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Now let's see where the first law fails. Finally, imagine a universe with three bodies $1$, $2$, and $3$ of equal finite mass m. Let $d$ and $C$ be positive non-zero constants with the appropriate units. Suppose the universe has a potential energy function $$V(x_1,x_2)=-C\left(\frac{x_1-x_3-d}{2}\right)^{4/3}\;,$$ so $1$ and $3$ exert equal and opposite forces on each other $$-F_{31}=\frac{\mathrm dV}{\mathrm dx_1}=-\frac{\mathrm dV}{\mathrm dx_3}=F_{13}\;.$$ Suppose $x_1(0)=d$ and $x_2(0)=d/2$ and $x_3(0)=0$. Further suppose that $v_1(0)=v_2(0)=v_3(0)=0$. Obviously we can satisfy all three of Newton's laws by taking as solutions $x_i(t)=x_i(0)$, however, instead suppose the particles move as $$x_1(t)=d+(Kt)^3,~~ x_2(t)=d/2$$ and $$x_3(t)=-(Kt)^3$$ for $$K= \sqrt{\frac{2C}{9m}}.$$ Then third law holds because $F_{ij}=-F_{ji}$ and the second holds because $ma_2(0)=0=F_2$ and

Now let's see where the first law fails. Finally, imagine a universe with three bodies $1$, $2$, and $3$ of equal finite mass m. Let $d$ and $C$ be positive non-zero constants with the appropriate units. Suppose the universe has a potential energy function $$V(x_1,x_2,x_3)=-C\left(\frac{x_1-x_3-d}{2}\right)^{4/3}\;,$$ so $1$ and $3$ exert equal and opposite forces on each other $$-F_{31}=\frac{\partial V}{\partial x_1}=-\frac{\partial V}{\partial x_3}=F_{13}\;.$$ Suppose $x_1(0)=d$ and $x_2(0)=d/2$ and $x_3(0)=0$. Further suppose that $v_1(0)=v_2(0)=v_3(0)=0$. Obviously we can satisfy all three of Newton's laws by taking as solutions $x_i(t)=x_i(0)$, however, instead suppose the particles move as $$x_1(t)=d+(Kt)^3,~~ x_2(t)=d/2$$ and $$x_3(t)=-(Kt)^3$$ for $$K= \sqrt{\frac{2C}{9m}}.$$ Then the third law holds because $F_{ij}=-F_{ji}$ and the second holds because $ma_2(0)=0=F_2$ and

Now let's see where the first law fails. Finally, imagine a universe with three bodies $1$, $2$, and $3$ of equal finite mass m. Let $d$ and $C$ be positive nonzero constants with the appropriate units. Suppose the universe has a potential energy function $V(x_1,x_2)=-C(\frac{x_1-x_3-d}{2})^{4/3}$, so $1$ and $3$ exert equal and opposite forces on each other $-F_{31}=\frac{dV}{dx_1}=-\frac{dV}{dx_3}=F_{13}$. Suppose $x_1(0)=d$ and $x_2(0)=d/2$ and $x_3(0)=0$. Further suppose that $v_1(0)=v_2(0)=v_3(0)=0$. Obviously we can satisfy all three of Newton's laws by taking as solutions $x_i(t)=x_i(0)$, however, instead suppose the particles move as $x_1(t)=d+(Kt)^3$, $x_2(t)=d/2$ and $x_3(t)=-(Kt)^3$, for $K= \sqrt{\frac{2C}{9m}}$. Then third law holds because $F_{ij}=-F_{ji}$ and the second holds because $ma_2(0)=0=F_2$ and

$ma_1=mK^36t=mK^26Kt=m\frac{2C}{9m}6(K^3t^3)^{1/3}=\frac{4C}{3}(\frac{2K^3t^3}{2})^{1/3}=\frac{4C}{3}(\frac{x_1(t)-x_2(t)-d}{2})^{1/3}=F_1$

$ma_3=-mK^36t=-mK^26Kt=-m\frac{2C}{9m}6(K^3t^3)^{1/3}=-\frac{4C}{3}(\frac{2K^3t^3}{2})^{1/3}=-\frac{4C}{3}(\frac{x_1(t)-x_2(t)-d}{2})^{1/3}=F_3.$

Now let's see where the first law fails. Finally, imagine a universe with three bodies $1$, $2$, and $3$ of equal finite mass m. Let $d$ and $C$ be positive non-zero constants with the appropriate units. Suppose the universe has a potential energy function $$V(x_1,x_2)=-C\left(\frac{x_1-x_3-d}{2}\right)^{4/3}\;,$$ so $1$ and $3$ exert equal and opposite forces on each other $$-F_{31}=\frac{\mathrm dV}{\mathrm dx_1}=-\frac{\mathrm dV}{\mathrm dx_3}=F_{13}\;.$$ Suppose $x_1(0)=d$ and $x_2(0)=d/2$ and $x_3(0)=0$. Further suppose that $v_1(0)=v_2(0)=v_3(0)=0$. Obviously we can satisfy all three of Newton's laws by taking as solutions $x_i(t)=x_i(0)$, however, instead suppose the particles move as $$x_1(t)=d+(Kt)^3,~~ x_2(t)=d/2$$ and $$x_3(t)=-(Kt)^3$$ for $$K= \sqrt{\frac{2C}{9m}}.$$ Then third law holds because $F_{ij}=-F_{ji}$ and the second holds because $ma_2(0)=0=F_2$ and

$$ma_1=mK^36t=mK^26Kt=m\frac{2C}{9m}6(K^3t^3)^{1/3}=\frac{4C}{3}\left(\frac{2K^3t^3}{2}\right)^{1/3}=\frac{4C}{3}\left(\frac{x_1(t)-x_2(t)-d}{2}\right)^{1/3}=F_1$$

$$ma_3=-mK^36t=-mK^26Kt=-m\frac{2C}{9m}6(K^3t^3)^{1/3}=-\frac{4C}{3}\left(\frac{2K^3t^3}{2}\right)^{1/3}=-\frac{4C}{3}\left(\frac{x_1(t)-x_2(t)-d}{2}\right)^{1/3}=F_3.$$

You cannot derive any of the laws from each other.

Imagine a universe with two bodies of equal finite mass m. One exerts a constant force on the other that pulls it towards the origin with a force proportional to how far away it is from the origin and the other exerts no force on the one. The one has motion $x(t)=100$ and the other has motion $x(t)=\sin(\omega t)$. The first two laws are satisfied, the third is not.

Now imagine a universe with three bodies of equal finite mass m. The first two exert a constant nonzero external force on the other, each of the forces are equal and opposite. The motions are $x(t)=100$ and $x(t)=50$ and $x(t)=0$. The first law is satisfied, as is the third. The second law is not.

Now imagine a universe with three bodies $1$, $2$, and $3$ of equal finite mass m. Let $d$ and $C$ be positive nonzero constants with the appropriate units. Suppose the universe has a potential energy function $V(x_1,x_2)=-C(\frac{x_1-x_3-d}{2})^{4/3}$, so $1$ and $3$ exert equal and opposite forces on each other $F_{13}=\frac{dV}{dx_1}=-\frac{dV}{dx_3}=F_{31}$. Suppose $x_1(0)=d$ and $x_2(0)=d/2$ and $x_3(0)=0$. Further suppose that $v_1(0)=v_2(0)=v_3(0)=0$. Obviously we can satisfy all three of Newton's laws by taking as solutions $x_i(t)=x_i(0)$, however, instead suppose the particles move as $x_1(t)=d+(Kt)^3$, $x_2(t)=d/2$ and $x_3(t)=-(Kt)^3$, for $K= \sqrt{\frac{2C}{9m}}$. Then third law holds because $F_{ij}=-F_{ji}$ and the second holds because $ma_2(0)=0=F_2$ and

You cannot derive any of the laws from each other. In particular, for each law there is a possible universe where one law fails and the other two hold.

So let's see where the third law fails. Imagine a universe with two bodies (with positions $x_1$ and $x_2$) of equal finite mass ($0< m_1=m_2 <\infty$). One exerts a constant force on the other $F_{12}$ that pulls it towards the origin with a force proportional to how far away it is from the origin ${F}_{12}=-m_2\omega^2{x}_2$ and the other exerts no force on the one ${F}_{21}= 0$. The motions are ${x}_1(t)=100$ and ${x}_2(t)=\sin(\omega t)$. The first two laws are satisfied (${F}_{21}= 0$ and $m_1$ is at rest and stays at rest, ${F}_{21}=m_1 a_1$, ${F}_{12}=m_2 a_2$), but since $ F_{12}+ F_{21}\neq 0$ the third law is not satisfied

Let's see where the second law fails. Now imagine a universe with three bodies of equal finite mass $0< m_1=m_2=m_3 < \infty$. The first two exert a constant nonzero external force on the other, each of the forces are equal and opposite $ F_{12}=- F_{21} \neq 0$. All other forces are zero $ F_{13}= F_{31}= F_{23}= F_{32}= 0.$ The motions are $x_1(t)=100$ and $x_2(t)=50$ and $x_3(t)=0$. The first law is satisfied ($ F_{13}= F_{23}= 0$, $m_3$ is at rest and stays at rest), as is the third ($F_{ij}+ F_{ji}= 0$). The second law is not ($ F_{12}+ F_{32}= F_{12}+ 0= F_{12}\neq 0 =m_2 a_2$).

Now let's see where the first law fails. Finally, imagine a universe with three bodies $1$, $2$, and $3$ of equal finite mass m. Let $d$ and $C$ be positive nonzero constants with the appropriate units. Suppose the universe has a potential energy function $V(x_1,x_2)=-C(\frac{x_1-x_3-d}{2})^{4/3}$, so $1$ and $3$ exert equal and opposite forces on each other $-F_{31}=\frac{dV}{dx_1}=-\frac{dV}{dx_3}=F_{13}$. Suppose $x_1(0)=d$ and $x_2(0)=d/2$ and $x_3(0)=0$. Further suppose that $v_1(0)=v_2(0)=v_3(0)=0$. Obviously we can satisfy all three of Newton's laws by taking as solutions $x_i(t)=x_i(0)$, however, instead suppose the particles move as $x_1(t)=d+(Kt)^3$, $x_2(t)=d/2$ and $x_3(t)=-(Kt)^3$, for $K= \sqrt{\frac{2C}{9m}}$. Then third law holds because $F_{ij}=-F_{ji}$ and the second holds because $ma_2(0)=0=F_2$ and