Unit 4 · Hints & solutions

The force is radial, and its signed magnitude depends only on $r$: $\mathbf F=F(r)\hat{\mathbf r}$. A radial velocity-dependent drag shares the zero-torque property but is not this conservative position-force model.

$\dot{\mathbf L}=\mu\dot{\mathbf r}\times\dot{\mathbf r}+\mathbf r\times\mathbf F=0+F(r)\mathbf r\times\hat{\mathbf r}=0$. The first term is zero because the vectors are parallel, and the second because the force is radial.

For nonzero constant $\mathbf L$, this equation confines all positions to the fixed plane through the origin normal to $\mathbf L$. When $\mathbf L=0$, no unique normal is defined; motion is radial along a line away from collision singularities.

$\mathbf L=3(2,0,0)\times(1,2,0)=(0,0,12)\,\mathrm{kg\,m^2\,s^{-1}}$. Thus $dS/dt=12/(2\times3)=2\,\mathrm{m^2\,s^{-1}}$.

$\mu=(2m)(3m)/(5m)=6m/5$. It is less than either constituent mass. The total mass is $5m$ and describes the separate center-of-mass motion.

$\mathbf r_1=\mathbf R+(m_2/M)\mathbf r$, $\mathbf r_2=\mathbf R-(m_1/M)\mathbf r$. On expansion, the mixed coefficient is $m_1m_2/M-m_2m_1/M=0$. The relative squared coefficient is $(m_1m_2^2+m_2m_1^2)/M^2=\mu$. Therefore $T=M\dot{\mathbf R}^2/2+\mu\dot{\mathbf r}^2/2$.

$V=Cr^3/3+C_0$. Dotting Newton’s law with velocity gives $d(\mu v^2/2)/dt=F\dot r=-dV/dt$. Thus $E=\mu v^2/2+Cr^3/3+C_0$ is constant.

Zero torque, constant angular momentum, planarity for nonzero angular momentum, and constant areal velocity survive. A potential $V(r,t)$ gives $\dot E=V_t$, so energy generally does not remain constant.

$\mu=m_2/(1+m_2/m_1)\simeq m_2[1-m_2/m_1+\cdots]$. The center of mass lies close to the heavy body, whose displacement is $m_2/M$ times the relative displacement.

No. At an apsis $\ell=\mu rv$, so different apsidal radii have different speeds. Along a general orbit the radial velocity also contributes to total speed without contributing to angular momentum.

The internal forces are equal and opposite. Thus $m_1\ddot{\mathbf r}_1+m_2\ddot{\mathbf r}_2=0=M\ddot{\mathbf R}$. Therefore $\mathbf R=\mathbf R_0+\mathbf V_0t$.

Total energy is $E_{\rm lab}=M\dot{\mathbf R}^2/2+E_{\rm rel}$. Both terms are separately constant for the isolated model, but they are equal only in the center-of-mass frame with the same potential zero.

Also $\dot{\hat{\boldsymbol\theta}}=-\dot\theta\hat{\mathbf r}$. Differentiating $\mathbf v=\dot r\hat{\mathbf r}+r\dot\theta\hat{\boldsymbol\theta}$ gives radial terms $\ddot r-r\dot\theta^2$ and angular terms $\dot r\dot\theta+\dot r\dot\theta+r\ddot\theta$. Hence $\mathbf a=(\ddot r-r\dot\theta^2)\hat{\mathbf r}+(2\dot r\dot\theta+r\ddot\theta)\hat{\boldsymbol\theta}$.

$E=\mu\dot r^2/2+\mu r^2\dot\theta^2/2+V$. With $\dot\theta=\ell/(\mu r^2)$, the angular term becomes $\ell^2/(2\mu r^2)$. Thus $V_{\rm eff}=V+\ell^2/(2\mu r^2)$ and $\mu\ddot r=-V_{\rm eff}^{\prime}$.

No. It is the angular kinetic energy expressed at fixed angular momentum, and its derivative appears after rearranging the radial acceleration. The inertial-frame force remains $-V^{\prime}\hat{\mathbf r}$.

$V_{\rm eff}^{\prime}=-1/r^3+1/r^2=0$ gives $r_c=1$. Then $E_c=1/2-1=-1/2$. The curvature is $3-2=1>0$, so the circle is radially stable.

$-4r^2=9-18r$ gives $4r^2-18r+9=0$. Thus $r_\pm=(18\pm\sqrt{324-144})/8=(9\pm3\sqrt5)/4$. The bounded allowed region lies between these roots.

Only $\dot r=0$. The tangential speed is $\ell/(\mu r)$, which remains nonzero at a finite radius. A radial turning point is an apsis, not complete rest.

Since $V_{\rm eff}^{\prime}(r_c)=0$, $\mu\ddot\eta=-V_{\rm eff}^{\prime\prime}(r_c)\eta$ to first order. Positive curvature gives oscillations, negative curvature exponential growth, and zero curvature requires higher-order analysis.

$\kappa r_c=\ell^2/(\mu r_c^3)$, so $r_c=[\ell^2/(\mu\kappa)]^{1/4}$. At the circle, $V_{\rm eff}^{\prime\prime}=4\kappa>0$.

Adding a constant changes the sign of $E$ without affecting forces or paths. In an oscillator potential, all finite-energy motion is bounded although the conventional energies are positive. Accessible radii must be found from $E\geq V_{\rm eff}$.

$V_{\rm eff}=\ell^2/(2\mu r^2)-a/r^3\to-\infty$ as $r\to0$. The attraction dominates. Some accessible inward trajectories can reach the center even with nonzero angular momentum; a barrier maximum may separate them from outer trajectories.

$\dot r=\pm\sqrt{2(E-V_{\rm eff})/\mu}$, so $t-t_0=\pm\sqrt{\mu/2}\int_{r_0}^r ds/\sqrt{E-V_{\rm eff}(s)}$. Choose the plus branch for outward motion and minus for inward motion, changing branch at ordinary turning points.

A double root can be a circular state or an asymptotic approach to an unstable circle. Near a maximum at the separatrix energy, $E-V_{\rm eff}\propto(r-r_c)^2$, so the time integral diverges logarithmically. It is not an ordinary finite-time bounce.

If $\ell=0$, then $\dot\theta=0$ away from the center, so angle cannot parametrize the changing radius. The relations containing $1/\ell$ are invalid. Use the radial Newton or energy equation instead.

$dr/d\theta=-u^{\prime}/u^2$ and $\dot\theta=\ell u^2/\mu$. Multiplication cancels $u^2$, giving $\dot r=-\ell u^{\prime}/\mu$.

$\ddot r=-\ell^2u^2u^{\prime\prime}/\mu^2$ and $r\dot\theta^2=\ell^2u^3/\mu^2$. Substitution into $\mu(\ddot r-r\dot\theta^2)=F(1/u)$ gives $u^{\prime\prime}+u=-\mu F(1/u)/(\ell^2u^2)$.

$F(1/u)=-ku^2$, so $-\mu F/(\ell^2u^2)=+\mu k/\ell^2$. Attraction produces a positive constant with this convention.

$u^{\prime\prime}+u=0$ implies zero force. Also $r\cos\theta=1/C$, or $x=1/C$. The physical angular branch requires $\cos\theta>0$ for $C>0$.

The sum is $C$. Rearranging Binet gives $F=-\ell^2u^2C/\mu=-\ell^2C/(\mu r^2)$, an attractive inverse-square force for $C>0$.

$v_r=-\ell u^{\prime}/\mu$, $v_\theta=\ell u/\mu$. Thus $E=\mu(v_r^2+v_\theta^2)/2+V=\ell^2(u^{\prime2}+u^2)/(2\mu)+V(1/u)$.

$u_0=1/2$ and $u_0^{\prime}=-\mu\dot r_0/\ell=3/2$. Positive $u^{\prime}$ corresponds to decreasing radius in the chosen positive angular direction.

$\Delta\theta=2\pi/\nu$ for $e>0$, $\nu>0$. Closure requires a rational relation between this advance and $2\pi$, subject to the orbit’s allowed radial range.

$\nu=\sqrt{0.96}$. The exact advance relative to $2\pi$ is $2\pi(1/\sqrt{0.96}-1)\simeq0.1296$ rad. The first-order estimate is $\pi(0.04)=0.1257$ rad, showing the size of the approximation error.

$dt=\mu r(\theta)^2d\theta/\ell$. Integrating from a known initial angle and time gives the temporal position along the orbit. The orbit equation alone does not imply uniform angular motion.

Physical radius must be positive. Regions with $u<0$ are not the same physical polar branch, and $u=0$ is an asymptote at infinity. Initial data and accessible angles determine which interval describes the trajectory.

A bounded orbit stays within a finite radial range. A closed orbit retraces after a finite repeat. A bounded precessing rosette can fail to close because radial and angular cycles are incommensurate.

$V^{\prime}=Cr^s$ and $V^{\prime\prime}=Cs r^{s-1}$. At a circle, $V_{\rm eff}^{\prime\prime}=V^{\prime\prime}+3V^{\prime}/r=C(s+3)r^{s-1}$. It is positive for $s>-3$, negative for $s<-3$, and marginal for $s=-3$.

$s+3=1/2>0$, so circles are linearly stable. The ratio is $\omega_r/\Omega=1/\sqrt2$, giving periapsis advance per radial cycle $2\pi\sqrt2$ in the near-circular limit.

Within the smooth attractive central-potential setting admitting stable circular and nearby bounded motions, only $V=-k/r$ and $V=\kappa r^2/2$, up to constants and with positive coefficients, have every bounded non-collision orbit closed. Other potentials may have particular closed or circular orbits.

The local ratio controls infinitesimal oscillations around a circle. Finite-amplitude apsidal advance can change with energy and angular momentum. Universal closure requires the entire bounded family to satisfy the condition, not just its linearized limit.

$\ddot{\mathbf r}+\omega^2\mathbf r=0$ with $\omega^2=\kappa/\mu$ gives $\mathbf r=\mathbf a\cos\omega t+\mathbf b\sin\omega t$. Every component and velocity repeats after $2\pi/\omega$, and the path is a linear image of a circle: an ellipse or a degenerate line.

For the isotropic oscillator the ellipse is centered on the force origin. For inverse-square attraction the origin is at a focus. Only in the circular case do center and focus coincide.

For a symmetric radial cycle, the first is twice the second. Near a stable circle they are $2\pi\Omega/\omega_r$ and $\pi\Omega/\omega_r$. State which convention is meant by “apsidal angle.”

For $F=-C/r^3$, $V_{\rm eff}=[\ell^2/\mu-C]/(2r^2)$. At the tuned value $\ell^2=\mu C$, the effective potential is flat. There is no radial restoring force; a small radial velocity causes drift. It is not ordinary oscillatory stability.

Yes. Its circle at $r_c=3\mu a/\ell^2$ is closed, but $V_{\rm eff}^{\prime\prime}=-3a/r_c^5<0$. Radial disturbances include a growing exponential, so the circle is unstable.

$s=1$ gives $\omega_r/\Omega=2$. Hence two radial cycles occur in one $2\pi$ revolution; successive closest approaches are separated by $\pi$.

$\Delta\theta=2\int_{r_-}^{r_+}\ell\,dr/[r^2\sqrt{2\mu(E-V_{\rm eff})}]$. A regular bounded orbit closes when $\Delta\theta/(2\pi)$ is rational, allowing integer numbers of radial and angular cycles to coincide.

$u=\mu k/\ell^2+C\cos\theta+D\sin\theta$. Combining the trigonometric terms and choosing periapsis as the polar axis gives $r=p/(1+e\cos\theta)$, with $p=\ell^2/(\mu k)$ and dimensionless eccentricity $e\geq0$.

$u^{\prime2}+u^2=(1+2e\cos\theta+e^2)/p^2$. With $\ell^2/(\mu p)=k$, $E=k(1+2e\cos\theta+e^2)/(2p)-k(1+e\cos\theta)/p=k(e^2-1)/(2p)$. Thus $e^2=1+2E\ell^2/(\mu k^2)$.

$a=(3+9)/2=6$, $e=(9-3)/(9+3)=1/2$, and $p=a(1-e^2)=6(3/4)=4.5$.

$E_c=-\mu k^2/(2\ell^2)=-1/8$. Lower energy would make $e^2<0$ and lies below the effective-potential minimum. At $E_c$ the circular radius is $p=4$.

$e^2=1+2(-1/4)=1/2$, so $e=1/\sqrt2<1$: an ellipse. Its semimajor axis is $a=-k/(2E)=2$ and semi-latus rectum is $p=1$.

$E=\mu v^2/2-k/r$. Setting $E=0$ gives $v_{\rm esc}=\sqrt{2k/(\mu r)}$. Circular speed is $\sqrt{k/(\mu r)}$, so escape speed is larger by $\sqrt2$, subject to collision and model qualifications.

$\ell^2=\mu ka(1-e^2)$. Thus increasing eccentricity at fixed $a$ decreases angular momentum, while $E=-k/(2a)$ remains fixed. The circle has the maximum angular momentum for that energy.

$E=k/(8r_0)-k/r_0=-7k/(8r_0)$, so $a=4r_0/7$. Also $\ell^2=\mu kr_0/4$, giving $e^2=1-7/16=9/16$, hence $e=3/4$. Since $a(1+e)=r_0$, launch is apoapsis; periapsis is $r_0/7$.

$1+2\cos\theta_\infty=0$ gives $\theta_\infty=2\pi/3=120^\circ$. The physical branch is $|\theta|<120^\circ$ around periapsis.

At finite $r$, $\mu v^2/2=k/r>0$. The total vanishes because $V=-k/r$. Speed tends to zero only as $r\to\infty$.

One obtains $(1-e^2)x^2+y^2+2pex-p^2=0$. Completing the square shifts $x$ by $pe/(1-e^2)=ae$. Hence the center is $(-ae,0)$ and the force center remains the focus at the origin.

Formally $\ell=0$ gives $e=1$ for any energy, which loses the radial motion information. The derivation assumed nonzero angular momentum. Radial collision or escape must instead be treated using $E=\mu\dot r^2/2-k/r$.

$dS=|\mathbf r\times d\mathbf r|/2$, so $dS/dt=|\mathbf r\times\dot{\mathbf r}|/2=\ell/(2\mu)$. Constant central-force angular momentum gives $\Delta S=\ell\Delta t/(2\mu)$ for any interval.

The equal-area law follows from zero torque for any central force. Elliptical Kepler orbits and the $a^3$ period law require the inverse-square interaction.

$\tau=2\pi\mu ab/\ell$. Squaring and using $b^2=a^2(1-e^2)$ and $\ell^2=\mu ka(1-e^2)$ gives $\tau^2=4\pi^2\mu a^3/k$. For gravity $k/\mu=G(m_1+m_2)$, giving $\tau^2=4\pi^2a^3/[G(m_1+m_2)]$.

$\tau_2/\tau_1=(a_2/a_1)^{3/2}=9^{3/2}=27$. The comparison assumes the same total gravitating mass.

$v_p/v_a=r_a/r_p=(1+e)/(1-e)=(4/3)/(2/3)=2$. The angular-speed ratio is $4$.

The periods are equal by $\tau^2=4\pi^2\mu a^3/k$, and relative energies are equal to $-k/(2a)$. Their angular momenta differ through $\ell^2=\mu ka(1-e^2)$.

The reduction uses $\mathbf r=\mathbf r_1-\mathbf r_2$, so its semimajor axis is $a=a_1+a_2$. In the center-of-mass frame $a_1=(m_2/M)a$ and $a_2=(m_1/M)a$. Substituting only one of them into the relative third law changes the inferred mass.

With $GM_\odot=4\pi^2\,\mathrm{AU^3\,yr^{-2}}$, $M/M_\odot=27/9=3$. This infers only the total mass.

Generally no. The conserved areal rate is measured from the force center, which is a focus for inverse-square attraction. A line from the geometric ellipse center does not satisfy the same torque argument except in special cases such as a circle.

$\Delta t=0.30\times120=36$ days for the specified forward sector. The result is independent of where the sector lies, provided its swept area fraction is the stated one.

From periapsis, $S=ab(\psi-e\sin\psi)/2$. Equating with $\ell(t-t_p)/(2\mu)$ and using $\tau=2\pi\mu ab/\ell$ gives $2\pi(t-t_p)/\tau=\psi-e\sin\psi$.

Zero torque fixes the plane and angular momentum. Inverse-square Binet motion produces a conic, and negative energy makes it an ellipse with the origin at a focus. Angular momentum gives constant area rate. Dividing ellipse area by that rate and eliminating angular momentum yields the third law.

$\mathbf A=\mathbf p\times\mathbf L-\mu k\hat{\mathbf r}$, with $\mathbf p=\mu\dot{\mathbf r}$ and $\mathbf L=\mathbf r\times\mathbf p$. Reversing the cross product changes its sign and does not preserve this convention.

$\dot{\hat{\mathbf r}}=\dot{\mathbf r}/r-\mathbf r\dot r/r^2$. Since $\dot r=(\mathbf p\cdot\hat{\mathbf r})/\mu$, this is $[\mathbf p-(\mathbf p\cdot\hat{\mathbf r})\hat{\mathbf r}]/(\mu r)$.

With $\dot{\mathbf L}=0$, $\dot{\mathbf A}=\dot{\mathbf p}\times\mathbf L-\mu k\dot{\hat{\mathbf r}}$. The force term is $-k\hat{\mathbf r}\times(\mathbf r\times\mathbf p)/r^2=k[\mathbf p-(\mathbf p\cdot\hat{\mathbf r})\hat{\mathbf r}]/r=\mu k\dot{\hat{\mathbf r}}$. The terms cancel.

$(\mathbf p\times\mathbf L)\cdot\mathbf L=0$ by perpendicularity. Also $\hat{\mathbf r}\cdot\mathbf L=0$ since $\mathbf L=\mathbf r\times\mathbf p$. Their difference therefore has zero dot product with $\mathbf L$.

$(\mathbf p\times\mathbf L)\cdot\mathbf r=\mathbf L\cdot(\mathbf r\times\mathbf p)=\ell^2$. Hence $Ar\cos\theta=\ell^2-\mu kr$, giving $r=[\ell^2/(\mu k)]/[1+(A/\mu k)\cos\theta]$. Therefore $e=A/(\mu k)$.

$A^2=p^2\ell^2+\mu^2k^2-2\mu k\ell^2/r$. Factor the momentum and potential terms as $2\mu\ell^2[p^2/(2\mu)-k/r]$. Thus $A^2=\mu^2k^2+2\mu E\ell^2$.

$\mathbf L=(0,0,2)$, $\mathbf p\times\mathbf L=(2,0,0)$, and $\mathbf A=(1,0,0)$. Thus $e=1$. Energy is $1/2-1/2=0$, so the nonradial orbit is parabolic, with periapsis along $+x$.

$\mathbf A=0$. Its magnitude gives $e=0$, but no direction can be assigned. A circle has no distinguished periapsis or major-axis direction.

$\tan(\chi/2)=1$, so $\chi/2=\pi/4$ in the physical range and $\chi=\pi/2=90^\circ$.

They obey $\mathbf A\cdot\mathbf L=0$ and $A^2=\mu^2k^2+2\mu E\ell^2$. These remove two independent scalar freedoms for a generic nonradial orbit, leaving five orbital constants. Initial phase along the orbit is still required.

The residual is $\dot{\mathbf A}=\delta F\hat{\mathbf r}\times\mathbf L$. It is generally nonzero, so the eccentricity vector need not remain fixed even while angular momentum and the orbital plane do.

Calculate $\mathbf p=\mu\mathbf v$, $\mathbf L=\mathbf r\times\mathbf p$, $E=p^2/(2\mu)-k/r$, and $\mathbf A=\mathbf p\times\mathbf L-\mu k\hat{\mathbf r}$. For $\ell\ne0$, the plane is normal to $\mathbf L$, eccentricity is $A/(\mu k)$, periapsis points along nonzero $\mathbf A$, and the semi-latus rectum is $\ell^2/(\mu k)$. The given state fixes the phase and direction of motion.