Angular momentum and central force

[spacetimedude]

Homework Statement

A particle of mass m moves under the influence of a central force
F(r)=-mk[(3/r^2)-2a/r^3]rhat

Show that if the particle is moving in a circular orbit of radius a, then its angular momentum is L=mh=m√(ka)

Homework Equations

L=mvr for circular orbit

The Attempt at a Solution

From the equation, we extract the second derivative of the position vector r''=-k[(3/r^2)-2a/r^3]rhat. We integrate this in order to find the velocity to use in L=mvr.
(At this point, can I dot product both sides by rhat to get rid of the vector?)
r''=-k[(3/r^2)-2a/r^3]
Integrating:
r'=-k[-3/r+a/r^2]=-k[(-3r+a)/r^2]
Using the angular momentum equation L=mvr=m*r'*r:
L=m*-k[(-3r+a)/r^2]*r.
We have r=a, so
L=m*(-k[(-3a+a)/a^2]*a)=m2k.

So obviously, the integration is wrong because I haven't included the integrating constant and also the answer we get is wrong. And intuitively by substituting in r=a, we are saying that r doesn't change.
How could I correctly approach this problem?

Thank you!

 

[BvU]

can I dot product both sides by rhat

That would be strange: $\vec L \equiv \vec r\times \vec p$ so dot product would be zero !

What is essential for a circular orbit ?

 

[spacetimedude]

That would be strange: $\vec L \equiv \vec r\times \vec p$ so dot product would be zero !

What is essential for a circular orbit ?

dL/dt=0 because there is no external force acting on the particle.

 

[BvU]

That's right, but: think back a lot further, when you first learned about circular motion. This is really a very easy exercise.
Note that the radius of the orbit is given!

 

[spacetimedude]

I've found the answer by using the equation of orbital motion -r=(r^4F(r))/(h^2m) where h=|L|/m. Rearranging the equation and setting r=a, I got L=msqrt(ka)

Thank you :)

If you don't mind me asking another question (using the same central force):

The particle is projected from point A, at distance a from the centre of force O, in a direction perpendicular to OA, its projection velocity being half of the velocity required for a circular orbit of radius a.
Compute the projection velocity and use it to determine the angular momentum.

The equation I have from the notes is velocity=sqrt(-aF(a)/m) assuming that r=a. Since the projection velocity is half that, v(projection)=(1/2)(sqrt(-aF(a)/m)). Then using the fact that v=h/a, we solve for h, and then use h=L/m to compute for L.
Are the steps correct?

 

[haruspex]

I've found the answer by using the equation of orbital motion -r=(r^4F(r))/(h^2m) where h=|L|/m. Rearranging the equation and setting r=a, I got L=msqrt(ka)

As BvU indicated, it's a bit simpler than that. Just write that the force equals the necessary centripetal force.
The second part seems even simpler. You have already found the angular momentum for a circular orbit at radius a. You are told the initial velocity is now half the velocity that corresponds to that angular momentum. So how large is the angular momentum?

 

[spacetimedude]

As BvU indicated, it's a bit simpler than that. Just write that the force equals the necessary centripetal force.
The second part seems even simpler. You have already found the angular momentum for a circular orbit at radius a. You are told the initial velocity is now half the velocity that corresponds to that angular momentum. So how large is the angular momentum?

Would the angular momentum be just 1/2 of that of the angular momentum we found before? So (m/2)(sqrt(ka))

 

[haruspex]

Would the angular momentum be just 1/2 of that of the angular momentum we found before? So (m/2)(sqrt(ka))

Looks right to me. The only thing that bothers me is that it seems too easy.

 

[spacetimedude]

Looks right to me. The only thing that bothers me is that it seems too easy.

Great! Thank you :) It gets a bit trickier. I was able to derive the radial equation of motion as r''-(9/4)ka/r^3=-3k/r^2 using the equation r''-h^2/r^3=Fr(r)/m and rearranging a bit.

Then the next question asks to define u(θ)=1/r(t) and show that r''-(9/4)ka/r^3=-3k/r^2 could be expressed as u''(θ)+9u(θ)=12/a.
Do I just let r(t)=1/u(θ), differentiate it twice to find r''(t) in terms of u(θ) to substitute into the equation of motion? Even though u and r have different values, namely θ and t, am I allowed to just use chain rule? For example, would r'(t)=-u'(θ)/u(θ)^2? I get an unusual coefficient for r'' if I do it that way.

The question after that asks to solve the equation obtained (u''(θ)+9u(θ)=12/a) and show that the particle follows the orbit:
r=3a/(r-cos(3θ)).
I am not quite sure how to start this one.
Any help will be appreciated as always!

 

[haruspex]

Do I just let r(t)=1/u(θ), differentiate it twice to find r''(t) in terms of u(θ)

Yes, but you have to be consistent about the independent variable with respect to which you are differentiating. If wrt t then terms like dθ/dt will appear: (d/dt)u = (du/dθ)(dθ/dt). So you need to find a relationship between r, θ and t which will provide a substitution for dθ/dt.

solve the equation obtained (u''(θ)+9u(θ)=12/a)

You could try multiplying through by u'. Another approach is to drop the inhomogeneous term on the right of the equals and see if the equation reminds you of anything.