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Work in an inertial frame, take all masses to be constant, and assume the internal forces obey the pairwise relation from Newton's third law $F_{ij}=-F_{ji}$. For the angular momentum result, also assume these pair forces are central, so $F_{ij}$ is parallel to $r_i-r_j$ and its pair torque vanishes. The equations from Newton's second law areSumming them and pairing the internal forces gives

About a fixed point $a$, define the total angular momentum and external torque byDifferentiation produces no velocity term because ${\dot{r}}_i\times m_i{\dot{r}}_i=0$. Grouping internal terms into pairs givesEach final cross product is zero by centrality, proving $\boxed{d{L}_a/dt=G_a}$. Equal and opposite internal forces give momentum conservation when the total external force is zero, but do not in general make their internal torques cancel; the central-force assumption supplies that extra step.

Let $M={\sum }_i{m}_i$ and let the centre of mass be $R=M^{-1}{\sum }_i{m}_i{r}_i$, so $P=M\dot{R}$. For an arbitrary moving reference point $a(t)$, the same differentiation givesTaking $a=R$, the extra term is $-\dot{R}\times M\dot{R}=0$. Therefore the corresponding angular momentum about the centre of mass satisfiesIts definition using velocities relative to $\dot{R}$ agrees with the one above, since ${\sum }_i{m}_i(r_i-R)=0$ makes ${\sum }_i(r_i-R)\times m_i\dot{R}=0$. Thus this result remains true even when the centre of mass accelerates.

Finally take the common mass to be $m$ and $F_i=-k{\dot{r}}_i$, with constant $k$. About either a fixed point or the centre of mass,The angular-momentum equation becomes $\dot{L}=-(k/m)L$. Multiplying by the integrating factor $e^{kt/m}$ gives $\frac{d}{dt}(e^{kt/m}L)=0$, hence the angular momentum decay under uniform linear drag isThe common mass and common drag coefficient ensure that the same decay factor applies to every term.

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.

The center of mass isBy Newton's third law, ${\mathbf{F}}_{21}=-{\mathbf{F}}_{12}$, soThus $\dot{\mathbf{R}}$ is constant. For $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$,Thereforewhere $\mu$ is the reduced mass.

For a zero-natural-length spring, extension is the full distance between its endpoints, so the squared spring lengths are quadratic functions of the Cartesian coordinates. The Lagrangian isThe gravitational potential energy has positive derivative in the upward coordinate, giving downward forces. Applying the Euler-Lagrange equations to all four coordinates givesThe two spring forces are equal and opposite on the two masses for the connecting spring, as required by Newton's third law.

There is no unconditional logical dependence of Newton's first law on Newton's third law. There is, however, a dependence if one applies the first to an extended, isolated composite and interprets its motion as the motion of its centre of mass. The difference is between the behaviour of an elementary body with zero total force and a theorem about cancellation of internal forces. Isaac Newton's own first-law discussion includes projectiles and spinning bodies, so the composite-body issue is physically significant.^[ref]

For a constant positive mass in an inertial frame, Newton's second law gives $m\ddot{\mathbf{x}}=\mathbf{F}$. Setting the total force to zero yields uniform rectilinear motion; it uses no action–reaction assumption. The first law additionally identifies a common class of inertial frames in which such a statement applies. One cannot establish that class by arbitrarily straightening a single trajectory, and one cannot test the claim by defining every observed acceleration to be an unspecified force.

A concrete countermodel separates these issues. Suppose two bodies obey the second-law equations with internal forces ${\mathbf{F}}_{12}=\mathbf{f}$ and ${\mathbf{F}}_{21}=-2\mathbf{f}$, with nonzero constant $\mathbf{f}$. Every elementary body with zero total force still has constant velocity, but the two-body momentum satisfiesThus the elementary-body first law is compatible with failure of the third. Calling the pair isolated does not make its internal total force zero: that is precisely what has not been assumed. This countermodel establishes logical independence, not a proposal for a successful fundamental interaction.

Now strengthen the first-law claim: every isolated composite has an inertially moving centre of mass whenever the total external force vanishes. With $M={\sum }_i{m}_i$, Newton's second law givesFor an isolated pair, the strengthened claim requires ${\mathbf{F}}_{12}+{\mathbf{F}}_{21}=0$. This is the weak, equal-and-opposite form of Newton's third law. It does not by itself require collinearity with the bodies' separation; the central-force condition is stronger and bears on conservation of angular momentum. For many bodies, cancellation of the sum does not alone imply pairwise cancellation without assumptions about the decomposition into interactions.

The right conclusion is therefore conditional. A first law for individual force-free bodies does not presuppose the third law. A first law for all isolated material composites, combined with the usual centre of mass and Newton's second law, imports an internal-momentum-balance condition. In theories with dynamical fields, material bodies can exchange momentum with those fields; conservation of momentum then concerns the combined system and need not imply instantaneous pairwise action–reaction between particles.

The familiar result follows if the force laws have Galilean invariance; the three abstract Newton's laws of motion do not guarantee that invariance. Isaac Newton's Corollary V correctly identifies a central feature of ordinary Newtonian mechanics, but its general physical scope requires assumptions about interactions as well as about the relation between force and acceleration.^[ref]

Take a constant velocity $\mathbf{U}$ and a Galilean transformationThen ${\ddot{\mathbf{x}}}_i^{\prime }={\ddot{\mathbf{x}}}_i$, ${\mathbf{x}}_i^{\prime }-{\mathbf{x}}_j^{\prime }={\mathbf{x}}_i-{\mathbf{x}}_j$, and ${\dot{\mathbf{x}}}_i^{\prime }-{\dot{\mathbf{x}}}_j^{\prime }={\dot{\mathbf{x}}}_i-{\dot{\mathbf{x}}}_j$. Any interaction depending only on these relative quantities, with unchanged parameters, supplies the same forces for the transformed initial data. Newton's second law consequently gives the same relative accelerations; uniqueness of the resulting initial value problem supplies the same relative histories. An inverse-square central interaction is a standard example. Newton's third law helps establish the common centre of mass result, but the boost argument principally needs invariant interactions and the second law.

Here is a counterexample to deduction from the three laws alone. Give two equal masses a preferred inertial frame and write $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $\mathbf{V}=({\dot{\mathbf{x}}}_1+{\dot{\mathbf{x}}}_2)/2$. For fixed $k>0$ and speed scale $u_{\ast }>0$, specifyThe forces are equal, opposite and central, the centre of mass is inertial, and each body obeys $m{\ddot{\mathbf{x}}}_i={\mathbf{F}}_i$. An elementary body with no force is inertial. Nevertheless,Boosting an isolated pair from $\mathbf{V}=0$ to $\mathbf{V}=\mathbf{U}$ changes its internal oscillation frequency. All three motion laws hold, but the proposed universal boost equivalence fails. The extra preferred structure in the force law is what makes the failure possible.

A passive coordinate transformation of this same model remains possible: transform the preferred frame's description too. That mathematical freedom does not establish an active dynamical symmetry which changes the physical pair's velocity relative to the preferred structure while holding the laws' background data fixed. Confusing those operations would hide the counterexample.

Finally, a change to rotating coordinates introduces centrifugal forces and Coriolis forces; their dependence on position or velocity can change internal experiments. Constant translational velocity has no analogous effect when the interactions have Galilean invariance. Corollary V expresses that physical symmetry, with the interaction assumption made explicit, rather than a consequence of action–reaction alone.

There are two different issues: whether a law follows logically from the other laws, and whether an experiment can test it without assuming anything else. Newton's laws of motion have empirical content when combined with rules for identifying inertial frames, measuring mass and specifying forces. An experiment tests predictions of that combination. Calling every observed acceleration an otherwise unspecified force would remove the empirical constraint rather than verify Newton's second law. The distinction is central to the empirical testing of Newton's laws of motion.

Newton's first law contributes the claim that a common family of inertial frames exists: bodies with zero net external force have uniform straight motion in those frames. It is not enough to choose coordinates which straighten the trajectory of just one body; an admissible frame must accommodate many independent bodies and interactions. Once that frame and the ordinary constant-positive-mass form of Newton's second law are assumed,This conditional deduction does not establish the existence of the required frame, or independently identify the physical circumstances in which the net force vanishes. Defining an inertial frame and then counting a straight trajectory as proof of the definition would not be a test.

Nevertheless, the law of inertia has testable implications in a network of assumptions. Consider low-friction motion along an air track: determine positions at measured times, reduce the known residual influences, and ask whether the remaining acceleration approaches zero in the same laboratory frame for different bodies. Rotating the frame changes the description through Coriolis forces and centrifugal forces; this dependence must be accounted for. No finite experiment establishes that every unmeasured influence is absent, but varying the apparatus can discriminate an interaction from a frame effect. A failed prediction is evidence against some part of the package, not automatically the isolated verbal statement of Newton's first law.

Newton's second law is more than a definition if the force law and mass assignments constrain predictions before the new acceleration is measured. For instance, independently characterize a spring's reproducible extension response, specify the same interaction in several configurations, and compare the resulting accelerations for different bodies. If its strength and every mass are freely refitted to each outcome, agreement is empty. If they are fixed by an independently checked model and calibration, the law predicts held-out trajectories. Such a test remains conditional on the spring model, initial conditions and frame identification; this is an instance of the Duhem problem, not a reason to deny all empirical content.

Newton's third law supplies a different constraint. For two bodies exchanging forces, Newton's second law giveswhen there is no net external force. Equal-and-opposite interaction forces imply conservation of momentum for the pair. The two second-law equations alone do not require that cancellation: assigning ${\mathbf{F}}_{12}=\mathbf{f}$ and ${\mathbf{F}}_{21}=-2\mathbf{f}$ is algebraically consistent with both second-law equations while violating the third law. This is a logical countermodel, not a proposed isolated physical interaction. Measuring paired impulses or checking the pair's total momentum therefore tests an additional constraint, conditional on the assumed absence of momentum exchange with anything else.

The system boundary matters. In electromagnetism, matter can exchange momentum with the electromagnetic field. Instantaneous mechanical forces between two particles need not satisfy the simple pairwise statement, although total matter-plus-field momentum is conserved under the appropriate closed-system conditions. Likewise, Noether theorem can derive conservation of momentum from a translationally invariant action; that uses an extra symmetry assumption and does not derive the third law from the second alone.

Thus the first law is not an additional differential consequence once the full second-law framework is fixed, and the third law is not a consequence of that framework. All three can contribute to empirical tests, but none is tested in total isolation from frame, interaction and measurement assumptions. This preserves the difference between logical dependence and experimental testability.^[ref]

Newton's second law applies separately to each body's total force: zero force on $q_1$ means zero acceleration of $q_1$, while a nonzero force on a free $q_2$ gives nonzero acceleration of $q_2$. It does not assert that those two forces must be equal and opposite.

The conducting shell is a third interacting body. Its induced surface charges shield the cavity from the external charge, while charges on its outer surface can still exert a force on $q_2$. Reaction forces act on the shell and, if it is held fixed, on its supports. In the electrostatic limit the force balance of the complete isolated charge–shell system is ${\mathbf{F}}_1+{\mathbf{F}}_2+{\mathbf{F}}_{\mathrm{shell}}=0$. Thus the given situation is consistent with both Newton's second law and Newton's third law once the shell is included.

Shielding removes the cavity field due to external charges; it does not generally remove the force of the inner surface charges induced by an off-centre $q_1$. Zero total force on $q_1$ is an additional condition, satisfied for example when it is at the centre of a spherical cavity.

For a fixed set of bodies in an inertial frame, total linear momentum is constant if the resultant external force vanishes. Summing Newton's second law cancels internal force pairs by Newton's third law. For an open subsystem, momentum carried across its boundary must also be included.

For an equal-intake-fuel mass-loading rocket, enclose the original craft and the stationary material collected during $dt$. Its initial momentum is $mu$. At the end, the craft has mass $m-\mu dt$ and the exhaust has mass $2\mu dt$. To first order,The first term is rocket thrust; the second is the momentum cost of collecting stationary matter. It vanishes in the conventional rocket equation, which therefore has no analogous fixed exhaust-speed ceiling in Newtonian mechanics. Here the acceleration vanishes at $u_{\max }=2u_0$; from rest it stays positive below that value while fuel remains. Relativistic limits are outside this model.

For constant $\mu$, mass conservation gives $m=m_i-\mu t$. Separating variables,ThereforeThe formal limiting speed would require $t=m_i/\mu$ and zero remaining craft mass. A positive rocket dry mass and finite carried fuel terminate the burn earlier. Maintaining nonzero collection rate even at rest is an idealization explicitly prescribed by the problem.

For a material volume moving with the fluid, the Reynolds transport theorem gives, componentwise,For a scalar $q$ this is also ${\int }_{V(t)}(Dq/Dt+q\nabla \cdot \mathbf{u})dV$. Apply the component formula to momentum density $q=\rho u_i$. Constant density and incompressibility make the integrand $\rho D{u}_i/Dt$.

The traction $\mathbf{t}(\mathbf{n})$ is the force per unit area exerted by the material outside a surface on the material inside, with outward unit normal $\mathbf{n}$. Reversing the normal exchanges the two sides of the same internal surface. Thus $\mathbf{t}(-\mathbf{n})=-\mathbf{t}(\mathbf{n})$ expresses equal and opposite internal contact forces, the local form of Newton's third law.

With no body force, Newton's second law equates the rate of change of the material volume's total momentum to its surface force. Combining this balance with the transport theorem givesThis is a vector identity, since the component argument applies to each direction.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal dry-contact model, the actual static friction adjusts to the equations of motion, subject to a bound, rather than always taking its maximum value.^[ref]Here $N$ is the magnitude of the normal force and ${\mu }_s$ the coefficient of static friction. Equality means limiting static friction; it is a threshold, not an instruction to replace every unknown contact force by ${\mu }_{s}N$.

For a stipulated example, place a nonrotating $2\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take gravitational acceleration $g=9.8m/{\mathrm{s}}^2$ and coefficient of static friction ${\mu }_s=0.40$, with no other vertical force. The vertical force balance gives $N=mg=19.6\mathrm{N}$, so the largest allowed static friction magnitude is $7.84\mathrm{N}$. Choose right as positive and apply a horizontal force $F$. Remaining at rest requires Newton's second law to giveAt $F=2\mathrm{N}$, the static friction is $-2\mathrm{N}$; at $F=6\mathrm{N}$, it is $-6\mathrm{N}$. Both states have zero acceleration. At $F=10\mathrm{N}$, balancing would require $f_s=-10\mathrm{N}$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent acceleration requires a kinetic friction law; setting $f_s=-7.84\mathrm{N}$ after contact slip begins would still use the wrong contact state. At exactly $F=7.84\mathrm{N}$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.

The toy model used in kinetic friction makes the contact-state distinction explicit: $m=1\mathrm{kg}$, $N=10\mathrm{N}$, ${\mu }_s=0.5$ and coefficient of kinetic friction ${\mu }_k=0.3$. On stationary ground, an initially resting box pushed right with $4\mathrm{N}$ has $f_s=-4\mathrm{N}$. At $6\mathrm{N}$ it cannot remain at rest; once it slides right, $f_k=-3\mathrm{N}$ and $a=3m/{\mathrm{s}}^2$. At $5\mathrm{N}$, rest is allowed with $f_s=-5\mathrm{N}$, whereas an already right-sliding box has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. The applied force alone does not specify the friction branch: contact slip and motion history matter.

For a flat contact with two tangential directions, the bound applies to a vector norm, not separately to each component:For the stipulated $7.84\mathrm{N}$ bound, balancing horizontal components $4\mathrm{N}$ and $4\mathrm{N}$ requires a force magnitude $\sqrt{32}\mathrm{N}\simeq 5.657\mathrm{N}$, which is allowed. Components $6\mathrm{N}$ and $6\mathrm{N}$ require $\sqrt{72}\mathrm{N}\simeq 8.485\mathrm{N}$, which is forbidden even though each component separately is below $7.84\mathrm{N}$. This circular bound assumes the same coefficient of static friction in every tangential direction. Near impending contact slip, the limiting static friction opposes the relative contact motion that would otherwise begin. Once actual contact slip exists, kinetic friction instead opposes its tangential relative velocity, which need not have the direction of the body's center of mass motion.

The normal force must also come from the equations of motion. On horizontal ground, an upward pull of magnitude $P$ at angle $\alpha$ above horizontal gives, for a block remaining at rest with no other forces,The sine component unloads the contact and the cosine component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative normal force. For the stipulated $2\mathrm{kg}$ block and $\alpha =30^{\circ }$, the positive threshold is $P={\mu }_{s}mg/(\cos \alpha +{\mu }_s\sin \alpha )\simeq 7.3544\mathrm{N}$, with $N\simeq 15.9228\mathrm{N}$ and $|f_s|\simeq 6.3691\mathrm{N}$. A downward pull increases $N$ instead. An inclined plane gives another reason why $N$ is not always $mg$.

To solve a new problem, impose zero contact slip, find the required static friction and normal force from the equations of motion, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing rolling without slipping on stationary ground, a forward static friction can accelerate the vehicle. Its role is to prevent contact slip, rather than to oppose all motion of the whole body.

Nor does the word “static” by itself make friction do zero work. The contact power delivered to a body is $P_s={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, a dot product with its material contact-point velocity in the chosen inertial frame. At an ideal contact undergoing rolling without slipping on stationary ground, ${\mathbf{v}}_c=0$, so this power is zero. For a rigid body, adding the translational and rotational contributions gives ${\mathbf{f}}_s\cdot {\mathbf{V}}_{\mathrm{CM}}+{\mathbf{\tau }}_{\mathrm{CM}}\cdot \mathbf{\omega }={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, where ${\mathbf{\tau }}_{\mathrm{CM}}$ is the contact torque about the center of mass and $\mathbf{\omega }$ its angular velocity. Counting only the first term omits the rotational contribution.

At a moving contact with zero contact slip, the two material contact points share a velocity ${\mathbf{v}}_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact power $f_{s}U$ when the belt's rightward velocity $U$ and rightward static friction $f_s$ are positive. The belt receives $-f_{s}U$ by Newton's third law. These contact contributions sum to zero in the ideal no-slip model, while the block's kinetic energy can increase. Zero contact slip and zero work on each body are different conditions.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.