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If I take r as the radial vector of the moving object and v as the velocity vector of the moving object in central force field. Then r should be perpendicular to r×v . So that depicts that r.(r×v) = 0 . So how does this say that the particle is lying on a constant plane ?

Qmechanic
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Nobody recognizeable
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3 Answers3

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PICTORIAL ANSWER

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In the Figure we see 4 positions $\;\rm P_1,P_2,P_3,P_4\;$ of the moving particle with position vectors $\;\mathbf{r_1},\mathbf{r_2},\mathbf{r_3},\mathbf{r_4}$. Under the influence of a central force the angular momentum vector $\;\mathbf{L}=\mathbf{r}\boldsymbol{\times}m\mathbf{v}\;$ is constant. The vectors $\;\mathbf{r},\mathbf{v}\;$ are the position and the velocity vector of the particle at an arbitrary time moment.

So \begin{align} \mathbf{r_1}& \perp \mathbf{L_1}=\mathbf{r_1}\boldsymbol{\times}m\mathbf{v_1}=\mathbf{r}\boldsymbol{\times}m\mathbf{v}=\mathbf{L} \tag{01.1}\label{eq01.1}\\ \mathbf{r_2}& \perp \mathbf{L_2}=\mathbf{r_2}\boldsymbol{\times}m\mathbf{v_2}=\mathbf{r}\boldsymbol{\times}m\mathbf{v}=\mathbf{L} \tag{01.2}\label{eq01.2}\\ \mathbf{r_3}& \perp \mathbf{L_3}=\mathbf{r_3}\boldsymbol{\times}m\mathbf{v_3}=\mathbf{r}\boldsymbol{\times}m\mathbf{v}=\mathbf{L} \tag{01.3}\label{eq01.3}\\ \mathbf{r_4}& \perp \mathbf{L_4}=\mathbf{r_4}\boldsymbol{\times}m\mathbf{v_4}=\mathbf{r}\boldsymbol{\times}m\mathbf{v}=\mathbf{L} \tag{01.4}\label{eq01.4}\\ \end{align} that is the position vector $\;\mathbf{r_\jmath}(t_\jmath)\;$ at any time $\;t_\jmath\;$ is normal to the constant vector $\;\mathbf{L}$.

So all points $\;\rm P_\jmath\;$, all position vectors $\;\mathbf{r_\jmath}\;$ and consequently all velocity vectors $\;\mathbf{v_\jmath}\;$ lie on a plane perpendicular to $\;\mathbf{L}\;$ and we have plane motion.

Frobenius
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If the force on an object is radial, $\mathbf{F}\parallel\mathbf{r}$ so the angular momentum $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ has a vanishing time derivative, a sum of two cross products of parallel vectors viz. $\mathbf{v}\times\mathbf{p}+\mathbf{r}\times\mathbf{F}$. It is orthogonality to this conserved angular momentum that completes the proof.

J.G.
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  • So dL/dt becomes r×F. As v×mv is zero. Then what is the orthogonality and what does it explain. Please explain the matter. – Nobody recognizeable Aug 10 '18 at 06:03
  • @user187604 Vectors are orthogonal when their for product is $0$. The set of vectors orthogonal to a given non-zero element of $\mathbb{R}^3$ is a plane. (See if you can prove that when $\mathbf{L}=\mathbf{0}$ the motion is along a line.) – J.G. Aug 10 '18 at 06:17
  • it will be more helpful if you can simplify to me that how can we say the motion of the particle in an central force field is in a constant plane. If we only have r.(r×v). A pictorial answer or even a document would really help me out. – Nobody recognizeable Aug 10 '18 at 06:21
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Interestingly, this has nothing to do with 3D & cross product per se: We can define angular momentum $L^{ij}:=x^ip^j-x^jp^i$ in arbitrary spatial dimension $d$, and OP's title statement remains true.

This follows merely from the fact that a central force yields the equations of motion $\dot{\bf x} \parallel {\bf p}$ and $\dot{\bf p} \parallel {\bf x}$ (which in turn imply that the angular momentum $L^{ij}$ is conserved). Define $\pi$ to be the plane/line/point (through the origin) that is spanned by the initial position and momentum vectors. Deduce (from the equations of motion $\dot{\bf x} \parallel {\bf p}$ and $\dot{\bf p} \parallel {\bf x}$) that the point mass continues to be confined to this plane/line/point $\pi$ for all time $t$. $\Box$

Qmechanic
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