Central force and plane of motion

Question

So I am aware that for a given central force, that is $\vec F= F \vec e_r$, the motion lies in the plane. We prove this by computing the derivative of $\vec n$, where $\vec n= \vec r × \dot{\vec r}$, which gives $\vec 0$. How and why do we conclude from there that the motion lies on a plane?

Answer 1

Let as assume that $\vec{r}(t_0)$ and $\frac{d \vec{r}}{dt}(t_0)$ are linearly independent. Then they span a plane in $\mathbb R^3$, passing through $\vec{0}$.

We have at any time $t$:$$\frac{d}{dt}\left(\vec{r}\times\frac{d \vec{r}}{dt}\right) = \frac{d \vec{r}}{dt}\times\frac{d \vec{r}}{dt} + \vec{r}\times\frac{d^2 \vec{r}}{dt^2} = \vec{r}\times\frac{1}{m}\vec{F} = \vec 0$$ That means that $$ \vec{r}(t)\times\frac{d \vec{r}}{dt}(t) = \vec{r}(t_0)\times\frac{d \vec{r}}{dt}(t_0)$$ $$ \vec{r}(t) \cdot \left(\vec{r}(t_0)\times\frac{d \vec{r}}{dt}(t_0)\right) = 0$$ which means that $\vec{r}(t)$ lies in the plane spanned by $\vec{r}(t_0)$ and $\frac{d \vec{r}}{dt}(t_0)$.

If $\vec{r}(t_0)$ and $\frac{d \vec{r}}{dt}(t_0)$ are linearly dependent (but at least one is non-zero), then they define a line in $\mathbb R^3$. You can reduce the EOM to a one-dimensional equation and show that the motion lies on that line.

Answer 2

With

$\vec n = \vec r \times \dot{\vec r}, \tag 1$

we have

$\dot{\vec n} = \dot{\vec r} \times \dot{\vec r} + \vec r \times \ddot{\vec r} = \vec r \times \ddot{\vec r}, \tag 2$

since

$\dot{\vec r} \times \dot{\vec r} = 0, \tag 3$

always. Now $\ddot{\vec r}$, being the acceleration of the curve $\vec r(t)$, is proportional to the force

$\vec F = F\vec e_r, \tag 4$

indeed, if we consider $\vec r(t)$ to be the path of a particle of mass

$m \ne 0, \tag 5$

then

$m \ddot{\vec r} = \vec F = F\vec e_r; \tag 6$

that is,

$\ddot{\vec r} = \dfrac{F}{m} \vec e_r; \tag 7$

furthermore

$\vec e_r = \dfrac{\vec r}{\vert \vec r \vert}; \tag 8$

thus we have

$\ddot{\vec r} = \dfrac{F}{m} \dfrac{\vec r}{\vert \vec r \vert} = \dfrac{F}{m \vert \vec r \vert} \vec r; \tag 9$

now from (2),

$\dot{\vec n} = \vec r \times \ddot{\vec r} = \vec r \times \dfrac{F}{m \vert \vec r \vert} \vec r = \dfrac{F}{m \vert \vec r \vert} \vec r \times \vec r = 0, \tag{10}$

again since

$\vec r \times \vec r = 0. \tag{11}$

We thus see that $\vec n$ is a constant vector for any central force $F\vec e_r$; it follows that if for some $t_0$, $\vec r(t_0)$ and $\dot {\vec r}(t_0)$ are linearly independent, they are for all $t$ for which $\vec r(t)$ is defined since in this event

$\vec r(t) \times \dot {\vec r}(t) = \vec n(t) = \vec n(t_0) = \vec r(t_0) \times \dot {\vec r}(t_0) \ne 0, \tag{12}$

which of course implies the linear independence of $\vec r(t)$, $\dot{\vec r}(t)$; furthermore, both $\vec r(t)$ and $\dot{\vec r}(t)$ are normal to $\vec n(t)$ by virtue of the definition (1):

$\vec r(t) \cdot \vec n(t) = \vec r(t) \cdot (\vec r(t) \times \dot{\vec r}(t)) = 0 $ $\dot{\vec r}(t) \cdot (\vec r(t) \times \dot{\vec r}(t)) = \dot{\vec r}(t) \cdot n(t); \tag{13}$

since $\vec n(t)$ is fixed in time, so is the plane normal to $\vec n(t)$, in which both $\vec r(t)$ and $\dot{\vec r}(t)$ lie.

If $\vec r(t_0)$ and $\dot{\vec r}(t_0)$ are not linearly indpendent for some $t_0$, then

$n(t_0) = \vec r(t_0) \times \dot{\vec r}(t_0) = 0, \tag{14}$

and thus

$\vec r(t) \times \dot{\vec r}(t) = n(t) = 0; \tag{15}$

that is, $\vec r(t)$ and $\dot{\vec r}(t)$ remain parallel (linearly dependent) for all $t$; as such, we may write

$\dot{\vec r}(t) = a(t) \vec r(t) \tag{16}$

for some (continuous) scalar function $a(t)$; then

$\vec r(t) = \exp \left (\displaystyle \int_{t_0}^t a(s) \; ds \right) \vec r(t_0), \tag{17}$

is the unique solution to the differential equation (16) taking the value $\vec r(t_0)$ at $t = t_0$; we note that (16), (17) show that both $\vec r(t)$ and $\dot{\vec r}(t)$ lie in the one-dimensional subspace spanned by $\vec r(t_0)$ for all $t$.

Thus the motion $\vec r(t)$, in the event of central forces, is always constrained to lie in some plane containing $\vec r(t_0)$; if $\vec r(t_0)$ and $\dot {\vec r}(t_0)$ are linearly independent, the plane will be uniquely determined as the normal plane to $\vec n(t_0) = \vec r(t_0) \times \dot{\vec r}(t_0)$; if, on the other hand, $\vec r(t_0)$ and $\dot {\vec r}(t_0)$ are linearly dependent, any plane containing the vector $\vec r(t_0)$ will contain the motion $\vec r(t_0)$ for all times $t$.