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Differentiating $\mathbf{r}=r\hat{\mathbf{r}}$ twice and using $\dot{\hat{\mathbf{r}}}=\dot{\theta }\hat{\mathbf{\theta }}$ and $\dot{\hat{\mathbf{\theta }}}=-\dot{\theta }\hat{\mathbf{r}}$ gives the displayed acceleration. For a central force, its angular component vanishes, so $r^2\dot{\theta }$ is constant and expresses conservation of angular momentum.

For unit mass in a central force with potential $\Phi (r)$ and conserved angular momentum $L$, a radial turning point $r_{\ast }$ has zero radial speed and therefore satisfies

For an attractive central force of magnitude $c/r^5$, a circle of radius $R$ through the force center has polar equation $r=2R\cos \theta$. With specific angular momentum $h$, substitution into the Binet equation gives $c=8m{h}^2{R}^2$. Consequently $R$ is inversely proportional to the magnitude of the angular momentum, and the angular-time integral gives $T=2\pi R^2/|h|\propto R^3$. This is the finite travel time along the punctured circle from one approach to the singular center to the next; the force law alone does not specify continuation through the collision.

Conservation of angular momentum says that total angular momentum stays fixed when the net external torque about the chosen fixed inertial origin vanishes. Individual angular momenta and angular velocities can change. The radius change at constant angular momentum below makes this distinction visible: pulling two rotating point masses inward lowers the moment of inertia, raises the angular velocity, and supplies work that increases the kinetic energy.

Follow the same constant-mass point masses throughout, in an inertial frame, with all positions measured from a fixed origin $O$. If the internal pair forces are opposite and central, meaning directed along each pair's separation, thenThus zero net external torque throughout an interval implies constant total angular momentum throughout that interval. A vanishing external torque component, such as ${\tau }_z^{\mathrm{ext}}=0$, conserves the corresponding $L_z$ even when the other components change. This is the balance law for a fixed collection; angular-momentum flux must also be accounted for when matter crosses its boundary. ^[ref]

The central qualification is essential. Write ${\mathbf{F}}_{ij}$ for the internal force on member $i$ from member $j$. Under Newton's third law, ${\mathbf{F}}_{ji}=-{\mathbf{F}}_{ij}$, but the sum of their torques isThis vanishes when ${\mathbf{F}}_{ij}$ is parallel to ${\mathbf{r}}_i-{\mathbf{r}}_j$. Equal-and-opposite forces alone do not suffice: two opposite transverse forces with separated lines of action form a couple with nonzero torque. Cancellation of the internal pair torques, together with Newton's second law, proves the total balance above.

For one point mass under a central force about $O$, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ gives ${\mathbf{\tau }}_O=\mathbf{r}\times \mathbf{F}=0$. The force may still change the speed and kinetic energy; its zero torque preserves angular momentum. With nonzero ${\mathbf{L}}_O$, $\mathbf{r}\cdot {\mathbf{L}}_O=0$ keeps the motion in a fixed plane. Choosing the $z$ axis perpendicular to that plane, $r^2\dot{\theta }=L_z/m$ governs its angular progress. This removes an unknown from the central-force radial turning point calculation and helps reduce the Kepler orbit to a radial problem.

There is a historical geometric version of this result. In Book I, Section II, Proposition I of the Principia, Newton proves that motion under a central force toward a fixed centre stays in a plane and sweeps equal areas in equal times, using successive central impulses and equal triangle areas. ^[ref] In the modern notation above, constant areal velocity is $\dot{A}=L_z/(2m)$: the old geometric proof and the angular momentum calculation express the same invariant.

For potential energy $\Phi (r)=-mk/r^2$, the inward central force has magnitude $2mk/r^3$. With conserved specific angular momentum $h=r^2\dot{\theta }$, the radial equation is $\ddot{r}=(h^2-2k)/r^3$. At the critical value $h^2=2k$, this reduces to $\ddot{r}=0$. An inward initial radial velocity $-v$, with $v>0$ and $r(0)=a$, therefore gives $r=a-vt$ until collision at $t_{\ast }=a/v$.

For $h>0$, integration of $\dot{\theta }=h/(a-vt{)}^2$ gives $\theta -\theta (0)=(h/v)(1/r-1/a)$. Thus the orbit is a hyperbolic spiral and the particle completes infinitely many turns before the finite collision time. The tangential speed $h/r$ diverges. The potential energy and kinetic energy individually diverge with opposite signs, while their sum remains the finite conserved mechanical energy $m{v}^2/2$.

For a radial conservative force plus the magnetic-type force $\dot{\mathbf{r}}\times (\mu \mathbf{r}/r^3)$, the torque equals $\mu d\hat{\mathbf{r}}/dt$. Subtracting $\mu \hat{\mathbf{r}}$ from ordinary angular momentum therefore produces a constant vector. Its dot product with $\hat{\mathbf{r}}$ is $-\mu$, so motion lies on a cone about that vector. At $\mu =0$, ordinary planar central force motion is recovered. This is the classical radial magnetic monopole angular-momentum correction.

Constant areal velocity gives constant specific angular momentum $l=r^2\dot{\theta }$, hence zero transverse acceleration and a central force. For a focus-based ellipse $u=1/r=(1+\epsilon \cos \theta )/A$, one has $u_{\theta \theta }+u=1/A$. The Binet equation then gives the displayed radial acceleration, proving an attractive inverse-square law. An orbit's shape alone does not specify its acceleration without a time law.

Here inverse-square potential means the potential energy associated with an attractive inverse-square law for a central force: the potential energy itself varies as inverse distance. For $k>0$ and position $\mathbf{r}$ with $r=|\mathbf{r}|>0$,The gradient makes the distinction explicit: doubling $r$ halves the magnitude of potential energy but quarters the magnitude of force. The neighboring repulsive inverse-square potential instead has potential energy proportional to $r^{-2}$ and an inverse-cube force; the two naming conventions must be distinguished.

For Newton's law of universal gravitation, $k=GMm$, where $G$ is the gravitational constant. Treat the attracting mass $M$ as an isolated fixed point mass, or as a body with spherical symmetry and exterior test particle of constant mass $m\ll M$. The spherical shell theorem gives the same exterior Newtonian gravitational potential energy $U=-GMm/r$ without requiring uniform mass density. Here $r$ is measured from the centre, and the formula applies outside the body; it does not describe its interior. The Newtonian gravitational potential per unit mass is $\Phi =U/m=-GM/r$.^[ref]

The negative sign records binding relative to infinite separation, rather than a negative kinetic energy. Moving from $r_1$ to $r_2>r_1$ increases potential energy byBecause the force is a conservative force, this work depends on the endpoints, including for a curved path. Integrating $dV/dr=k/r^2$ with zero at infinity gives $V=-k/r$; adding a constant leaves the force unchanged but shifts the numerical escape energy threshold by that same constant. This is why the reference at infinity must accompany an escape velocity calculation.

For an exact synthetic comparison, stipulate the standard gravitational parameter $\mu =GM=8{\mathrm{m}}^3/{\mathrm{s}}^2$ and a $1\mathrm{kg}$ test particle. At $r=2\mathrm{m}$, $U=-4\mathrm{J}$ and the inward force has magnitude $2\mathrm{N}$; at $r=4\mathrm{m}$, $U=-2\mathrm{J}$ and the inward force has magnitude $0.5\mathrm{N}$. The increase is $2\mathrm{J}$ of potential energy. These stipulated parameters are an arithmetic example, not observations of a particular body. The escape-energy balance diagram uses the same parameters.

The relative displacement of a Kepler orbit sweeps equal areas about the attracting focus of a conic in equal times. A short interval near periapsis covers a wide angle at small separation; the same interval near apoapsis covers a narrow angle at large separation. The swept areas are equal even though the travelled distances are different. In particular, the sectors have curved ellipse boundaries: triangles formed by joining their endpoints only approximate the swept areas for short intervals.

Let $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ be the relative specific angular momentum. The areal velocity isA central force exerts zero torque about its centre, so conservation of angular momentum makes this rate constant. This explanation also applies to other central forces; the ellipse shape specifically comes from Newtonian gravity.

In the $a=1$, $e=0.6$, $\mu =1$ example, specific angular momentum is $h=\sqrt{\mu a(1-e^2)}=\sqrt{0.64}=0.8$, giving areal velocity $0.4$. At periapsis and apoapsis, the velocity is perpendicular to the relative displacement, so $h=rv$. The endpoint speeds are consequentlyThe fourfold speed difference is consistent with the fourfold separation difference. Away from these endpoints, $h/r$ gives only the transverse component of velocity; the full speed follows from the vis-viva equation.

To actually place equal-time markers, use Kepler's equation. Write mean anomaly as $\mathcal{M}=nt$, with $t=0$ at periapsis and mean motion $n=\sqrt{\mu /a^3}=1$. The different symbol $\mathcal{M}$ avoids confusing this angle with the mass $M$. Solve for eccentric anomaly $E$ in radians, then compute the relative position:This timing procedure is explained in NASA educational archive, How Orbital Motion is Calculated.

For the first interval $0\le t\le \pi /6$, the final eccentric anomaly is $E\approx 1.041495$. For the equal interval $\pi \le t\le 7\pi /6$ starting at apoapsis, it ends at $E\approx 3.471065$. Both swept areas are $\Delta A=(h/2)\Delta t=0.4\pi /6=\pi /15\approx 0.209440$, one twelfth of the entire ellipse's area $0.8\pi$. The diagram computes its markers from Kepler's equation, making visible how uniform time steps become uneven spatial steps.

For particles with the same charge-to-mass ratio, rotationally invariant mutual central forces, and a weak uniform magnetic field, a frame with the displayed angular velocity cancels the velocity-dependent magnetic Lorentz force against the Coriolis force. The remaining rotational and magnetic terms are quadratic in $\mathbf{B}$. Thus, to first order in the field, the rotating-frame equations coincide with the zero-field equations. This is the classical dynamical basis of Larmor precession; equality is an approximation rather than an exact removal of the magnetic field.

Isaac Newton made the greater contribution if the criterion is the scope and explanatory integration of mathematical natural philosophy; Galileo Galilei has the stronger claim if the criterion is the formative transformation of how terrestrial motion could be investigated. The first criterion gives a defensible overall verdict for Isaac Newton, but a serious comparison must show what each changed rather than award credit according to who worked later.

Galileo Galilei's contribution was not simply to attach numbers to familiar events. In the Third Day of Two New Sciences by Galileo Galilei, he defines uniformly accelerated motion by equal additions of speed in equal intervals of time, then derives consequences. Starting from rest, the modern notation is $v=at$; the mean speed over the interval is $at/2$, soHence distances in successive equal intervals have the ratios $1:3:5:7:\dots$. The novelty lies in making a precisely specified kind of motion the subject of mathematical demonstration, with observed falls assessed against its consequences. This is not the same as deriving the underlying physical cause of gravity.^[ref]

The reported apparatus shows how such an idealization could be investigated. A smooth bronze ball descended in a straight grooved wooden channel, raised at one end. Water discharged through a small opening into a vessel during the descent; its collected mass served as a measure of elapsed time. Repeating runs over different portions of the channel compared the predicted squared dependence of time on distance. The inclined plane slowed the event, while repetition and weighing made a temporal comparison possible without a modern stopwatch. The published report is evidence of Galileo Galilei's experimental presentation, not a surviving raw-data archive from which every reported precision can now be reconstructed.^[ref]

A necessary physical qualification is often missed. A ball in rolling without slipping is not a point particle sliding without friction. For a solid sphere on an ideal inclined plane, modern rigid body dynamics gives $a=5g\sin \alpha /7$, rather than $g\sin \alpha$. Both are constant accelerations, so both give $s\propto t^2$. The trial can therefore test the temporal form without directly establishing the free-fall coefficient. This distinction makes the experiment more intelligible: the evidence supports a specified relation under approximations, rather than every detail of a modern idealization.

The Fourth Day extends this approach to projectile motion. Combine uniform horizontal velocity $u$ with constant downward acceleration $g$, taking downward displacement positive. Then $x=ut$ and $y=g{t}^2/2$. Eliminating time givesa parabola. Actual projectiles experience air resistance, but the ideal curve organizes the effects that need separate treatment. Galileo Galilei thus gave a powerful example of how mathematical abstraction and physical observation could cooperate.^[ref]

Isaac Newton went further by connecting kinds of motion through a general account of force. Newton's second law links the change of momentum to the impressed force; for constant mass, its modern differential expression is $\mathbf{F}=m\mathbf{a}$. Newton's third law makes interaction reciprocal, enabling a consistent analysis of interacting bodies. The Philosophiae Naturalis Principia Mathematica by Isaac Newton develops consequences of central forces and then asks which laws fit astronomical phenomena. Mathematical deduction now connects measurable orbits to an underlying force law, rather than just describing each trajectory separately.^[ref]

A simple modern reconstruction shows the increase in explanatory reach. For a circular orbit of radius $r$ and orbital period $T$ about a much more massive central body, centripetal acceleration is $4{\pi }^{2}r/T^2$. Kepler's third law, $T^2\propto r^3$, therefore implies an acceleration proportional to $r^{-2}$. This circular argument is illustrative; Isaac Newton's treatment of elliptical orbits is more general. With Newton's law of universal gravitation,the two-body orbital period is tied to the semimajor axis $a$ and both masses. The same theory connects the Moon's orbital acceleration with terrestrial gravity, and supports further investigations of tides, comets and departures from a simple two-body orbit.^[ref]

This achievement depended on more than two individuals. Johannes Kepler's relations supplied crucial astronomical constraints, and both authors inherited geometry, observational records and technical practices. Nor should the Philosophiae Naturalis Principia Mathematica by Isaac Newton be described simply as the first application of symbolic calculus: its published arguments are largely geometrical. What distinguishes Isaac Newton is the demonstrative network uniting terrestrial and celestial problems, together with explicit idealizations whose failures generated further calculable questions.

On breadth, quantitative unification and the inference from motion to force, Isaac Newton's contribution was greater. Galileo Galilei's work supplied an exemplary way to mathematize terrestrial motion and test its idealized relations. Giving the former the overall verdict should therefore acknowledge the latter's distinct foundational achievement, not portray the Philosophiae Naturalis Principia Mathematica by Isaac Newton as a creation without predecessors.

For an attractive central force, the transverse acceleration vanishes. In polar coordinates,Thus the specific angular momentum $h=r^2\dot{\theta }$ is constant. Assume $h\ne 0$, so angle can parameterize the orbit, and put $u=1/r$. Primes denote angle derivatives. The chain rule givesThe outward radial component of the attractive force is $-f(u)$. Its Newton's second law equation is therefore $m(\ddot{r}-r{\dot{\theta }}^2)=-f(u)$, giving the Binet equationThe sign reflects $f$ being the positive inward force magnitude. Zero angular momentum produces radial motion and is outside this angular parametrization.

A central force has zero torque, so $h=r^2\dot{\theta }$ is constant and the angular momentum is $mh$. The areal velocity is $h/2$. Put $u=1/r$ and let primes denote angular derivatives. Then $\dot{r}=-h{u}^{\prime }$, $\ddot{r}=-h^2{u}^2{u}^{\prime \prime }$, and $r{\dot{\theta }}^2=h^2{u}^3$. The radial equation with inward force magnitude $f(u)$ is $m(\ddot{r}-r{\dot{\theta }}^2)=-f(u)$, giving the Binet equationChoose the angular origin along a diameter of a radius-$R$ circle through $O$. Its polar equation is $r=2R\cos \theta$, so $u=(2R{)}^{-1}\sec \theta$ on $-\pi /2<\theta <\pi /2$. Since $(\sec \theta {)}^{\prime \prime }+\sec \theta =2{\sec }^3\theta$, substitution gives $u^{\prime \prime }+u=8R^2{u}^3$ and henceThe force is attractive. For fixed force coefficient $c$ and mass, $R|mh|=\sqrt{mc/8}$, so $R$ is inversely proportional to the magnitude of the angular momentum. The time along the circle isThis circular orbit through an inverse-fifth-power singularity reaches the force center at the two endpoints of this angular interval. The integral is finite; continuation through the singular collision is not specified by the force law alone.

Apply the rotating-vector derivative rule first to the position vector and then to its velocity. Since $\mathbf{\omega }$ is constant,For example, the derivative of ${\mathbf{v}}^{\prime }$ contributes ${\mathbf{a}}^{\prime }+\mathbf{\omega }\times {\mathbf{v}}^{\prime }$, while that of $\mathbf{\omega }\times \mathbf{x}$ contributes the other two terms. This is the equation of motion in a rotating frame. The mutual central forces are unchanged in form because rotations preserve lengths and differences of position vectors. Newton's second law for particle $i$ becomesSet $\mathbf{\omega }=-e\mathbf{B}/(2m)$. The velocity-dependent Lorentz force is $e{\mathbf{v}}_i^{\prime }\times \mathbf{B}=-e\mathbf{B}\times {\mathbf{v}}_i^{\prime }$, and it cancels the Coriolis force $-2m\mathbf{\omega }\times {\mathbf{v}}_i^{\prime }=e\mathbf{B}\times {\mathbf{v}}_i^{\prime }$. The remaining terms combine to $m\mathbf{\omega }\times (\mathbf{\omega }\times {\mathbf{x}}_i)$ and are quadratic in the magnetic field. Neglecting them givesidentical to the zero-field equations to first order. The Larmor theorem requires the common charge-to-mass ratio; a mixture with different ratios cannot use one rotating frame to cancel every magnetic term.

The unit vectors of plane polar coordinates obey ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$, as follows by differentiating their Cartesian sine/cosine components. Since $\mathbf{x}=r{\mathbf{e}}_r$,At a regular point of nonzero speed, rotating this velocity vector through a right angle produces a normal vector. Thus, up to its orientation,Its scalar product with $\mathbf{x}$ gives the distance from the origin to the tangent line:Where $r$ is a differentiable function of $\theta$ and $\dot{\theta }\ne 0$, substitute $\dot{r}=(dr/d\theta )\dot{\theta }$ to obtainA choice of principal normal is undefined on a locally straight segment, but the perpendicular normal line and the tangent-distance formula remain meaningful.

For a central force, the torque vanishes. Use physical signed angular momentum $h=m{r}^2\dot{\theta }$, with $h\ne 0$. The preceding geometric formula implies $|h|=mpv$, where $v=|\dot{\mathbf{x}}|$. Conservation of energy gives $E=m{v}^2/2+V(r)$. Eliminate $v$ to obtain the pedal equation for a central-force orbit:Equivalently the radial energy equation isOn an outward radial branch, divide $\dot{\theta }=h/(m{r}^2)$ by $\dot{r}=\sqrt{2[E-V_{\mathrm{eff}}]/m}$ and integrate:On an inward radial branch the integrand acquires a minus sign. At a turning point the two branches must be joined; the quoted positive-root integral is a local branch representation, not a single global time-parametrized orbit.

For $V=c/r^2$ with $c>0$, defineA finite moving orbit has $E>0$. In the radial integral, set $s=r_{\min }/r$. Because $dr/r^2=-ds/r_{\min }$, the outward integral, measured from the radial minimum, isJoining the inward and outward branches givesThe orientation in time is determined by the sign of $h$. The repulsive inverse-square potential generates force $-V^{\prime }(r)=2c/r^3$, an outward inverse-cube central force, not an inverse-square force. The orbit approaches infinity along two asymptotic directions, turns once at $r_{\min }$ and escapes again. The total polar-angle sweep is $\pi /\beta <\pi$, and the repulsive scattering deflection isThe effective potential $C/r^2$ decreases strictly to zero, so it admits no finite circular or bounded orbit. The sketch shows this scattering in a repulsive inverse-square potential. In the radial case $h=0$, the formulas dividing by $h$ are inapplicable; the direction is fixed, the turning radius is $\sqrt{c/E}$ and the particle reverses on the same ray.

Let $\mu :P\to \mathbb{R}$ be the moment map for the circle group, assume $c$ is a regular value, and assume the circle action on $N={\mu }^{-1}(c)$ is free. The action is proper because the circle is compact. Then $N$ has dimension $2n-1$ and $P_c=N/SO(2)$ is a smooth manifold of dimension $2n-2$.

If ${\xi }_P$ generates the action, $\omega ({\xi }_P,v)=d\mu (v)=0$ for every $v\in TN$. In fact $(TN{)}^{\omega }$ is exactly the line spanned by ${\xi }_P$, since $TN=\ker d\mu$ is a hyperplane and $\omega$ is nondegenerate. Therefore the kernel of the restricted two-form is precisely the orbit direction. The restriction is invariant and horizontal, so it descends to a unique nondegenerate closed two-form ${\omega }_c$ withThis proves the Marsden-Weinstein theorem in the circle case. If the Hamiltonian $H$ is invariant, then $\{\mu ,H\}=0$: its Hamiltonian flow stays on $N$, projects to the quotient and is generated there by the descended function $H_c$. This is a reduced Hamiltonian system in $2n-2$ local variables. At a critical level or a nonfree orbit the quotient can be singular, so the smooth dimension assertion needs the stated hypotheses.

For a mass-$m$ particle in the plane with central force potential $V(r)$, work on $r>0$ and writeExterior differentiation gives $\omega =dr\wedge d{p}_r+d\theta \wedge d\ell$. Rotation translates $\theta$ and has moment map $\ell$. Fix $\ell ={\ell }_0$ and quotient out $\theta$ to obtain the planar rotational symplectic reductionThe reduced equations are $\dot{r}=p_r/m$ and ${\dot{p}}_r={\ell }_0^2/(m{r}^3)-V^{\prime }(r)$. The centrifugal term arises from the conserved angular momentum; it is not an extra force imposed by hand. Reconstruct the angle from $\dot{\theta }={\ell }_0/(m{r}^2)$. For ${\ell }_0\ne 0$ the origin cannot lie on the level; for ${\ell }_0=0$ the same local reduction holds away from the origin, while the full phase-space origin is a singular fixed orbit.

Write $k=GM$ and $r=|\mathbf{r}|$. The gravitational central force has zero torque, so the specific angular momentumis constant. For a nonradial orbit $h=|\mathbf{h}|>0$, motion lies in the plane normal to $\mathbf{h}$, and plane polar coordinates give $r^2\dot{\theta }=h$. With $u=1/r$ and primes denoting $\theta$ derivatives,The radial equation $\ddot{r}-r{\dot{\theta }}^2=-k/r^2$ is therefore the Binet equationIts general solution, after choosing the polar origin of angle, isThe origin is a focus of this conic section. To identify which conic section occurs, use the conserved specific energyA bound, nonradial orbit has $\epsilon <0$, hence $0\le e<1$ and is an ellipse. For example, putting $x=r\cos \theta$, $y=r\sin \theta$ in $r+ex=p$ givesan explicit ellipse equation. This is the intended low, bound Kepler orbit. The inverse-square equation alone does not imply an ellipse: an initially tangential speed larger than $\sqrt{2GM/r}$ gives $\epsilon >0$, $e>1$ and a hyperbolic Kepler orbit. Zero angular momentum instead gives radial motion. Those qualifications are necessary for the elliptic conclusion.

In a circular orbit, centripetal acceleration balances the gravitational acceleration:Here $\Omega$ is the positive angular speed; either sense of circulation has this period.

With the stated drag force, use the full angular momentum $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$. Gravitational torque is still zero, while the drag torque isTherefore $\boxed{\mathbf{L}(t)=\mathbf{L}(0)e^{-At/m}}$ for constant $A>0$. Its direction remains fixed, and its magnitude has exponential decay with time constant $m/A$; the orbit remains planar although its shape is no longer a fixed ellipse.

Outside a spherical planet, Newtonian mechanics gives the inverse-square central force equationThe satellite's mass cancels. The angular momentum is conserved, so the orbit lies in a plane. In that plane the general equations in polar coordinates areA synchronous circular orbit fixed above a surface point must be equatorial, prograde and have angular velocity $\omega =2\pi /T$. Substitution of constant $r$ into the radial equation givesThe radius is measured from the planet's centre; an external orbit requires it to exceed the planet's radius.

Now let the release speed be $v=(1+\eta )v_s$, with $0<\eta \ll 1$, in the same tangential direction. At release $\dot{r}=0$, butThus the satellite initially moves outward. Its specific total energy is $v^2/2-GM/r_s<0$ for this small increase, so the Kepler orbit is an ellipse, with the release point its pericentre. To make this explicit, put $b=v^2/v_s^2\in (1,2)$. The radial turning-point equation from energy and angular momentum conservation has rootsIts semi-major axis is $a=r_s/(2-b)>r_s$. The speed follows from the same conserved energy:It is greatest at pericentre and least at apocentre. In particular, the tangential speed at the two turning points satisfies $r_{\min }{v}_{\max }=r_{\max }{v}_{\min }$. The orbital period becomes $2\pi \sqrt{a^3/(GM)}>T$, so the orbit is no longer fixed over the same surface point.

A central force acts along the line joining the particle to a fixed centre. Taking that centre as the origin, its usual position-dependent form isPositive $f$ gives an outward force and negative $f$ an inward one. For a conventional central force the magnitude depends only on distance from the centre, not on direction. The angular-momentum conclusions below require only that the force remain radial about that fixed centre.

The particle's angular momentum about the force centre is $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$. A central force exerts zero torque about that centre, soBy conservation of angular momentum, $\mathbf{L}$ is a constant vector. If it is nonzero, the identity $\mathbf{r}\cdot \mathbf{L}=0$ places every point of the orbit in the fixed plane through the centre perpendicular to $\mathbf{L}$:If $\mathbf{L}=0$, position and velocity are parallel. Away from the force centre, ${\dot{\mathbf{e}}}_r=(\dot{\mathbf{r}}-\dot{r}{\mathbf{e}}_r)/r=0$, so the motion is radial along a fixed line, which lies in any plane containing that line. Thus the zero-angular-momentum case also gives planar motion. As usual, these assertions apply while the force law supplies a regular orbit; continuation through a singular force centre is a separate issue.