What is the new orbit if k is halved in a circular orbit with F(r)=-k/r^2?

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Homework Help Overview

The problem involves a particle in a circular orbit influenced by a force field described by F(r)=-k/r^2. The original poster questions the nature of the new orbit when the parameter k is halved, specifically how this affects the total energy and leads to a parabolic trajectory.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating the orbital velocity pre-halving and consider the implications of constant momentum after k is halved. There are attempts to equate forces and derive the orbital equation, with some questioning the role of energy in determining the new orbit.

Discussion Status

Some participants have provided insights into using energy considerations to understand the transition to a parabolic orbit. There is an ongoing exploration of different approaches, including angular momentum and energy conservation, but no consensus has been reached on the final interpretation or method.

Contextual Notes

Participants note the importance of showing work and adhering to a structured approach in their queries. There is also mention of the need for clarity regarding assumptions about energy changes when k is halved.

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Homework Statement


A particle in circular orbit moves in a force field given by F(r)=-k/r^2. If k is halved what will the new orbit be?


Homework Equations


The answer is the orbit becomes parabolic but how? how does the total energy become zero as for parabolic orbits total energy=0? help please


The Attempt at a Solution


 
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The particle was in a circular orbit in an inverse square central force system, so you should be able to calculate the particle's pre-halving orbital velocity. Now assume that the particle's momentum remains unchanged when k is halved to get the final answer.
 
I equated the given force to the centrifugal force as the orbit is circular. Then i get the orbital velocity. Now after halving k and using the fact that angular momentum is constant in central force motion am still unable to get the final answer. Help me see what am missing. How do I get equation of orbit as that of parabola??
 
You suggested in your original post an approach you didn't use in your second post: Energy. The total energy for a parabolic trajectory is zero. Beyond that, I find it rather difficult to comment since you have not shown your work.

In the future, please follow our template when asking for homework help and show your work in the form of mathematical expressions.
 
D H said:
You suggested in your original post an approach you didn't use in your second post: Energy. The total energy for a parabolic trajectory is zero. Beyond that, I find it rather difficult to comment since you have not shown your work.

In the future, please follow our template when asking for homework help and show your work in the form of mathematical expressions.

I tried like this: when k is halved, potential energy goes from -k/r to -k/2r (an
increase of k/2r) so that total energy is also increased by k/2r
(kinetic energy is unaffected). total energy before k halving K was
-k/2r. So it is now 0. As total energy goes to zero tha path is parabolic. How correct am I??
 
There you go. There is no need to use angular momentum at all for this problem.
 
Yes indeed, all it needed was a couple of bottles of beer to me,lol. Here's another problem am working on: Two particles moving under the influence of their mutual gravitational force describe circular orbits about one another with a period of T. If they are suddenly stopped and allowed to gravitate each other, show that they will collide with each other after a time T/4squareroot2. I believe that when the particles are stopped the angular momentum goes to zero and the head towards each other for collision. Maybe some energy conservation should lead me to the result. I will cry for help if I ciould not get it done by myself though hints are always welcome.
 
The resulting integral is doable but is rather nasty. A much easier approach: How long is it until closest approach if you cancel some, but not all, of the angular momentum? What happens when in the limit that angular momentum goes to zero?
 
I am now working on the Fourier transforms. Please help me how to find the Fourier transform of 1/(x square+ 4x + 13). The limit is minus negative to positive infinity so when I applied the formula the limit makes it all zero, which i do not think is the right answer. Shed some light where i made the mistake.
 

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