Proving time to origin for particle with inverse-square attractive force

[Tom McCurdy]

Energy Question -- Help Please

A particle of mass m moves on the x-axis under the influence of a forace of attraction towards the origin give by
$$F= \frac{-k}{x^2} i$$.

If the particle starts from rest at x=a prove that it will arrive at the origin in a time given by

$$\frac{1}{2}\pi a \sqrt{\frac{ma}{2k}}$$

Our teacher explained it to me in class but unforutently he wasn't able to finish it. I was wondering if someone could step me through this proof please.

 

[Tom McCurdy]

I tried divding by mass and integradting twice to get distance but it didn't work

 

[HallsofIvy]

The differential equation $m\frac{dv}{dt}= -\frac{k}{x^2}$ can't just be "integrated twice" because you don't know x as a function of t.

However, since t does not appear explicitely in the equation, there is a standard "trick" for reducing to a first order equation which can be integrated:
$\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}$ by the chain rule.
But $\frac{dx}{dt}= v$ so the equation becomes:
$$mv\frac{dv}{dx}= -\frac{k}{x^2}$$

That's a "separable" differential equation. We separate it to get
$$m v dv= -\frac{kdx}{x^2}$$
and integrate to get
$$\frac{1}{2}m v^2= \frac{k}{x}+ C$$
When t=0, v= 0 and x= a so we have $0= \frac{k}{a}+ C$ or $C= -\frac{k}{a}$ which gives

$$v^2= \frac{2k}{m}(\frac{1}{x}-\frac{1}{a})$$
$$= \frac{2k}{ma}\frac{a-x}{x}$$
Which reduces to
$$v= \frac{dx}{dt}= \sqrt\(\frac{2k}{ma}\frac{a-x}{x}\)}$$
or
$$\sqrt{\frac{x}{a-x}}dx= \sqrt{\frac{2k}{ma}}dt$$

To integrate that let u= (a-x)^1/2 so that du= (1/2)(a-x)^-1/2dx and u²= a-x so x= a- u². The equation becomes
$$2\sqrt{a-u^2}du= \sqrt{\frac{2k}{ma}}dt$$.

The left hand side is now a fairly standard trigonometric substitution:
Let u= √(a) sin(θ) so that du= √(a) cos(θ)dθ and √(a- u^2) becomes √(a) cos(θ) so the equation, in terms of θ is:
$$2a cos^2\theta dx= \sqrt{\frac{2k}{ma}} dt$$

To integrate that, use the trig identity cos²θ= (1/2)(1+ cos(2θ)) so that the equation becomes:
$$a(1+ cos(2\theta))d\theta= \sqrt{\frac{2k}{ma}}dt$$

That can be integrated directly to get

$$a(\theta+ \frac{1}{2}sin(2\theta)= \sqrt{\frac{2k}{ma}}t+ C$$

When t= 0, x= a so that u= 0 and θ= 0. The equation becomes
a(0+ 0)= 0+ C so C= 0.

When x= 0, $u= \sqrt{a}$ and $\theta= \frac{\pi}{2}$. Of course, in that case $2\theta= \pi$ so $sin(2\theta)= 0$ and our formula becomes
$$\frac{a\pi}{2}= \sqrt{\frac{2k}{ma}}T$$
and
$$T= \frac{\pi a}{2}\sqrt{\frac{ma}{2k}}$$
as advertised!

Wow! I hope you teacher had some simpler way of doing that!

 

[Tom McCurdy]

Thank you so much for your response... this looks almsost identical to the way he showed in class today... Its alright that i didn't get it done in time, only a few kids in my class had it done. This is very hard for me as I am just taking Calc BC this year, your response is very useful to me.