The central-force problem, and the two conservation laws that solve it

Idea. Differentiate, then let the identities do the rearranging. Everything here is pure kinematics: it is true of any curve, which is why it can be used for any force.

1. \(r^2 = \vec r\cdot\vec r\). Differentiating, with the product rule for the dot product, \(2r\dot r = 2\vec r\cdot\dot{\vec r}\). Divide by 2. \(\square\)

2. By the quotient rule and part 1, \[ \frac{\mathrm d\hat r}{\mathrm dt} = \frac{\dot{\vec r}}{r} - \frac{\dot r\,\vec r}{r^2} = \frac{(\vec r\cdot\vec r)\,\dot{\vec r} - (\vec r\cdot\dot{\vec r})\,\vec r}{r^3} . \] BAC–CAB with \(\vec a = \vec r\), \(\vec b = \dot{\vec r}\), \(\vec c = \vec r\) reads \(\vec r\times(\dot{\vec r}\times\vec r) = \dot{\vec r}(\vec r\cdot\vec r) - \vec r(\vec r\cdot\dot{\vec r})\), which is the numerator. Finally \(\vec r\times(\dot{\vec r}\times\vec r) = -\vec r\times(\vec r\times\dot{\vec r}) = (\vec r\times\dot{\vec r})\times\vec r\) by antisymmetry, twice. \(\square\)

3. Lagrange's identity with \(\vec a = \vec c = \vec r\) and \(\vec b = \vec d = \dot{\vec r}\): \(\lVert\vec r\times\dot{\vec r}\rVert^2 = r^2\lVert\dot{\vec r}\rVert^2 - (\vec r\cdot\dot{\vec r})^2 = r^2\lVert\dot{\vec r}\rVert^2 - r^2\dot r^2\). Divide by \(r^2\) and rearrange. With \(\vec L = \mu\,\vec r\times\dot{\vec r}\), multiply by \(\tfrac12\mu\): \(\tfrac12\mu\lVert\dot{\vec r}\rVert^2 = \tfrac12\mu\dot r^2 + \lVert\vec L\rVert^2/2\mu r^2\). \(\square\)

4. By part 2, \(\mathrm d\hat r/\mathrm dt = (\vec L/\mu)\times\vec r/r^3 = \vec 0\), so \(\hat r\) is constant on every interval where \(\vec r\ne\vec 0\), and \(\vec r = r(t)\hat r_0\). \(\square\)

Check. Dimensions in part 2: \(\mathsf{L\cdot LT^{-1}\cdot L}/\mathsf{L^3} = \mathsf{T^{-1}}\), the dimension of the derivative of a unit vector ✓. Magnitude: \(\vec r\times\dot{\vec r}\perp\vec r\), so \(\lVert\mathrm d\hat r/\mathrm dt\rVert = \lVert\vec r\times\dot{\vec r}\rVert\,r/r^3 = L/\mu r^2\), which in the plane is \(|\dot\theta|\) — matching \(\mathrm d\hat r/\mathrm dt = \dot\theta\,\hat\theta\) from the polar-coordinates slide ✓. Special case: uniform circular motion at radius \(R\) and speed \(v\) gives \(\lVert\dot{\vec r}\rVert^2 = 0 + (Rv)^2/R^2 = v^2\) in part 3 ✓.

Another way

Part 3 is the Pythagorean theorem: \(\dot{\vec r}\) splits into its component \(\dot r\) along \(\hat r\) and a component perpendicular to \(\hat r\) of length \(\lVert\vec r\times\dot{\vec r}\rVert/r\) (the area corollary: \(\lVert\vec r\times\dot{\vec r}\rVert = r\,\lVert\dot{\vec r}\rVert\sin\varphi\)).

Idea. The relative coordinate obeys a one-body law with mass \(\mu\) in a field of strength \(k = Gm_\oplus m_{\text{Moon}}\), so the combination that sets the period is \(k/\mu = G(m_\oplus+m_{\text{Moon}})\), not \(Gm_\oplus\).

1. \(x = 0.0123002\) (the Moon is \(1/81.30\) of the Earth), and \(\mu/m_{\text{Moon}} = m_\oplus/(m_\oplus+m_{\text{Moon}}) = 1/(1+x) = 0.98785\). A \(1.2\%\) correction: far larger than for any planet about the Sun.

2. From \(\vec r_1 = \vec R + (m_2/M)\vec r\), the Earth's centre is at distance \((m_{\text{Moon}}/M)\,a = a\,x/(1+x) = 3.84399\times10^{8}\times0.0123002/1.0123002 = 4.6707\times10^{6}\) m from the centre of mass: \(4671\) km, which is \(1700\) km below the Earth's surface. The Earth wobbles about a point inside itself once a month.

3. Circular orbit, \(T = 2\pi\sqrt{a^3/(k/\mu)}\).

• With \(GM_\oplus\) alone: \(a^3 = 5.6799\times10^{25}\ \mathrm{m^3}\), \(a^3/GM_\oplus = 1.42498\times10^{11}\ \mathrm{s^2}\), \(T = 2\pi\times3.77489\times10^{5}\) s \(= 2.37183\times10^{6}\) s \(= 27.452\) d.
• With \(GM_\oplus(1+x)\): divide by \(\sqrt{1.0123002} = 1.006131\): \(T = 27.284\) d.

Observed: \(27.3217\) d. The naive value is \(0.48\%\) long; including the Moon's mass overshoots, to \(0.14\%\) short. What is left is not a two-body effect: it is the Sun. To first order in \(r/R\) (unit 04's Taylor expansion of the Sun's field about the Earth, at distance \(R\)), the Sun's pull on the Moon minus its pull on the Earth is the tidal acceleration \((GM_\odot/R^3)\big[3(\vec r\cdot\hat R)\hat R - \vec r\big]\), whose radial part is \(n'^2r(3\cos^2\varphi - 1)\), with \(n'^2 = GM_\odot/R^3 = (2\pi/T_{\text{year}})^2\) and \(\varphi\) the angle between the directions of the Moon and the Sun. Over a month \(\varphi\) runs through every value, \(\cos^2\varphi\) averages to \(\tfrac12\), and the radial part averages to \(+\tfrac12n'^2r\), outwards. Against the Earth–Moon attraction \(n^2r\), \(n = 2\pi/T_{\text{month}}\), that is a weakening by the fraction \(\tfrac12(T_{\text{month}}/T_{\text{year}})^2 = 0.28\%\), which at fixed \(a\) lengthens the period by half that, \(0.14\%\) — the size of the residual. The step not proved here is replacing the tidal force by its monthly average; justifying it is unit 10's secular-perturbation argument.

Check. Dimensions: \(\sqrt{\mathrm{m^3}/(\mathrm{m^3\,s^{-2}})} = \mathrm s\) ✓. Limit: as \(x\to0\) both periods agree and \(\mu\to m_{\text{Moon}}\) ✓. Size: the fractional change in \(T\) is \(\approx x/2 = 0.62\%\), and \(27.452\times(1-0.0061) = 27.284\) ✓.

Idea. The same orbit can be read two ways: through \(V_{\mathrm{eff}}\) in the radial variable, or through the linear equation in Cartesian coordinates. The second gives the shape outright, and the first tells you how \(r\) oscillates along it.

1. \(V_{\mathrm{eff}}(r) = \tfrac12\kappa r^2 + L^2/2\mu r^2\). It tends to \(+\infty\) at both ends, as \(r\to0\) (barrier) and as \(r\to\infty\) (spring). So \(\{V_{\mathrm{eff}}\le E\}\) is a bounded interval \([r_-,r_+]\) for every \(E\) above the minimum, and there are no unbounded orbits at all — in contrast to the inverse square, whose \(V_{\mathrm{eff}}\to0\) at infinity.

2. \(V_{\mathrm{eff}}' = \kappa r - L^2/\mu r^3 = 0\) at \(r_c^4 = L^2/\mu\kappa\). There \(\omega_\theta = \dot\theta = L/\mu r_c^2 = \sqrt{\kappa/\mu}\). And \(V_{\mathrm{eff}}''(r_c) = \kappa + 3L^2/\mu r_c^4 = \kappa + 3\kappa = 4\kappa\), so \(\omega_r = \sqrt{4\kappa/\mu} = 2\omega_\theta\).

3. The motion is planar (the force is central); in that plane \(\mu\ddot x = -\kappa x\) and \(\mu\ddot y = -\kappa y\). By unit 06 each has solution space spanned by \(\cos\omega t,\sin\omega t\), \(\omega = \sqrt{\kappa/\mu}\), so \((x,y)^{\!\top} = M\,(\cos\omega t,\ \sin\omega t)^{\!\top}\) with \(M = [\vec\alpha\ \vec\beta]\), where \(\vec\alpha = \vec r(0)\) and \(\vec\beta = \dot{\vec r}(0)/\omega\). Then \(L = \mu(x\dot y - y\dot x) = \mu\omega\det M\), so \(L\ne0\) iff \(M\) is invertible. Write the singular value decomposition \(M = U\Sigma W^{\!\top}\) with \(U,W\) orthogonal and \(\Sigma = \mathrm{diag}(\sigma_1,\sigma_2)\), \(\sigma_i > 0\). \(W^{\!\top}\) maps the unit circle to itself, \(\Sigma\) maps it to \(\{X^2/\sigma_1^2 + Y^2/\sigma_2^2 = 1\}\), and \(U\) rotates (or reflects) that set about the origin. The image is an ellipse with semi-axes \(\sigma_1,\sigma_2\), centred at the origin. \(\square\)

4. On a centred ellipse \(r\) is largest at the two ends of the major axis and smallest at the two ends of the minor axis, which are \(90^\circ\) apart. So the apsidal angle is \(\Delta\theta = \pi/2\), and \(r\) runs through two full oscillations per revolution — exactly \(\omega_r = 2\omega_\theta\) from part 2, now for every orbit and not just the near-circular ones.

Check. Dimensions: \([\kappa] = \mathsf{MT^{-2}}\), so \(\sqrt{\kappa/\mu}\) is a frequency ✓; \(r_c^4 = L^2/\mu\kappa\) has dimension \(\mathsf{M^2L^4T^{-2}}/\mathsf{M^2T^{-2}} = \mathsf{L^4}\) ✓. Special case: \(\sigma_1 = \sigma_2\) gives a circle of radius \(\sigma_1\), and then \(r_c = \sigma_1\) ✓. The period \(2\pi/\omega\) is the same for every orbit — the harmonic oscillator's isochronism (unit 00), in two dimensions.

Idea. Fix \(L\) so that \(r_0\) is a critical point, then compute the second derivative there. The ratio \(\omega_r/\omega_\theta\) turns out independent of \(r_0\) and \(c\), which is mechanical similarity at work: a power law has no preferred length.

1. \(V_{\mathrm{eff}}'(r) = c\,r^n - L^2/\mu r^3\). Choosing \(L^2 = \mu c\,r_0^{\,n+3}\) makes \(r_0\) a critical point, so \(\dot r\equiv0\), \(r\equiv r_0\) solves the radial equation. Any \(r_0 > 0\) will do.

2. \(V_{\mathrm{eff}}''(r) = c\,n\,r^{n-1} + 3L^2/\mu r^4\). At \(r_0\), with the \(L^2\) of part 1, \[ V_{\mathrm{eff}}''(r_0) = c\,n\,r_0^{\,n-1} + 3c\,r_0^{\,n-1} = c\,(n+3)\,r_0^{\,n-1} , \] positive iff \(n > -3\). A critical point with positive second derivative is a strict local minimum; for \(n < -3\) it is a strict local maximum. At \(n = -3\) exactly, \(V_{\mathrm{eff}}\) is \((L^2/\mu - c)/2r^2\), identically zero for this \(L\): every radius is neutral, and the circle is not a strict minimum. \(\square\)

3. Put \(r = r_0 + \xi\). Taylor's theorem gives \(V_{\mathrm{eff}}'(r_0+\xi) = V_{\mathrm{eff}}''(r_0)\,\xi + O(\xi^2)\), so to first order \(\mu\ddot\xi = -V_{\mathrm{eff}}''(r_0)\,\xi\): unit 06's oscillator with \(\omega_r^2 = c(n+3)r_0^{\,n-1}/\mu\). The circular angular velocity is \(\omega_\theta = L/\mu r_0^2\), so \(\omega_\theta^2 = \mu c\,r_0^{\,n+3}/\mu^2r_0^4 = c\,r_0^{\,n-1}/\mu\). Dividing, \(\omega_r^2/\omega_\theta^2 = n + 3\). From pericentre to apocentre is half a radial period, \(\pi/\omega_r\), during which \(\theta\) advances by \(\omega_\theta\,\pi/\omega_r = \pi/\sqrt{3+n}\) (to leading order in the amplitude, since \(\dot\theta = L/\mu r^2\) differs from \(\omega_\theta\) by \(O(\xi)\)).

4. \(n = -1\): \(\Delta\theta = \pi/\sqrt2 = 127.28^\circ\). From one pericentre to the next is \(254.56^\circ\), short of a full turn, so the pericentre regresses by about \(105.44^\circ\) each radial period. Whether it closes depends on the exact apsidal angle, which is only approximately \(\pi/\sqrt2\) and varies continuously with the amplitude (P5): at the energies where it is a rational multiple of \(\pi\) the orbit closes, but those are exceptions, and a typical near-circular star traces a rosette filling an annulus.

Check. Dimensions: \([c\,r^{n-1}/\mu] = \mathsf{MLT^{-2}L^{-n}}\cdot\mathsf{L^{n-1}}/\mathsf M = \mathsf{T^{-2}}\) ✓ (since \([c] = \mathsf{MLT^{-2}}\mathsf{L^{-n}}\)). Limits: \(n = -2\) gives \(\omega_r = \omega_\theta\), \(\Delta\theta = \pi\), the Kepler ellipse ✓; \(n = 1\) gives \(\omega_r = 2\omega_\theta\), \(\Delta\theta = \pi/2\), the centred ellipse of P3 ✓; \(n\to-3^+\) gives \(\omega_r\to0\), marginal ✓.

Idea. If every bounded orbit closes, the apsidal angle can take only rational multiples of \(\pi\); being continuous in \(E\), it must then be constant. So its value near the circular orbit (P4) must equal its value at the other end of the energy range, and each end can be computed in closed form because one term of the potential dominates there. This is Arnold's argument.

1. From the quadrature slide, \(\Phi = \int_{r_-}^{r_+}L\,\mathrm dr\big/r^2\sqrt{2\mu(E - V(r)) - L^2/r^2}\). With \(x = L/r\), \(\mathrm dx = -L\,\mathrm dr/r^2\), and \(L^2/r^2 = x^2\); reversing the limits absorbs the sign, and \(2\mu(E - V) - x^2 = 2\mu(E - W(x))\). The turning points \(r_\pm\) become \(x_1 = L/r_+\), \(x_2 = L/r_-\). \(\square\)

2. \(\alpha > 0\), \(a > 0\): \(W(x) = x^2/2\mu + aL^\alpha x^{-\alpha}\). As \(E\to\infty\), the large root has \(x_2^2/2\mu = E - aL^\alpha x_2^{-\alpha}\), so \(x_2^2/2\mu E\to1\), and the small root \(x_1\to0\). Put \(x = x_2y\): \[ \Phi = \int_{x_1/x_2}^{1}\frac{x_2\,\mathrm dy}{\sqrt{2\mu E - x_2^2y^2 - 2\mu aL^\alpha x_2^{-\alpha}y^{-\alpha}}} . \] Divide numerator and denominator by \(x_2\): the denominator becomes \(\sqrt{2\mu E/x_2^2 - y^2 - \varepsilon_E\,y^{-\alpha}}\) with \(\varepsilon_E = 2\mu aL^\alpha x_2^{-\alpha-2}\to0\). By the lemma the limit is \(\int_0^1\mathrm dy/\sqrt{1-y^2} = \pi/2\). \(\square\)

3. \(-2 < \alpha < 0\): write \(a = -b\), \(b > 0\), \(\beta = -\alpha\in(0,2)\). Then \(V(L/x) = -bL^{-\beta}x^\beta\) and bound orbits have \(E < 0\). By the lemma, \(\lim_{E\to0^-}\Phi(E)\) is the integral at \(E = 0\): \(2\mu(0 - W(x)) = 2\mu bL^{-\beta}x^\beta - x^2 = x_2^{\,2-\beta}x^\beta - x^2\) with roots \(0\) and \(x_2\). With \(x = x_2y\), \[ \Phi(0) = \int_0^1\frac{\mathrm dy}{\sqrt{y^\beta - y^2}} = \int_0^1\frac{\mathrm dy}{y^{\beta/2}\sqrt{1 - y^{2-\beta}}} . \] Now \(z = y^{(2-\beta)/2}\), \(\mathrm dz = \tfrac{2-\beta}{2}\,y^{-\beta/2}\mathrm dy\), maps \([0,1]\) to \([0,1]\): \[ \Phi(0) = \frac{2}{2-\beta}\int_0^1\frac{\mathrm dz}{\sqrt{1-z^2}} = \frac{2}{2-\beta}\cdot\frac\pi2 = \frac{\pi}{2+\alpha} . \qquad\square \]

4. After \(m\) radial periods the radius has turned through \(2m\Phi\); the orbit closes iff \(2m\Phi = 2\pi j\) for some integers, i.e. iff \(\Phi/\pi\in\mathbb Q\). If every bounded orbit closes, \(\Phi\) is continuous (lemma) with values in the countable set \(\pi\mathbb Q\); a continuous non-constant function on an interval takes every value between two of its values, uncountably many, so \(\Phi\) is constant. Near the bottom of the well P4 gives \(\Phi\to\pi/\sqrt{3+n} = \pi/\sqrt{\alpha+2}\).

• \(\alpha > 0\): equate with part 2, \(\sqrt{\alpha+2} = 2\), so \(\alpha = 2\): \(V = \tfrac12\kappa r^2\).
• \(-2 < \alpha < 0\): equate with part 3, \(\sqrt{\alpha+2} = \alpha + 2\), so \(\alpha+2 = 1\), \(\alpha = -1\): \(V = -k/r\).

Conversely both do close: P3 for the harmonic force, the Binet slides for the inverse square. \(\square\)

Check. The two limit formulas agree with each other and with P4 at the survivors: \(\alpha = 2\) gives \(\pi/\sqrt4 = \pi/2\) ✓, \(\alpha = -1\) gives \(\pi/\sqrt1 = \pi/(2-1) = \pi\) ✓. A number for a non-survivor, \(\alpha = -\tfrac12\) (\(n = -\tfrac32\)): near-circular \(\Phi = 180^\circ/\sqrt{1.5} = 146.97^\circ\), at \(E\to0^-\) \(\Phi = 180^\circ/1.5 = 120^\circ\); evaluating the integral numerically at half the well's depth gives \(137.0^\circ\), in between, as a continuous non-constant \(\Phi\) must.

What is not proved here

The restriction to power laws. For a general \(V\) the near-circular condition only forces \(V_{\mathrm{eff}}\) to look like a power law near every circular orbit, and upgrading that to the global statement needs an expansion to third order in the amplitude — Goldstein, Classical Mechanics, §3.6 (Appendix A of the 2nd edition).

The exponents left out

The statement excludes \(\alpha = 0\) and \(\alpha\le-2\). For \(\alpha\le-2\) (P4's \(n\le-3\)) the circular orbit is not a strict minimum of \(V_{\mathrm{eff}}\) (P4, part 2) and there are no bounded orbits around it to close. The case \(\alpha = 0\) is the logarithmic potential \(V = c\ln r\), \(c > 0\), P4's \(n = -1\). Part 2's rescaling works for it unchanged: with \(x = x_2y\) and the root condition \(2\mu E = x_2^2 + 2\mu c\ln(L/x_2)\), the radicand divided by \(x_2^2\) is \(1 - y^2 + (2\mu c/x_2^2)\ln y\), and \(2\mu c/x_2^2\to0\) as \(E\to\infty\), so by the same lemma \(\Phi\to\pi/2\). Near the circle P4 gives \(\pi/\sqrt2\ne\pi/2\), so not every orbit closes. (Numerically, with \(\mu = L = c = 1\), \(\Phi\) falls from \(127.3^\circ\) near the circle to \(94.4^\circ\) at \(E = 10\) and \(90.4^\circ\) at \(E = 100\).) So among all power laws only \(n = -2\) and \(n = 1\) survive.

Idea. Angular momentum makes the problem planar and supplies the change of variable \(t\mapsto\theta\); \(u = 1/r\) makes the orbit equation linear; energy fixes the one remaining constant; areal velocity turns the shape into a period.

1. The force \(-k\hat r/r^2\) is central, so \(\dot{\vec L} = \vec r\times\vec F = \vec 0\). Then \(\vec L\cdot\vec r = (\vec r\times\vec p)\cdot\vec r = 0\) for all \(t\), and \(\vec L\) is a fixed nonzero vector, so \(\vec r(t)\in\vec L^\perp\), a plane. In polar coordinates in that plane \(\vec L = \mu r^2\dot\theta\,\hat z\), so \(\dot\theta = L/\mu r^2\) has the fixed sign of \(L\ne0\) and never vanishes: \(\theta\) is strictly monotone in \(t\), hence a legitimate independent variable.

2. Radial equation (polar acceleration, radial component): \(\mu(\ddot r - r\dot\theta^2) = -k/r^2\), i.e. \(\mu\ddot r = -k/r^2 + L^2/\mu r^3\). With \(u = 1/r\): \(\dot r = -(L/\mu)u'\) as in rung 3, and \(\ddot r = -(L/\mu)u''\dot\theta = -(L^2/\mu^2)u^2u''\). Substituting, \(-(L^2/\mu)u^2u'' = -ku^2 + (L^2/\mu)u^3\); dividing by \(-(L^2/\mu)u^2\), \[ u'' + u = \frac{\mu k}{L^2} = \frac1p . \] The operator \(\mathcal D u = u'' + u\) has characteristic polynomial \(\lambda^2+1\), so \(\cos\theta,\sin\theta\in\ker\mathcal D\); they are independent, and \(\dim\ker\mathcal D = 2\) because by unit 01's uniqueness theorem a solution is fixed by \((u(0),u'(0))\in\mathbb R^2\) (unit 06's evaluation map is an isomorphism). The constant \(1/p\) is a particular solution, so \(u = 1/p + C\cos(\theta-\theta_0)\). Choose the axis so \(\theta_0 = 0\) and \(C\ge0\), and set \(e = Cp\): \(r = p/(1+e\cos\theta)\). \(\square\)

3. At \(\theta = 0\), \(r = p/(1+e)\) is a minimum of \(r\) so \(\dot r = 0\), and \(E = L^2/2\mu r^2 - k/r\). Using \(L^2 = \mu kp\): \[ E = \frac{k(1+e)^2}{2p} - \frac{k(1+e)}{p} = \frac{k(1+e)}{2p}(e - 1) = \frac{k(e^2-1)}{2p} = \frac{\mu k^2(e^2-1)}{2L^2} . \] So \(e^2 = 1 + 2EL^2/\mu k^2\), and \(e\ge0\) takes the positive root. \(E < 0\) gives \(e^2 < 1\). \(\square\)

4. Second law. The area swept between \(t\) and \(t+\Delta t\) is the triangle \(\tfrac12\lVert\vec r(t)\times\vec r(t+\Delta t)\rVert\) up to \(O(\Delta t^3)\); expanding \(\vec r(t+\Delta t) = \vec r + \Delta t\,\dot{\vec r} + O(\Delta t^2)\) and using \(\vec r\times\vec r = \vec 0\) gives \(\dot A = \tfrac12\lVert\vec r\times\dot{\vec r}\rVert = L/2\mu\), constant. (Or: \(\dot A = \tfrac12r^2\dot\theta\) from \(A = \int\tfrac12r^2\,\mathrm d\theta\).) Third law. For \(e < 1\) the orbit is an ellipse with \(a = p/(1-e^2)\), \(b = a\sqrt{1-e^2}\), so \(b^2 = ap\), and area \(\pi ab\). One period sweeps it at rate \(L/2\mu\): \(T = 2\pi\mu ab/L\). From \(L^2 = \mu kp = \mu kb^2/a\), \(T = 2\pi\mu ab\sqrt{a}/(b\sqrt{\mu k}) = 2\pi\sqrt{\mu a^3/k}\). \(\square\)

Check. Dimensions: \(p = L^2/\mu k\) is \(\mathsf{M^2L^4T^{-2}}/\mathsf{M\cdot ML^3T^{-2}} = \mathsf L\) ✓; \(\mu a^3/k\) is \(\mathsf{T^2}\) ✓. Limits: \(E = -\mu k^2/2L^2\) (bottom of the well) gives \(e = 0\), a circle of radius \(p\) ✓; \(E\to0^-\) gives \(e\to1\) and \(a = -k/2E\to\infty\) ✓. Number: for Halley's comet, the deck's running example, \(a = 17.8325\) AU gives \(T = 1.000019\times17.8325^{3/2} = 75.305\) yr ✓.

Idea. The slide proved \(\dot{\vec A} = \vec 0\) with BAC–CAB and no coordinates. Here the same fact comes out of the polar components, where the inverse square enters through the radial equation. Two routes to one result is the check.

1. As in rung 3, \(\vec p\times\vec L = \mu L(r\dot\theta\,\hat r - \dot r\,\hat\theta)\), and \(\mu Lr\dot\theta = L\cdot L/r\). Subtract \(\mu k\hat r\): \(\vec A = (L^2/r - \mu k)\,\hat r - \mu L\dot r\,\hat\theta\). \(\square\)

2. Product rule, with \(\dot{\hat r} = \dot\theta\hat\theta\), \(\dot{\hat\theta} = -\dot\theta\hat r\): \[ \dot{\vec A} = -\frac{L^2\dot r}{r^2}\hat r + \Big(\frac{L^2}{r} - \mu k\Big)\dot\theta\,\hat\theta - \mu L\ddot r\,\hat\theta + \mu L\dot r\dot\theta\,\hat r . \] \(\hat r\)-component: \(-L^2\dot r/r^2 + \mu L\dot r\dot\theta = -L^2\dot r/r^2 + L^2\dot r/r^2 = 0\), using \(\mu\dot\theta = L/r^2\). \(\hat\theta\)-component: \((L^2/r - \mu k)\dot\theta - \mu L\ddot r = L^3/\mu r^3 - kL/r^2 - \mu L\ddot r\). The radial equation \(\mu\ddot r = -k/r^2 + L^2/\mu r^3\) — the one line where the inverse square is used — makes \(\mu L\ddot r = -kL/r^2 + L^3/\mu r^3\), and the component is zero. \(\square\)

3. The components are orthogonal, so \[ \lVert\vec A\rVert^2 = \Big(\frac{L^2}{r} - \mu k\Big)^2 + \mu^2L^2\dot r^2 = \mu^2k^2 + 2\mu L^2\Big(\frac{\mu\dot r^2}{2} + \frac{L^2}{2\mu r^2} - \frac kr\Big) = \mu^2k^2 + 2\mu L^2E . \qquad\square \]

4. At perihelion \(\dot r = 0\) and \(r = p/(1+e)\), so \(\vec A = (L^2(1+e)/p - \mu k)\hat r = \mu k\big((1+e) - 1\big)\hat r = \mu ke\,\hat r\), using \(L^2/p = \mu k\). \(\square\)

Check. Part 3 with part 4: \(\mu^2k^2e^2 = \mu^2k^2 + 2\mu EL^2\) is exactly \(e^2 = 1 + 2EL^2/\mu k^2\) ✓. Dimensions: \(L^2/r\) and \(\mu k\) are both \(\mathsf{M^2L^3T^{-2}}\) ✓. Special case: a circular orbit has \(\dot r = 0\) and \(L^2 = \mu kr\), so \(\vec A = \vec 0\), \(e = 0\) ✓. Failure case: with a general \(f(r)\) the radial equation reads \(\mu\ddot r = f + L^2/\mu r^3\), and the \(\hat\theta\)-component becomes \(-(f r^2 + k)L/r^2 = -\mu(fr^2 + k)\dot\theta\), matching the slide's \(\dot{\vec A} = -\mu(fr^2+k)\,\mathrm d\hat r/\mathrm dt\) ✓.

Idea. Escaping needs energy; falling in needs the angular momentum removed, and the Earth's orbital angular momentum is large. That asymmetry is the whole problem.

1. Circular: \(v_c^2/r = GM_\odot/r^2\), \(v_c = \sqrt{GM_\odot/\mathrm{AU}} = 29.785\ \mathrm{km\,s^{-1}}\). Escape: \(E = 0\), \(v_e = \sqrt{2GM_\odot/\mathrm{AU}} = 42.122\ \mathrm{km\,s^{-1}}\). The ratio is \(\sqrt2\) exactly, for any inverse-square field at any radius: both are \(\sqrt{GM/r}\) times a pure number.

2. Moving along the Earth's velocity, the probe's heliocentric speed is \(v_c + v_\infty\), and it escapes iff that is at least \(v_e\): \(v_\infty = 12.337\ \mathrm{km\,s^{-1}}\).

3. \(v_0 = \sqrt{v_\infty^2 + 2GM_\oplus/r_0} = \sqrt{(12\,337)^2 + 1.17737\times10^{8}} = 16.430\ \mathrm{km\,s^{-1}}\). Circular speed at \(r_0\) is \(\sqrt{GM_\oplus/r_0} = 7.673\ \mathrm{km\,s^{-1}}\), so the burn is \(8.757\ \mathrm{km\,s^{-1}}\). (The \(v_\infty\) "bought" by the burn is much larger than the burn would suggest, because a burn deep in a potential well adds kinetic energy where the speed is already high.)

4. From rung 3 with \(r_a = 1\) AU and \(r_p = R_\odot\): \(v_a = \sqrt{2GM_\odot R_\odot/\big(\mathrm{AU}(\mathrm{AU}+R_\odot)\big)} = 2.866\ \mathrm{km\,s^{-1}}\). The probe must lose \(29.785 - 2.866 = 26.919\ \mathrm{km\,s^{-1}}\) relative to the Earth — more than twice the \(12.337\) needed to leave the solar system. The Sun is the hardest place in the solar system to reach, because the Earth's angular momentum has to be cancelled almost entirely.

Check. Dimensions throughout are \(\sqrt{\mathrm{m^3s^{-2}/m}} = \mathrm{m\,s^{-1}}\) ✓. Part 4's limit \(R_\odot\to0\) gives \(v_a\to0\): to reach a point mass you must stop dead, \(L = 0\) ✓. Part 3's limit \(v_\infty\to0\) gives \(v_0 = \sqrt2\times7.673 = 10.851\ \mathrm{km\,s^{-1}}\), escape from low orbit ✓.

Idea. The asymptotes of the conic are where \(r\to\infty\); the angle between them is fixed by \(e\), and \(e\) is fixed by \(E\) and \(L\), which are fixed by conditions at infinity. The answer depends on \(k\) only through \(k^2\).

1. \(E = \tfrac12\mu v_\infty^2\) (as \(V\to0\)) and \(L = \mu v_\infty b\) (the initial line is at distance \(b\) from the centre, and \(\lVert\vec r\times\vec v\rVert = v\times\) that distance). Then \(e^2 - 1 = 2EL^2/\mu k^2 = \mu v_\infty^2\cdot\mu^2v_\infty^2b^2/\mu k^2 = (\mu v_\infty^2b/k)^2\).

2. \(f = -k/r^2\) with \(k < 0\) gives \(u'' + u = \mu k/L^2 = -1/|p|\). So \(u = -1/|p| + C\cos\theta\) and \(u > 0\) needs \(C|p| > 1\); writing \(e = C|p|\), \(r = |p|/(e\cos\theta - 1)\). Energy at \(\theta = 0\) (where \(u\) is largest, so \(r\) least and \(\dot r = 0\)): \(E = L^2(e-1)^2/2\mu p^2 + |k|(e-1)/|p| = \tfrac{|k|}{2|p|}(e-1)\big((e-1)+2\big) = \tfrac{|k|}{2|p|}(e^2-1)\), the same relation as before, so the same \(e\).

3. Attractive: \(r\to\infty\) where \(\cos\theta_\infty = -1/e\), with \(\theta_\infty\in(\pi/2,\pi)\); the particle comes in along \(-\theta_\infty\) and leaves along \(+\theta_\infty\), sweeping \(2\theta_\infty\), and its direction of travel turns by \(\Theta = 2\theta_\infty - \pi\). Then \(\cos\theta_\infty = \cos(\pi/2 + \Theta/2) = -\sin(\Theta/2)\), so \(\sin(\Theta/2) = 1/e\). Repulsive: \(r\to\infty\) where \(\cos\theta_\infty = 1/e\), \(\theta_\infty\in(0,\pi/2)\); the particle sweeps only \(2\theta_\infty\) and turns by \(\Theta = \pi - 2\theta_\infty\), so \(\cos\theta_\infty = \sin(\Theta/2) = 1/e\) again. In both cases \(\tan(\Theta/2) = 1/\sqrt{e^2-1} = |k|/\mu v_\infty^2b\). \(\square\)

4. \(|k| = 2\times79\times1.439965 = 227.514\) MeV fm. \(\mu/m_\alpha = m_{\mathrm{Au}}/(m_\alpha + m_{\mathrm{Au}}) = 0.98009\), so \(E = 0.98009\times7.69 = 7.5369\) MeV. Then \(\tan(\Theta/2) = |k|/2Eb = 227.514/(2\times7.5369\times20) = 0.75467\), \(\Theta = 74.08^\circ\). Head-on (\(L = 0\)), all of \(E\) becomes potential energy: \(d = |k|/E = 30.19\) fm, about four times the gold nucleus's radius — which is why the formula held in Geiger and Marsden's experiment. (\(\Theta\) is the turning of the relative velocity, the centre-of-mass angle; the gold recoils slightly and the laboratory angle is \(72.97^\circ\).)

5. \(k/\mu = GM_\odot\) and \(\tan(\Theta/2) = GM_\odot/c^2R_\odot = 2.1225\times10^{-6}\), so \(\Theta = 4.2450\times10^{-6}\) rad \(= 0.876''\). General relativity gives twice this, \(1.75''\); unit 37 derives it, and the factor of two is what the 1919 eclipse expedition measured.

Check. Dimensions: \(k/\mu v^2b\) is \(\mathsf{ML^3T^{-2}}/(\mathsf{M\,L^2T^{-2}\,L}) = 1\) ✓. Limits: \(b\to\infty\) gives \(\Theta\to0\), no deflection ✓; \(b\to0\) gives \(\Theta\to\pi\), straight back for repulsion ✓. The \(b\) for \(90^\circ\) is \(|k|/2E = d/2 = 15.09\) fm ✓, consistent with part 4.

Idea. The energy fixes \(a\), and \(a\) fixes the speed at every radius. Each burn is the difference between the speed the next orbit needs at that radius and the speed the current one has.

1. Energy per unit mass on the orbit is \(\tfrac12v^2 - GM_\oplus/r\), and the first-law slide gives \(E/\mu = -GM_\oplus/2a\). Equate: \(v^2 = GM_\oplus(2/r - 1/a)\). \(\square\)

2. \(a = 2.44675\times10^{7}\) m, \(e = (r_2 - r_1)/(r_2 + r_1) = 0.72327\). Circular speeds \(v_1 = \sqrt{GM_\oplus/r_1} = 7672.6\ \mathrm{m\,s^{-1}}\), \(v_2 = 3074.7\ \mathrm{m\,s^{-1}}\). On the ellipse by vis-viva: at perigee \(v_p = 10\,072.1\ \mathrm{m\,s^{-1}}\), at apogee \(v_a = 1617.4\ \mathrm{m\,s^{-1}}\). So \(\Delta v_1 = 2399.5\ \mathrm{m\,s^{-1}}\), \(\Delta v_2 = 1457.2\ \mathrm{m\,s^{-1}}\), total \(3856.7\ \mathrm{m\,s^{-1}}\). Transfer time \(\pi\sqrt{a^3/GM_\oplus} = 1.9044\times10^{4}\) s \(= 5.290\) h, about 5 h 17 min.

3. Kinetic energy added: \(\tfrac12(1200)\big[(v_p^2 - v_1^2) + (v_2^2 - v_a^2)\big] = 2.9649\times10^{10}\) J. The burns are instantaneous, so each changes \(T\) and not \(V\), and the total equals \(E_2 - E_1 = GM_\oplus m(1/2r_1 - 1/2r_2)\) — unit 04's \(2.965\times10^{10}\) J ✓.

Check. Apse condition: \(r_1v_p = 6.8198\times10^{10} = r_2v_a\ \mathrm{m^2s^{-1}}\) ✓ (Kepler's second law). Dimensions of vis-viva: \(\mathrm{m^3s^{-2}\cdot m^{-1}} = \mathrm{m^2s^{-2}}\) ✓. Limits: \(r_2\to r_1\) gives \(a\to r_1\), \(v_p\to v_1\), both burns \(\to0\) ✓; \(r = a\) in vis-viva gives \(v^2 = GM/a\), circular ✓.

Idea. A \(1/r^3\) force is invisible to the Binet equation's structure except that it shifts the natural frequency in \(\theta\) away from 1. Frequency not equal to 1 means the ellipse does not come back to itself: it turns.

1. \(f(1/u) = -ku^2 - cu^3\), so the Binet right side is \((\mu/L^2u^2)(ku^2 + cu^3) = \mu k/L^2 + (\mu c/L^2)u\), and \(u'' + \gamma^2u = \mu k/L^2\) with \(\gamma^2 = 1 - \mu c/L^2 \in (0,1)\). By unit 06 the homogeneous solutions are \(\cos\gamma\theta,\sin\gamma\theta\); a particular solution is the constant \(\mu k/\gamma^2L^2\). Aligning the axis, \(u = (1 + e\cos\gamma\theta)/p'\) with \(p' = \gamma^2L^2/\mu k\).

2. \(r\) is least where \(\cos\gamma\theta = 1\), at \(\theta = 2\pi j/\gamma\). Consecutive perihelia are \(2\pi/\gamma > 2\pi\) apart: one full turn plus \(\delta = 2\pi(1/\gamma - 1)\), an advance. With \(\varepsilon = \mu c/L^2\) small, \(1/\gamma = (1-\varepsilon)^{-1/2} = 1 + \varepsilon/2 + O(\varepsilon^2)\), so \(\delta = \pi\varepsilon + O(\varepsilon^2) = \pi\mu c/L^2\).

3. Orbits per century: \(36\,525/87.969 = 415.20\). So \(\delta = 43''/415.20 = 0.10356''\) per orbit \(= 5.021\times10^{-7}\) rad. Then \(\mu c/L^2 = \delta/\pi\), and with \(L^2 = \mu kp\) (to leading order), \(c/k = p\,\delta/\pi\). Here \(p = a(1-e^2) = 5.5460\times10^{10}\) m, so \(c/k = 8.86\times10^{3}\) m. And \(6GM_\odot/c_{\text{light}}^2 = 6\times1476.6 = 8.860\times10^{3}\) m: the same, to the accuracy of the \(43''\). General relativity's correction, which unit 37 derives, advances the perihelion — to first order — exactly as this extra attraction does with \(c/k = 6GM_\odot/c_{\text{light}}^2\).

Check. Dimensions: \([c] = \mathsf{MLT^{-2}\cdot L^3}\), so \([\mu c/L^2] = \mathsf{M\cdot ML^4T^{-2}}/\mathsf{M^2L^4T^{-2}} = 1\) ✓, and \([c/k] = \mathsf L\) ✓. Limit: \(c\to0\) gives \(\gamma\to1\), \(\delta\to0\), \(p'\to p\), the closed Kepler ellipse ✓. Sign: an extra attraction gives an advance; a repulsive \(c < 0\) gives \(\gamma > 1\) and a regression ✓.

Idea. The areal velocity turns time into an integral over angle. For the ellipse that integral leads to Kepler's equation, which cannot be inverted in closed form (unit 09); for the parabola it is a cubic polynomial in \(\tan(\theta/2)\). And here only \(t\) at a known \(\theta\) is wanted, so no inversion is needed at all.

1. \(e = 1\) in \(r = p/(1+e\cos\theta)\), and perihelion \(q = p/2\), so \(p = 2q\). Dividing \(p = L^2/\mu k\) by \(\mu\) twice: \(p = h^2/GM_\odot\), so \(h = \sqrt{2GM_\odot q}\).

2. Kepler's second law in the form \(r^2\dot\theta = h\) gives \(\mathrm dt = r^2\,\mathrm d\theta/h\). With rung 3, \(\mathrm dt = (q^2/h)\sec^4(\theta/2)\,\mathrm d\theta\). Put \(\tau = \tan(\theta/2)\): \(\mathrm d\theta = 2\cos^2(\theta/2)\,\mathrm d\tau\) and \(\sec^2(\theta/2) = 1+\tau^2\), so \(\mathrm dt = (2q^2/h)(1+\tau^2)\,\mathrm d\tau\). Integrating from perihelion (\(\tau = 0\)), \[ t = \frac{2q^2}{h}\Big(\tau + \frac{\tau^3}{3}\Big) = \frac{2q^2}{\sqrt{2GM_\odot q}}\Big(\tau + \frac{\tau^3}{3}\Big) = \sqrt{\frac{2q^3}{GM_\odot}}\Big(\tau + \frac{\tau^3}{3}\Big) . \qquad\square \]

3. \(r = R\) where \(1 + \cos\theta = 2q/R\), i.e. \(\cos^2(\theta/2) = q/R\), so \(\tau_R^2 = \sec^2 - 1 = (R-q)/q\). The radial motion is symmetric about perihelion (the quadrature slide's reflection argument), so \[ t_{\text{in}} = 2\sqrt{\frac{2q^3}{GM_\odot}}\,\tau_R\Big(1 + \frac{\tau_R^2}{3}\Big) = 2\sqrt{\frac{2q^3}{GM_\odot}}\sqrt{\frac{R-q}{q}}\cdot\frac{2q + R}{3q} = \frac{2\sqrt2}{3\sqrt{GM_\odot}}(R+2q)\sqrt{R-q} . \] With \(R = 1\) AU and \(q\) in AU, \(\sqrt{\mathrm{AU}^3/GM_\odot} = T_\oplus/2\pi\) where \(T_\oplus = 1.000019\) yr is the period of a 1 AU orbit (the third-law example), so \(t_{\text{in}} = \dfrac{\sqrt2}{3\pi}(1+2q)\sqrt{1-q}\) yr, to the \(2\times10^{-5}\) of that factor. \(\square\)

4. \(g(q) = (1+2q)\sqrt{1-q}\): \(g'(q) = 2\sqrt{1-q} - (1+2q)/2\sqrt{1-q} = 0\) iff \(4(1-q) = 1 + 2q\), \(q = \tfrac12\), and \(g\) rises before and falls after, so this is the maximum: \(g(\tfrac12) = 2\sqrt{1/2} = \sqrt2\), \(t_{\text{in}} = 2/3\pi\) yr \(= 0.21221\) yr \(= 77.5\) days. No comet on a parabola spends more than about eleven weeks inside the Earth's orbit. For Halley, \(g(0.586025) = 2.17205\times0.64341 = 1.39752\), \(t_{\text{in}} = 0.20970\) yr \(= 76.6\) days, against \(78.1\) days for the true ellipse (computed with unit 09's Kepler equation): the parabolic model is \(1.9\%\) short, because a bound comet near perihelion moves slightly slower than a parabolic one at the same \(r\).

Check. Dimensions: \(\sqrt{\mathrm{m^3}/\mathrm{m^3s^{-2}}} = \mathrm s\) ✓. Limits: \(q\to R\) gives \(t_{\text{in}}\to0\), the comet just touches the circle ✓; \(q\to0\) gives \(\tfrac{\sqrt2}{3\pi}\) yr \(= 54.8\) days, a comet plunging straight at the Sun — less than the maximum, because it rushes through the inner region ✓. Small \(\theta\): \(t\approx\sqrt{2q^3/GM}\,\theta/2\), and directly \(\dot\theta = h/q^2\) at perihelion gives \(t\approx q^2\theta/h = \theta\sqrt{q^3/2GM}\), the same ✓.