# Trajectory of a particle under the given force

• Abhishek11235
The first term on the left is not a vector, the other two are.1. Yes, that is intentional. The vector on the right hand side is ω2.

#### 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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Hello Abhishek, Abhishek11235 said:
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.

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
Chandra Prayaga said:
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##

@Chandra Prayaga : sorry I mistakenly took you for the thread starter
@Abhishek11235: that's not Binet, that's Newton ! Benefit from the 'hints' in post #4 ## What is a trajectory of a particle under the given force?

The trajectory of a particle under the given force refers to the path that a particle follows when acted upon by a specific force. It describes the position of the particle at different points in time as it moves through space.

## How is the trajectory of a particle under the given force calculated?

The trajectory of a particle can be calculated using the equations of motion, specifically the equations for position, velocity, and acceleration. These equations take into account the force acting on the particle, as well as its initial position, velocity, and acceleration.

## What factors affect the trajectory of a particle under the given force?

The trajectory of a particle is affected by the magnitude and direction of the force acting on it, as well as the particle's initial position, velocity, and mass. Other factors such as air resistance and friction can also impact the trajectory.

## Can the trajectory of a particle under the given force be changed?

Yes, the trajectory of a particle can be changed by altering the force acting on it. Increasing or decreasing the force, or changing its direction, will result in a different trajectory for the particle.

## How does the trajectory of a particle under the given force relate to its motion?

The trajectory of a particle is directly related to its motion. The shape and direction of the trajectory determine how the particle will move through space, and the equations of motion can be used to predict its future position and velocity.