Escape Velocity and Particle Position

  • Thread starter Cider
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In summary: The expression represents the force on the particle, which is F=m a. The velocity is found by solving the differential equation for r.
  • #1
Cider
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Homework Statement


A particle of mass m moving in one dimension has potential energy [itex]U(x)=U_0[2(\frac{x}{a})^2-(\frac{x}{a})^4][/itex], where [itex]U_0[/itex] and a are positive constants.

What is the function x(t) for the position of the particle when it starts at the origin with a velocity in the positive direction equal to the escape velocity?

Homework Equations


The escape velocity is [itex]$\sqrt{\frac{2U_0}{m}}$[/itex] which escapes at x=a.

The Attempt at a Solution


I was able to determine the function for the force easily, but I ran into many problems with logs when trying to integrate it twice and ended up with logarithms that could not be paired up so that I could actually solve for x.

Having a fair amount of trouble getting the latex code to work on these forums properly... Looks like it's because of Chrome.
 
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  • #2
What exactly is the question?
 
  • #3
Sorry, I had been having so much trouble when trying to get it done with Chrome (restarting the post multiple times) that I forgot to include that part. It's there now.
 
  • #4
Could you show us what you tried and got? It's hard to see where the problem is if all you do is give a vague description of what you tried.
 
  • #5
[itex]$(\frac{1}{a^2})(x \ln (x)-x-\frac{1}{2}(x\ln(x^2-a^2)+a\ln(x^2-a^2)-2x))+C$[/itex] all over ma^2 I believe. It's difficult getting this stuff to show up correctly on the forums :s But basically something that I can't really continue with.
 
  • #6
Cider said:
[itex]$(\frac{1}{a^2})(x \ln (x)-x-\frac{1}{2}(x\ln(x^2-a^2)+a\ln(x^2-a^2)-2x))+C$[/itex] all over ma^2 I believe. It's difficult getting this stuff to show up correctly on the forums :s But basically something that I can't really continue with.

What is this expression supposed to represent (position, velocity, force, the numer of apples in a carton of width [itex]x[/itex])? And how did you get it?

Posting an expression without any work shown or description of what the expression is and where it came from doesn't help us to see where you are going wrong.

In any case, can you think of a conserved quantity that will help you get an expression for velocity? Can you determine position as a function of time from that expression?
 
  • #7
I figured out what I was doing wrong. The correct method was the second I tried, but I had represented something in a way that prevented me from seeing the ease with which I could have solved it. Basically the process would use E=1/2 mv^2 + U(r), which I can simplify so that I get a first order differential equation that requires some simple partial fractions to integrate. Thanks though.
 

What is position-dependent potential?

Position-dependent potential refers to a type of potential energy that varies with the position of an object. This means that the potential energy of the object changes as it moves through different positions within a given system.

What factors can affect position-dependent potential?

The factors that can affect position-dependent potential include the type of force acting on the object, the distance between the object and the source of the force, and the properties of the object itself (such as mass or charge).

How is position-dependent potential different from position-independent potential?

Position-dependent potential is dependent on the position of the object, whereas position-independent potential is not affected by the position of the object. In other words, position-independent potential is a constant value, while position-dependent potential changes based on the location of the object.

What are some real-world examples of position-dependent potential?

Some real-world examples of position-dependent potential include gravity, electric potential, and magnetic potential. In each of these cases, the potential energy of an object changes based on its position in relation to the source of the force (e.g. the Earth in the case of gravity).

How is position-dependent potential used in scientific research?

Position-dependent potential is used in many areas of scientific research, such as studying the movement of particles in a system, calculating the energy of a system, and understanding the behavior of waves. It is also an important concept in fields such as physics, chemistry, and engineering.

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