Would just like a hint on how to start this pls.

 

A particle P is attracted towards a fixed point O in an intertial frame by a force of magnitude [math]\lambda r^{-3}[/math] per unit mass where r=OP. The particle is set in motion at a distance a from O with speed v perpendicular to OP.

 

b)If [math]\lambda >a^{2}v^{2}[/math] show that [math]u=r^{-1}=a^{-1}\cosh(\omega\theta)[/math] where [math]\omega[/math]should be determined , and that the particvle reaches O in time [math]\frac{a^2}{\sqrt(\lambda-a^{2}v^{2})}[/math]

 

cheers

Write down a differential equation for r(t), using Newton's Second Law (F=ma). You can assume that motion is in a plane (since it is a central force) but should consider axial and radial motion (the equation for one of which is trivial but gives you an expression for angular momentum conservation). Convert (the non-trivial equation) to one concerning u(theta) (which makes it easier to slove) and solve it.

 

If you feel ambitious, solve it for a central force of the form [math]\lambda r^{-3} +\kappa r^{-2}[/math]. Once you have the equation for u, you will see that this is the most general central force with an orbit equation which can be solved in the time frame of an exam (since the r^-2 and r^-3 are the only forces which will give a linear differential equation).

This is what i have got so far.

 

this is the general equation of motion after writing in axial and radial motion

 

m(r'' - h^2/r^3)=F®

 

where h = r^2 times theta' = constant

 

writing u = 1/r

 

u get that diff equation in the form

 

2nd diff. u wrt theta + u = -F(1/u)/(m h^2 u^2)

 

substituting all the required values you get

 

2nd diff of u wrt theta + u = lamda *u /h^2

 

so .. 2nd diff of u wrt theta +u(1- lambda/h^2)=0

 

I do not know where the condition lambda >a^2 * v^2 comes in..

different cases for value of lambda... since the solution we require is of the form cosh something.. ill just assume that the solutions to the auxilary equation are real solutions. i.e (lambda/h^2 - 1 ) > 0... --> lambda > h^2

 

so the general solution i get is

 

[math]u=r^{-1}=Ce^{\omega\theta}+De^{-\omega\theta}[/math]

 

where [math]\omega = \sqrt{(\frac{\lambda}{h^2}-1)}[/math]

 

dont know where to go after that

All i have to do now is to show that [math]C+D=(2a)^{-1}[/math]

 

but i am not sure about the whole solution , or where the condition

[math]\lambda>a^2v^2[/math] comes in

The angular momentum (per unit mass) is h=va (since it is constant you just take this from initial conditions). The constraint [math]\lambda > v^2a^2[/math] is then just insiting that the object under the square root is positive.

 

At time [math]t=0, \; \theta=0[/math] and [math]u=1/a[/math], so [math]1/a = C+D[/math]. the other 2 comes from the definition of cosh.

oops sorry i got it wrong....

 

Dont we have to show that [math]C=D=1/2a[/math]

 

as

 

[math]\tfrac{1}{2a}e^{\omega\theta}+\tfrac{1}{2a}e^{-\omega\theta}[/math]

[math]=\frac{1}{a}[\frac{1}{2}(e^{\omega\theta}+e^{-\omega\theta}][/math]

[math]=\frac{1}{a}\cosh(\omega\theta)[/math]

Well, you could impose your boundary condition on the velocity (or [math]\frac{du}{d \theta}[/math]) to give you the extra constraint, but it is fairly obvious that C=D from the symmetry of the problem. The boundary conditions don't specify whether your initial velocity is clockwise or anti-clockwise because the whole problem is necessarily symmetric under [math]\theta \to -\theta[/math].

could you explain why is obvious from the symmetry of the problem that C=D. also i got no idea what boundary condition to put on [math]\frac{du}{d\theta}[/math]

OK, your general radial solution was:

 

[math]u(\theta) = C e^{\omega \theta} +D e^{-\omega \theta} ~~~(1)[/math]

 

Differentiating wrt [math]\theta[/math] gives:

 

 

[math]u^{\prime}(\theta) = \omega ( C e^{\omega \theta} -D e^{-\omega \theta}) ~~~(2)[/math]

 

 

Your boundary conditions are [math]u(0)=1/a[/math] and [math]u^{\prime}(0)=0[/math] since start a distance a out, with zero velocity parallel to OP.

 

[math]u(0)=C+D=1/a[/math]

[math]u^{\prime}(0)=C-D=0[/math]

 

Thus C=D=1/2a.

 

 

that was the mathsy was to do it, but you could think more physically. The general solution can be rewritten as:

 

[math]u(\theta)=E \cosh (\omega \theta) + F \sinh (\omega \theta)[/math]

 

where E and F are constants.

 

Now since the force is a central force, pointing in towards the centre of the system, it doesn't care if you go round clockwise or anticlockwise. Whether you choose to define increasing theta in a clockwise or anticlockwise sense is entirely up to you, and therefore the answer can't depend on your choice. This symmetry is only broken by the choice of initial conditions (as to which way the object is moving). But this initial condition, having no radial part, cannot effect the radial equation (other than through the angular momentum). therefore you r answer must be symmetric in replacing theta with -theta. Since sinh is anti-symmetric about 0, its coefficient F=0. Thus E=1/a.

cheers mate. uve really helped me out ....

 

What are u anyway? u in research?

 

Also for the second part of the question. u need to find at what time it reaches O... but that means r = 0.... but it cant cos then r^-1 would not be defined.... do i have to think of it as in limits?

For the last bit, you should use [math]\frac{d \theta}{dt}=hu^2 = \frac{h}{a^2}\cosh^2 (\omega \theta)[/math]. The object hits O as [math]\theta \to \infty[/math] so we can use the above equation to write [math] t = \int_{0}^{\infty} \frac{a^2}{h \cosh^2(\omega \theta)} d\theta[/math].

 

Integrate this....

i am not sure if that can be done analytically. tried to do that in maple, and it didnt give me anything useful

tut tut! You should do integrations yourself, not using maple....

 

[math]\int_0^{\infty} \frac{1}{\cosh^2 (\omega \theta)} d\theta = \frac{1}{\omega} \left\{ \tanh \omega \theta \right\}_0^{\infty} = \frac{1}{\omega}[/math]

 

so [math]t=\frac{a^2}{h \omega}[/math] which , along with h=av gives the right answer.....

hehe... it was just a quick fix... i would have tried doing it properly afterwards..

 

ur a genius by the way

 

cheers

 

[edit]oh yah ..... just remembered that derivative of tanh is sech squared....[/edit]

'Derivative'. The word is 'derivative'.

 

....and you're welcome.