eff}}(r)\), at which \(\dot r=0\), are the turning points or apsides of the orbit.
(a) Classification of orbits
Consider an effective potential of the general shape shown in Figure 6.3, with a minimum \(V_0\) at \(r=r_0\) and a maximum \(V_{\max}\) at a larger radius, beyond which it decreases without limit. The nature of the orbit depends on where the energy lies relative to \(V_0\) and \(V_{\max}\), as shown in Figure 6.4.
Figure 6.3 An effective potential with a minimum \(V_0\) and a maximum \(V_{\max}\).Figure 6.4 The possible types of motion in the effective potential of Figure 6.3 for different energies.
If \(E<V_0\) (panel a), the only allowed region is \(r\geq a\), outside the turning point \(a\) on the falling branch; the particle comes in to \(r=a\) and then recedes to infinity, so the orbit is unbounded. If \(E=V_0\) (panel b), the particle can sit at the bottom of the well, \(r=r_0\), where \(dV_{\text{eff}}/dr=0\) and \(d^2V_{\text{eff}}/dr^2>0\); the orbit is a stable circle of radius \(r_0\). With the same energy an unbounded motion in the outer region \(r\geq a\) is also possible. If \(V_0<E<V_{\max}\) (panel c), there are three turning points \(a\), \(b\) and \(c\). A particle in the well oscillates radially between \(a\) and \(b\) while it revolves about the centre, so the orbit is bounded and lies in the annulus \(a\leq r\leq b\); a particle outside the barrier, \(r\geq c\), is unbounded. If \(E=V_{\max}\) (panel d), the point \(b\) at the maximum is a position of unstable equilibrium: the particle may move on an unstable circular orbit of radius \(b\), or approach it from \(r=a\) (or from outside) in a motion that takes an infinite time to reach \(r=b\). Finally, if \(E>V_{\max}\) (panel e), there is a single turning point \(a\) on the inner wall, and the particle comes in from infinity, is turned back at \(r=a\) and escapes.
A bounded orbit is in general not closed: during one radial oscillation from \(a\) to \(b\) and back, the angle \(\theta\) advances by an amount that need not be a rational multiple of \(2\pi\), and the orbit then fills the annulus densely. Only for the inverse-square and the linear (Hooke) forces are all bounded orbits closed (Bertrand’s theorem).
(b) Circular orbits and their stability
A circular orbit of radius \(r_0\) corresponds to an extremum of the effective potential,
which reproduces the condition of Section 6.2(b), and its energy is \(E=V_{\text{eff}}(r_0)\). The orbit is stable if \(r_0\) is a minimum, \(V''_{\text{eff}}(r_0)>0\), and unstable if it is a maximum. For a small radial disturbance at fixed \(L\) the radius oscillates about \(r_0\) with angular frequency
as for any one-dimensional potential well (Chapter 5). Since \(V''_{\text{eff}}=3L^2/mr^4+V''\) and \(L^2/mr_0^3=V'(r_0)\), the stability condition can be expressed in terms of the potential alone:
Hence circular orbits in an attractive power-law force \(F\propto-r^{-n}\) are stable if and only if \(n<3\). The inverse-square law (\(n=2\)) and the linear force (\(n=-1\)) give stable orbits, while the inverse-cube and steeper forces do not. The ratio of the frequency of radial oscillation to the orbital frequency is \(\omega_r/\omega_0=\sqrt{3-n}\), since \(m\omega_0^2=V'(r_0)/r_0=K/r_0^{n+1}\). For \(n=2\) this ratio is \(1\): the radius completes one oscillation per revolution and the slightly perturbed orbit closes, as an ellipse with the centre of force at a focus must. For the linear force it is \(2\), two radial oscillations per revolution, which describes an ellipse centred on the centre of force (Example 6.1).
Exam tip. For a circular orbit set \(V'_{\text{eff}}(r_0)=0\); for stability check \(V''_{\text{eff}}(r_0)>0\); for the frequency of small radial oscillations use \(\omega_r^2=V''_{\text{eff}}(r_0)/m\), always at fixed angular momentum.
6.4 The differential equation of the orbit
The radial equation determines \(r(t)\). Often we need instead the shape of the orbit, \(r(\theta)\). Since \(L=mr^2\dot\theta\), time derivatives can be converted into derivatives with respect to \(\theta\) by
This is the differential equation of the orbit, also called Binet’s equation. It can be used in two directions. Given the force, it yields the orbit \(u(\theta)\); given the orbit, it yields the force law that produces it, by computing \(d^2u/d\theta^2+u\) and multiplying by \(-L^2u^2/m\). The orbit equation is symmetric under \(\theta\to-\theta\), so an orbit is symmetric about every apsidal line.
The energy can be expressed in the same variables. Since \(\dot r^2=(L^2/m^2)(du/d\theta)^2\),
This first integral of Binet’s equation is often the quickest route to the energy of a given orbit. Solving the energy equation for \(dr/dt\) and dividing by \(\dot\theta\) also gives the orbit in integral form,
with \(k=GMm\) for gravitation and \(k=Ze^2/4\pi\epsilon_0\) for an electron attracted by a nucleus of charge \(Ze\). The effective potential is
\[V_{\text{eff}}(r)=\frac{L^2}{2mr^2}-\frac kr.\]
It tends to \(+\infty\) as \(r\to0\) and to zero from below as \(r\to\infty\). Setting \(dV_{\text{eff}}/dr=-L^2/mr^3+k/r^2=0\) gives a single extremum at
and since \(V''_{\text{eff}}=3L^2/mr^4-2k/r^3\) equals \(k/r_0^3=m^3k^4/L^6>0\) there, it is a minimum. Figure 6.5 shows the four types of motion. For \(E=-mk^2/2L^2\) the particle moves on the circle \(r=r_0\). For \(-mk^2/2L^2<E<0\) it oscillates radially between the two turning points \(r_{\min}\) and \(r_{\max}\); we shall see that the orbit is an ellipse. For \(E=0\) and for \(E>0\) there is a single turning point and the motion is unbounded, along a parabola and a hyperbola respectively.
Figure 6.5 The effective potential of the Kepler problem, with the energies of circular, elliptic, parabolic and hyperbolic orbits.
(a) The orbit is a conic section
With \(F(1/u)=-ku^2\), Binet’s equation becomes
\[\frac{d^2u}{d\theta^2}+u=\frac{mk}{L^2}.\]
Putting \(y=u-mk/L^2\) reduces it to \(d^2y/d\theta^2+y=0\), whose solution is \(y=A\cos(\theta-\theta_0)\). Choosing the axis so that \(\theta_0=0\),
which is the polar equation of a conic section with one focus at the centre of force. The constant \(\alpha\) is the semi-latus rectum (the value of \(r\) at \(\theta=\pi/2\)), and \(e\geq0\) is the eccentricity. The point \(\theta=0\) of closest approach is the pericentre (perihelion for the Sun, perigee for the Earth). The radius of the circular orbit with the same angular momentum is \(r_0=\alpha\).
(b) Energy and eccentricity
The eccentricity is fixed by the energy. At the pericentre \(\theta=0\) we have \(\dot r=0\) and \(r_p=\alpha/(1+e)\), so
The same result follows by substituting \(\dot r=(eL/m\alpha)\sin\theta\) and \(\alpha/r=1+e\cos\theta\) into the energy equation at a general point. The shape of the orbit is thus decided by the sign of \(E\):
Energy
Eccentricity
Orbit
\(E>0\)
\(e>1\)
Hyperbola
\(E=0\)
\(e=1\)
Parabola
\(-\dfrac{mk^2}{2L^2}<E<0\)
\(0<e<1\)
Ellipse
\(E=-\dfrac{mk^2}{2L^2}\)
\(e=0\)
Circle
Bound orbits (\(E<0\)) are ellipses, with the circle as the special case \(e=0\), and unbound orbits are parabolas or hyperbolas. A parabola is the borderline case of a particle that just reaches infinity with zero speed.
(c) Geometry of the elliptic orbit
For \(E<0\) let the ellipse have semi-major axis \(a\) and semi-minor axis \(b=a\sqrt{1-e^2}\), with the centre of force at the focus \(S=(ae,0)\) measured from the centre of the ellipse (Figure 6.6). The pericentre and apocentre distances are
At the apsides the velocity is perpendicular to the radius vector, so \(L=mv_{\min}r_{\max}=mv_{\max}r_{\min}\) and \(v_{\max}/v_{\min}=r_{\max}/r_{\min}=(1+e)/(1-e)\). The planet moves fastest at perihelion and slowest at aphelion.
(d) Circular, parabolic and escape speeds
For a circular orbit of radius \(r\) the vis-viva equation with \(a=r\) gives \(v_c=\sqrt{k/mr}\), which is \(\sqrt{GM/r}\) for gravitation. A parabolic orbit has \(E=0\), so its speed at distance \(r\) is
the escape speed from that distance; at the Earth’s surface \(v_{\text{esc}}=\sqrt{2GM/R}=\sqrt{2gR}\approx11.2\;\text{km s}^{-1}\) (Chapter 4). A body at distance \(r\) moving with speed \(v\) is therefore on an ellipse if \(v<v_{\text{esc}}\), on a parabola if \(v=v_{\text{esc}}\) and on a hyperbola if \(v>v_{\text{esc}}\), whatever the direction of motion. For a circular orbit the kinetic energy is \(\frac12mv_c^2=k/2r\), the potential energy is \(-k/r\), and the total energy is \(-k/2r\), so that
The closure of every Kepler ellipse reflects a conserved quantity beyond \(E\) and \(\mathbf L\). For \(\dot{\mathbf p}=-k\hat{\mathbf r}/r^2\) and constant \(\mathbf L=m\mathbf r\times\dot{\mathbf r}\),
is a constant of the motion. It lies in the plane of the orbit, because both \(\mathbf p\times\mathbf L\) and \(\hat{\mathbf r}\) are perpendicular to \(\mathbf L\). Taking the scalar product with \(\mathbf r\) and using
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