# Path of a particle in central force motion

A particle of mass $$m$$ moves in a plane, and is attracted to the origin with a force proportional to its distance $$r$$ from the origin. We are given the force $$F(r) = -k^2mr, \enspace(k>0)$$

and two acceleration components $$\ddot{x}=-k^2x,\enspace\ddot{y}=-k^2y$$. The problem is to show that the path of the particle is an ellipse, if it satisfies the initial conditions $$x(0)=a, \dot{x}(0)=0, y(0)=0, \dot{y(0)}=b$$, where $$a,b>0$$.

Solution: Let $$x=y_1, y_2 = y, y_3=x', y_4 = y'$$. The system of two equations in two variables $$\ddot{x}=-k^2x,\enspace\ddot{y}=-k^2y$$

becomes a system of four equations in four variables. I can post the full solution but in brief, the solutions are $$x=a\cos kt, \enspace y=\frac{b}{k}\sin kt$$

and therefore $$\frac{x^2}{a^2}+\frac{y^2k^2}{b^2}=\cos^2kt+\sin^2kt=1.$$

Every position of the particle affected by the force within the plane can be described by this equation. This equation implies the minor axis is $$b/k$$.

Is there a physical interpretation as to why the central force $$F(r)$$ determines the minor axis of the ellipse ( the path of the particle )? It seems to me, that a stronger force would "squeeze" the path of the particle, as in making the minor axis smaller.

• The only reason one axis depends on k and the other does not is because you have ${\dot x} = 0$ and ${\dot y} = b$ as initial conditions. If both of them were nonzero, you'd have k terms in both axes of the ellipse (and it would be rotated so that x and y axes wouldn't be your major and minor axes.) Mar 31 at 18:20

how do you know which axis is minor and which major? Your solution indeed gives an ellipse, with the two axes equal to $$a$$ and $$b/k$$ (note that $$a$$ and $$b$$ have different dimensions, $$a$$ is length and $$b$$ is velocity, and thus cannot be directly compared). If there are no numbers in the problem, you cannot determine which of the axes is larger.

Yet, you can consider a limiting case. Say, you have a very strong force, which would mean $$k$$ is very large. Then, the $$b/k$$ axis becomes very short, and the ellipse becomes more and more squeezed upon further increase of $$k$$.

Let's write the equation of the ellipse as being: $$\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1,$$ where $$\alpha$$ and $$\beta$$ represent the distances that the ellipse cuts the $$x$$ and $$y$$ axes respectively. I interpret the question then as being: why is $$\alpha = a$$ independent of $$k$$, while $$\beta = b/k$$ dependent on $$k$$?

As alluded to in the other answer, this is because of the initial conditions. While $$a$$ represents the initial position, $$b$$ represents the initial velocity. As a result, both $$b$$ and $$k$$ are required to specify the point at which the particle cuts the $$y-$$axis. For a fixed force (i.e. a fixed $$k$$) the harder you throw the particle, the further it goes before cutting the $$y-$$axis (i.e., the larger $$\beta$$ is). Similarly, for a fixed initial velocity $$b$$, the stronger the force is, the sooner it will cut the $$y-$$axis, and so on.

However, this is completely dependent on the initial conditions. Indeed, if you started with these initial conditions:

$$x(0) = 0 \quad\quad\quad \dot{x}(0) = a \quad\quad\quad y(0) = b \quad\quad\quad \dot{y}(0) = 0,$$

the solution would be: $$\frac{x^2 k^2}{a^2} + \frac{y^2}{b^2} = 1.$$

Alternatively, if you instead started with $$x(0) = a \quad\quad\quad \dot{x}(0) = 0 \quad\quad\quad y(0) = b \quad\quad\quad \dot{y}(0) = 0,$$ the solution would be:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$$ and in this case neither of the axes depend on the value of $$k$$!

The question forms one of the well explored topics in the two and three dimensional harmonic oscillator. This webpage addresses the connection between a mass attached to a spring undergoing two dimensional motion in the plan to the elliptic orbits of objects around a planet due to gravitational force.

If we consider the Newton's second law equation of motion under gravity $$\ddot{\vec{r}}=\mu \frac{\vec{r}}{r^3}$$ where $$0 < \mu$$, then the conservation of energy implies that $$m\frac{\vec{r}\cdot\vec{r}}{2} - \frac{\mu}{r} = E := -\frac{\mu}{2}.$$ Applying a (brilliantly intuited) change of variables from time $$t$$ to a novel notion of time $$s$$ given by $$\frac{ds}{dt}=\frac{1}{r}$$ and denoting $$t' := \frac{ds}{dt} = r$$ we obtain $$\vec{r}' = \frac{d\vec{r}}{ds} = r\dot{r}$$, which implies that $$\mu(t' - 1)^2 + m\vec{r}'\cdot \vec{r}' = 1$$. Further, it can be shown (after some manipulation) that $$\vec{r}''' = -k\vec{r}'$$ where $$0 < k$$, so that the dynamical equation $$\vec{r}'' = -k\vec{r} + \vec{c}$$ where $$\vec{c}$$ is a constant (the Runge–Lenz) vector, recovers the two dimensional harmonic oscillator (two dimensional spring-mass motion). The variable transform connects the two phenomena together and has been a topic of research earlier and is likely still being explored.