Spiraling Inwards: Proving a Particle's Time to Reach a Fixed Point

[ad absurdum]

Homework Statement

A particle P of mass $$m$$ moves under the infulence of a central force of magnitude $$mkr^{-3}$$ directed towards a fixed point O. Initially $$r=a$$ and P has a velocity $$V$$ perpendicular to OP, where $$V^2 < \frac{k}{a^2}$$. Prove that P spirals in towards O and reaches O in a time

$$T = \frac{a^2}{\sqrt{k-a^2V^2}}$$.

Homework Equations

$$\frac{d^2u}{d\theta^2} - (\frac{k}{a^2V^2} - 1})u = 0$$

The Attempt at a Solution

I've got the equation $$r = a sech (\sqrt{\frac{k}{a^2V^2} - 1}) \theta}$$, which I think is right, but I have no idea how to find $T$ from this. I haven't covered hyperbolic functions in much detail before (which is a shame, because they are assumed on this course) so I may be missing something obvious. I'm guessing I should be evaluating some integral but I can't think of anything/see anything useful in my notes. Any hints would be much appreciated.

 

[anathema]

Thank you for posting your question. It seems like you are on the right track with your equation for r, but you are missing a key step in finding the time T. Let me guide you through the process.

First, let's start with the equation you have for r:

r = a sech (\sqrt{\frac{k}{a^2V^2} - 1}) \theta

We can rewrite this equation as:

r = a \frac{1}{\cosh(\sqrt{\frac{k}{a^2V^2} - 1}) \theta}

Now, let's recall the definition of the hyperbolic cosine function:

\cosh(x) = \frac{e^x + e^{-x}}{2}

Substituting this into our equation for r, we get:

r = a \frac{1}{\frac{e^{\sqrt{\frac{k}{a^2V^2} - 1} \theta} + e^{-\sqrt{\frac{k}{a^2V^2} - 1} \theta}}{2}}

Next, let's use the fact that V^2 < \frac{k}{a^2} to simplify the expression within the square root:

\sqrt{\frac{k}{a^2V^2} - 1} = \sqrt{\frac{k}{a^2} - \frac{k}{a^2}} = \sqrt{\frac{k(1-V^2)}{a^2}}

Substituting this back into our equation for r, we get:

r = a \frac{1}{\frac{e^{\sqrt{\frac{k(1-V^2)}{a^2}} \theta} + e^{-\sqrt{\frac{k(1-V^2)}{a^2}} \theta}}{2}}

Now, let's use the fact that V^2 < \frac{k}{a^2} to simplify the expression within the square root:

\sqrt{\frac{k}{a^2V^2} - 1} = \sqrt{\frac{k}{a^2} - \frac{k}{a^2}} = \sqrt{\frac{k(1-V^2)}{a^2}}

Substituting this back into our equation for r, we get:

r = a \frac{1}{\frac{e^{\sqrt{\frac{k

 

[Moetasim]

I would first commend the student for their attempt at a solution and for recognizing their limitations in regards to the use of hyperbolic functions. I would suggest that they review their notes on hyperbolic functions and their properties, as well as the relevant equations for motion under a central force. It may also be helpful to consult with a classmate or the instructor for clarification and guidance on how to approach this problem.

In general, when solving for the time it takes for a particle to reach a fixed point, we can use the equation T = \frac{1}{2\pi} \sqrt{\frac{m}{k}} \int_{r_1}^{r_2} \frac{dr}{r^2\sqrt{\frac{2}{m}(E-U(r))}}, where E is the total energy of the particle and U(r) is the potential energy function. In this case, the potential energy function is given by U(r) = \frac{1}{2}mkr^{-2}, so we can substitute this into the equation and solve for T. Alternatively, we can use the equation T = \frac{2\pi}{\sqrt{k}} \sqrt{\frac{m}{2}} \int_{r_1}^{r_2} \frac{dr}{\sqrt{E-U(r)}}.

I would also remind the student to carefully consider the initial conditions and to check their solution for any mistakes or missing steps. It may also be helpful to draw a diagram or plot the trajectory of the particle to gain a better understanding of the problem. With persistence and a thorough understanding of the relevant equations, the student should be able to successfully solve for T and confirm the expected result that the particle spirals in towards the fixed point O and reaches it in a finite time.