# Trajectory of a particle under the given force

## Main Question or Discussion Point

A particle of mass m in xy plane is attracted toward the origin with the force
\begin{align}\vec{f} = - \frac{k^{2} m}{r^{6}}\vec{r}\end{align} where $\vec r$ is position vector of particle measured from origin. If it starts at position $(a,0)$ with speed $$v=\frac{k}{\sqrt{2} a^{2}}$$ perpendicular to x axis show that trajectory of given particle is
$$\vec r= a cosΘ$$
The equation I got is Binet equation which I can't solve for r. Thanks for help

The Binet equation I got is:

$$\ddot r - \vec r w^2 = \vec f$$

Here f is same as ($1$)

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BvU
Homework Helper
2019 Award
Hello Abhishek, The equation I got is Binet equation
Pity you don't post it: then I could compare it with mine, which looks reasonable. I can't solve it offhand either, but when I know the answer I can check that it satisfies the equation (didn't check the constants, though)

• berkeman
Chandra Prayaga
Couple of clarifications needed.
1. In:

$$\vec r=acosΘ$$

The left hand side is a vector. Is that intentional? If so, what is the vector on the right hand side

2. Similarly, in

$\ddot r-\vec r$ω2=$\vec f$

The first term on the left is not a vector, the other two are.

BvU
Homework Helper
2019 Award
1. No. $|\vec r| = a\cos\theta$ is the r-component of a polar vector. The right hand side is a scalar. At $t=0$ you have $\vec r = (a,0)\$ (both in polar and in cartesian)

2. No again. The double derivative of a time dependent vector is a vector, in this case an acceleration vector $\ddot{\vec r} \equiv \vec a$ .

You edited in your result for the Binet equation. Good (it looks simple). But given the expression for $\vec F$, I find it hard to believe. Can you show the derivation ?

(edit: this looks more like a warbled Newton equation if you ask me -- see here
and a bit further down that link also has an example with $|\vec F|\propto r^{-3}$, so just follow the same path ... )

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• Abhishek11235
Couple of clarifications needed.
1. In:

$$\vec r=acosΘ$$

The left hand side is a vector. Is that intentional? If so, what is the vector on the right hand side

2. Similarly, in

$\ddot r-\vec r$ω2=$\vec f$

The first term on the left is not a vector, the other two are.
No. It is my typing mistake.
It should be $\ddot {\vec r}-\vec r$ω2=$\vec f$

And for other it is:
$r= acos\theta$

BvU
@Abhishek11235: that's not Binet, that's Newton ! Benefit from the 'hints' in post #4 