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By Newton's third law, the force on the second particle is $-F_{12}$. For $M=m_1+m_2$ and centre of mass $R=(m_1{r}_1+m_2{r}_2)/M$, Newton's second law gives $M\ddot{R}=F_{12}-F_{12}=0$. ThusThe centre moves on a straight line with constant velocity; zero velocity is allowed. For the relative position $r=r_1-r_2$, subtraction of the two equations givesHence the two-body problem reduces to a single particle of reduced mass $\mu$, together with the elementary centre motion. Reconstruct the positions as $r_1=R+(m_2/M)r$, $r_2=R-(m_1/M)r$.

Yes: the two equal masses can follow the same fixed circle in diametrically opposite positions. Put the stationary centre of mass at the circle's center and choose tangential velocities with the same sense of rotation. If the circle has radius $a$, the separation is $2a$ and the mutual gravitational force is directed toward that center, with magnitude $G{m}^2/(4a^2)$. The circular orbit condition is $ma{\omega }^2=G{m}^2/(4a^2)$, so an equal-mass circular binary hasThese initial data maintain the opposite positions and give the same constant angular speed for both particles.

A central force has no angular component, so the supplied polar acceleration formula and Newton's second law give $d(r^2\dot{\theta })/dt=0$. Thus $l=r^2\dot{\theta }$ is the conserved specific angular momentum. For a nonradial orbit, $l\ne 0$ allows $\theta$ to be used as a parameter. With $u=1/r$ and a prime denoting $d/d\theta$,Also $r{\dot{\theta }}^2=l^2{u}^3$. The radial equation $\ddot{r}-r{\dot{\theta }}^2=-f(r)$ therefore becomes the Binet equationFor a purely radial trajectory $l=0$, the angular parameter is unavailable, and one instead uses the direct radial equation. The orbit in part (b) is nonradial.

The PDF gives initial position $(x^{\prime },y^{\prime },z^{\prime })=(1,0,0)$ and initial rotating-frame velocity zero. Use those values; the TeX erroneously places dots over the position coordinates.

For rotation about the third axis, the supplied acceleration relation and Newton's second law give the equations of motion in a rotating frameThus $z^{\prime }=0$. Set $\zeta =x^{\prime }+i{y}^{\prime }$ to combine the planar equations:The two characteristic roots are $i\omega$ and $-3i\omega$. From $\zeta (0)=1$ and $\dot{\zeta }(0)=0$,The triple-angle formulas therefore give the particularly simple astroid motion of a harmonic oscillator in a rotating frameThe orbit satisfies $|x^{\prime }{|}^{2/3}+|y^{\prime }{|}^{2/3}=1$; the zero speed at each cusp is consistent with the smooth time-parametrized motion.

Its rotating-frame speed isHence, for $\omega >0$, the first maximum occurs at $t=\pi /(4\omega )$, andThe maximum repeats periodically. For a signed negative angular frequency, use $\pi /(4|\omega |)$ and $3|\omega |/2$; at $\omega =0$ the particle remains at rest.

Use zero-relative-velocity mass shedding: the released sand initially shares the balloon's vertical velocity, rather than being actively ejected. Let $q=M+m(t)$ and let a positive mass $\delta m$ be shed in a short interval $dt$. The momentum of the balloon and that newly released sand changes byThe leading external impulse is $(T-qg)dt$. Newton's second law applied to this combined material system therefore givesWriting $\frac{d}{dt}(qv)=T-qg$ for the remaining balloon alone would omit the outgoing momentum flux. A nonzero relative ejection velocity would instead produce a rocket equation thrust term.

Initial mechanical equilibrium gives $T=(M+m_0)g$. Write $\lambda =m_0/t_0$, so $m=m_0-\lambda t$ up to depletion. ThenIntegrating from rest,Since $A/t_0=1+M/m_0$ and $A/(A-t_0)=1+m_0/M$, the final upward speed isThe acceleration is nonnegative throughout release, consistent with the upward motion.

For inferring an inverse-square force from Kepler laws, first use the area swept by an infinitesimal radius sector:Kepler's second law makes the areal velocity constant, so $l=r^2\dot{\theta }$ is constant. Then $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$, meaning that the acceleration has no transverse component. Newton's second law therefore gives a central force directed along the Sun–planet line.

For the radial dependence, set $u=1/r$. Since $\dot{\theta }=l{u}^2$,The radial acceleration is the Binet equation expressionThe focus-based ellipse has $u=(1+\epsilon \cos \theta )/A$, hence $u_{\theta \theta }+u=1/A$. ThusFor a planet of mass $m$, the force is $F_r=-m{l}^2/(A{r}^2)$: it is attractive and obeys the inverse-square law. This deduction uses both the shape of the orbit and its constant areal velocity; the orbit shape alone would not determine its time-dependent acceleration.

For any vector $\mathbf{a}$, differentiating its components and the rotating basis gives the rotating-frame derivative formulaApply it twice to the position $\mathbf{x}$, with constant angular velocity $\mathbf{\omega }$. The inertial acceleration iswhere dots denote rotating-frame derivatives. Newton's second law therefore becomesThe last terms are the Coriolis force and the force corresponding to centrifugal acceleration.

For the particle on a uniformly rotating inclined plane, choose orthonormal vectors ${\mathbf{e}}_1$ along the horizontal $\xi$ axis and ${\mathbf{e}}_2$ up the slope, and set ${\mathbf{e}}_3={\mathbf{e}}_1\times {\mathbf{e}}_2$, the upward normal vector. Then ${\mathbf{e}}_z=\sin \theta {\mathbf{e}}_2+\cos \theta {\mathbf{e}}_3$ andHere $N$ is the normal reaction. The marble is modeled as the smooth sliding particle specified by the printed equations, without an additional rolling constraint. The rotating-frame Coriolis acceleration and centrifugal acceleration are respectivelyThe normal rotating-frame acceleration is zero, so the normal reaction is not generally $mg\cos \theta$; it satisfiesProjecting the same equation onto the two tangential axes gives exactly

Multiply the equations by $\dot{\xi }$ and $\dot{\eta }$ and add. The Coriolis force terms cancel, as expected because this force does no work relative to the rotating frame. The normal reaction also does no work along the plane. The conserved quantity is rotating-frame kinetic energy plus gravitational and centrifugal potential energy:This need not be the conserved inertial energy, since the externally driven rotating plane can exchange energy with the particle.

In classical charged particle in a uniform magnetic field motion, the Lorentz force and Newton's second law giveAssume $\mathbf{B}\ne 0$ and choose an orthonormal coordinate system with $\mathbf{B}=B{\mathbf{e}}_z$, $B=|\mathbf{B}|$. Write $\Omega =qB/m$, retaining the sign of the electric charge. The velocity equations areThus the parallel velocity is constant, and the perpendicular velocity rotates at constant angular frequency. Choosing the time origin and perpendicular axes conveniently givesIntegrating the velocity, one obtainsThe first two coordinates describe a circle of radius $m{v}_{\perp }/(|q|B)$, while the third advances uniformly. This is helical motion in a uniform magnetic field: the helix has axis parallel to $\mathbf{B}$ through the guiding centre. Its cyclotron frequency isand its signed angular velocity vector about that axis is $-q\mathbf{B}/m$. The minus sign follows from the ordering $\mathbf{v}\times \mathbf{B}$ in the Lorentz force. The helix degenerates to a circle when $v_{\parallel }=0$ and to a straight line when $v_{\perp }=0$. If $q=0$ or $B=0$, there is only straight-line motion. In a relativistic extension with fixed Lorentz factor $\gamma$, the frequency becomes $|q|B/(\gamma m)$; the classical result above uses $m\mathbf{v}$ as momentum.

Use Newton's second law, $F=ma$. Acceleration has physical dimension $L{T}^{-2}$, so multiplying by mass gives the dimension of forceThe bracket notation records a physical dimension, rather than the numerical value or the choice of units.

For $B_0\ne 0$, Newton's second law givesDefine the exponential relaxation time $\tau =mR/(B_0^2{L}^2)$ and terminal velocity $v_{\infty }={\mathcal{E}}_0/(B_{0}L)$. Solving the first-order equation for $v=\dot{s}$ and then integrating givesBoth prescribed initial values hold. The current approaches zero as the bar approaches terminal velocity. If $B_0=0$, there is no magnetic force and instead $\dot{s}=v_0$, $s=s_0+v_{0}t$; the constant current is ${\mathcal{E}}_0/R$.

The rotating-frame derivative formula is $(d\mathbf{A}/dt{)}_S=(d\mathbf{A}/dt{)}_{S^{\prime }}+\mathbf{\omega }\times \mathbf{A}$. Applying it twice to the position vector, with constant angular velocity, givesSolving for ${\mathbf{a}}_{S^{\prime }}$ identifies the displayed terms: its left side is the acceleration relative to the rotating reference frame, the first right-hand term is the inertial acceleration, $-2\mathbf{\omega }\times {\mathbf{v}}_{S^{\prime }}$ is the Coriolis acceleration, and $-\mathbf{\omega }\times (\mathbf{\omega }\times \mathbf{r})$ is the outward centrifugal acceleration.

For a particle of mass $m$, replace ${\mathbf{a}}_S$ by the actual force divided by $m$ using Newton's second law. The remaining terms then appear as apparent forces in $S^{\prime }$. There is no Euler acceleration because $\mathbf{\omega }$ is constant and no translational acceleration term because the common origin is fixed in the inertial frame.

The dimensions are $[m]=\mathsf{M}$, $[\alpha ]=\mathsf{M}{\mathsf{T}}^{-1}$, $[u_0]=\mathsf{L}{\mathsf{T}}^{-1}$, and $[g]=\mathsf{L}{\mathsf{T}}^{-2}$. Under the stated model, $m,\alpha ,u_0,g$ are the only parameters determining the ascent time. If $m^p{\alpha }^q{u}_0^r{g}^s$ is dimensionless, then $p+q=0$, $r+s=0$, and $-q-r-2s=0$, leaving one independent dimensionless variable. One convenient choice isThus dimensional analysis fixes the form, but Newton's second law must determine the function $f$.

Take upward velocity as positive. Throughout ascent, Newton's second law with linear drag gives $m\dot{v}=-mg-\alpha v$. An integrating factor or direct solution of this linear ordinary differential equation givesThe first zero of this velocity is the highest point, soThis vertical ascent under linear drag has $T<u_0/g$ for $u_0>0$, since $\log (1+\lambda )<\lambda$. The limit $\alpha \to 0$ recovers the no-drag ascent time $u_0/g$; $u_0=0$ gives $T=0$.

Write the density and pressure as ${\rho }_0+{\rho }^{\prime }$ and $p_0+p^{\prime }$, and let the small fluid velocity be the gradient of an acoustic velocity potential, $\mathbf{u}=\nabla \varphi$. The linear homentropic acoustic equations arewhere the adiabatic sound speed of a perfect gas is $c_0^2=\gamma p_0/{\rho }_0$. Absorbing a function of time into $\varphi$, the momentum equation givesEliminating $p^{\prime }$ and ${\rho }^{\prime }$ gives the acoustic wave equation

Let the shell displacement be $a(t)-a_0=\mathrm{Re}(\eta e^{-i\omega t})$ and put $k=\omega /c_0$, $\theta =k{a}_0$. The spherically symmetric solution regular at the centre isThe kinematic boundary condition ${\varphi }_r(a_0,t)=\dot{a}(t)$ givesSince $p^{\prime }=-{\rho }_0{\varphi }_t$, the pressure amplitude on the shell is thereforeThe radial Newton's second law per unit area isso its complex amplitude satisfiesFor a nonzero mode, cancellation of $\eta$ and multiplication by $a_0^2/(m{c}_0^2)$ yield the frequency equation for spherically symmetric vibration of a gas-filled elastic shell:The spectrum is discrete and contains infinitely many positive solutions. Indeed, for sufficiently large $n$, $\theta \cot \theta -1$ runs from $+\infty$ to $-\infty$ on $(n\pi ,(n+1)\pi )$, forcing at least one root there. These high modes are predominantly standing acoustic modes of the gas, while the lowest mode is predominantly the shell's radial breathing mode; the coupling shifts both families.

Label the two atoms in primitive cell $j$ by $u_{2j}$ and $u_{2j+1}$, with $j$ understood modulo $N$ because of the periodic boundary condition. Choose the spring between them to have constant $\lambda$. Newton's second law gives

For a longitudinal normal mode, putPeriodicity requires $e^{2iNka}=1$, so $k=\pi r/(Na)$. Since the primitive-cell length is $2a$, wavevectors differing by $\pi /a$ describe the same mode, and the first Brillouin zone may be chosen asThe amplitudes obeySetting its determinant to zero gives the alternating-spring chain dispersion relationThe lower sign is the acoustic phonon branch and the upper sign is the optical phonon branch.

At the edge of the Brillouin zone,so the frequency band gap isIt vanishes at $\alpha =1$, when the apparent two-atom primitive cell can be reduced to one atom.

Near the centre of the Brillouin zone, a Taylor expansion givesandThe group velocity of the acoustic branch at long wavelength is the speed of sound,

The Lorentz force and Newton's second law giveWriting $\Omega =qB/m$, the component equations areThe initial velocity therefore givesand integration with the initial position givesThis is helical motion in a uniform magnetic field: a helix of radius $v/\Omega$ around the line $x=1+v/\Omega$, $y=0$, with its axis parallel to the magnetic field.

By Newton's second law, force is mass times acceleration. Hence its dimensions are

Measure displacement downward from the release point and take downward speed $v\ge 0$. Newton's second law with quadratic drag givesWritingseparation of variables gives the terminal velocity solutionThe distance fallen by time $t$ is thereforeSetting $y(t)=h$ and solving for $t$ yields

Let $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$ and $r=|\mathbf{r}|$. Newton's second law for the two planetesimals givesSubtracting yields the forced relative two-body problemThe internal gravitational forces cancel from the centre of mass equation but add in the relative equation; external forces affect the internal orbit through their difference in acceleration.

The work done along the trajectory isBy Newton's second law, $F=m\ddot{x}$, soThis is the work-energy theorem: the work equals the change in kinetic energy.

A force is conservative when its work between two points is independent of the path, equivalently when it is the negative gradient of a potential energy. Here a potential with $V(0)=0$ isConservation of energy for $x>0$ givesThus

If $v_0>v_e$, the particle escapes with asymptotic speed $\sqrt{v_0^2-v_e^2}$; the graph decreases from $v_0$ to this horizontal asymptote. At the escape velocity $v_0=v_e$, it decreases as $v_e{e}^{-x/(2\lambda )}$ and approaches zero only at infinity. If $v_0<v_e$, it reaches the turning pointwhere $v=0$, then returns on the negative branch of the same curve and oscillates symmetrically about the origin.

For oscillation put $w=1-v_0^2/v_e^2$. One quarter of the period iswhere $u=e^{-x/\lambda }$. Therefore

Equating the satellite's mean density inside $r_t$ with the host's mean density inside orbital radius $R$ givesThus a measured tidal radius and independently estimated satellite mass infer $M(<R)\simeq m(R/r_t{)}^3$. Measurements from satellites over a range of $R$ trace the host's enclosed-mass profile and hence its gravitational potential. Globular clusters are more often tidally limited: their stellar extent can approach the Jacobi boundary, whereas dwarf galaxies commonly occupy extended dark-matter haloes and their observed stars may substantially underfill it. Orbital eccentricity, mass loss, and nonequilibrium structure complicate this inference.

Let the separation vector point from the host to the satellite. Subtracting the two Newton's second law equations givesThe relative orbit is therefore a one-body Kepler orbit with gravitational parameter $G(M+m)$. A circular orbit of separation $R_0$ hasThe centre-of-mass condition gives $x_m=M{R}_0/(M+m)$ for the satellite's distance from the barycentre.

Consider the inner collinear equilibrium a distance $r_J$ toward the host from the satellite. Taking the positive axis from the host toward the satellite, differentiation of the gravitational plus centrifugal effective potential givesSince $r_J\ll R_0$,The constant host-gravity term cancels ${\Omega }^2{x}_m=GM/R_0^2$. The remaining equation isFor a low-mass satellite, the term proportional to $m{r}_J$ is negligible compared with $3M{r}_J$, and the Jacobi tidal radius isThe outer collinear point gives the same leading result.

Stars that cross the two nearby Lagrange points are no longer bound to the satellite. Small energy and angular-momentum offsets place them on slightly different host orbits, producing one leading and one trailing tidal tail. Differential orbital frequency stretches these streams around the host, while epicyclic motion can create density clumps. The tail's position and velocity structure retain information about the satellite orbit and host potential.

Take positive $y$ vertically downward. For a short string element, the transverse tension resultant is $T{y}_{xx}dx$, while gravity and linear drag contribute $\mu gdx$ and $-2k\mu y_{t}dx$. Newton's second law givesSince $T=\mu c^2$, the damped wave equation isChoosing upward displacement reverses the sign of the gravity term.