What Is the Motion of a Proton in This Potential Energy Function?

In summary, the conversation discusses a proton of mass m moving in one dimension with a potential energy function of U(x)=a/x^2 - b/x. The proton is released from rest at x0=a/b and it was shown that U(x) can be written in a different form. The speed of the proton as a function of position can also be written as V(x)=sqrt((2a/m(x0)^2)[(x0/x)-(x0/x)^2]) for x>x0. The first question asks for a qualitative description of the motion in terms of Kepler orbits, which the speaker was unable to provide due to lack of understanding. The second question asks to show that the speed of the proton is maximum at x
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1. Hi, I was hoping for some help with the following. This is the first question of this type I have attempted so, any help either with answering the parts of the question I could not do or if you notice any mistakes with the parts I have answered would be appreciated.

A proton of mass m moves in one dimension. Its potential energy function is: U(x)=a/x^2 - b/x where a and b are positive constants. The proton is released from rest at x0=a/b.
I also showed in earlier questions that that U(x) can be written in the following way:

U(x)=(a/(x0)^2)[(x0/x)^2-(x0/x)] and that v(x) the speed of the proton as a function of position can be written as:

V(x)=sqrt((2a/m(x0)^2)[(x0/x)-(x0/x)^2]) with x>x0

a. Give a short qualitative description of the motion in terms of the classification of Kepler orbits.

I don't know what a kepler orbit is and my textbook (or the Internet) was not much help so, I could not do this part. What is a Kepler orbit? How are they classified?

b. Show that at x=2x0 the speed of the proton is maximum and comute that speed.
My attempt:
Differentiating the first expression of U(x) with respect to x and setting the derivative equal to zero gives that:

2a/x=b therefore x=2a/b =2x0 as required.

Subbing this into v(x) gives v=sqrt(a/2m(x0)^2)

c. What is the force on the proton at the point x=2x0?
My answer: F=0.

d. Instead let the proton be released (from rest) at x1=3a/b. Derive an expression of v(x) for the new release point in terms of a,b and m, also give a qualitative description of the orbital motion.
My attempt: Initially U(x)=U(3a/b)= -b^2/9a = Total energy
So at any time: Kinetic energy + U(x) = -b^2/9a
Hence 0.5mv(x)^2 +a/x^2 – b/x = -b^2/9a

This is as far as I have got for this part of the question; I do not know how to get rid of the x’s. Any ideas? I also don’t know what it means by description of the orbital motion.

Was it ok for me to use the same potential energy function for a different initial condition?

e. for each release point, (x0 and x1) find the maximum and minimum x values reached during the motion.
I did not even know how to start this part of the question, any guidance would be welcome.
 
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  • #2
for d, could I just leave v(x) in terms of x?
 

FAQ: What Is the Motion of a Proton in This Potential Energy Function?

What is a potential energy function?

A potential energy function is a mathematical equation that describes the relationship between the position and potential energy of an object. It is used to calculate the potential energy of a system based on the position of its components.

How is potential energy function related to force?

According to the principle of conservation of energy, the total energy of a system remains constant. Therefore, the potential energy function of an object can be used to calculate the force acting on that object, as force is the rate of change of potential energy with respect to position.

What are the different types of potential energy functions?

There are several types of potential energy functions, such as gravitational potential energy, elastic potential energy, and electric potential energy. These functions differ based on the type of force acting on the object and the type of potential energy stored in the system.

How is potential energy function used in physics?

Potential energy functions are used in physics to understand the behavior of objects in a system. They are used to calculate the potential energy of an object and the force acting on it, which can then be used to predict the motion and behavior of the system.

Can potential energy function be negative?

Yes, potential energy function can be negative. This can occur if the reference point for the potential energy is set at a higher energy level than the current position of the object. In this case, the potential energy would be negative, indicating that the object has less potential energy at its current position compared to the reference point.

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