# Trajectory of a particle under the given force

Abhishek11235
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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Homework Helper
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
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.

Homework Helper
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
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##