Closed trajectory in a central field of force; find the mass

[Nathanael]

Homework Statement

A particle moves along a closed trajectory in a central field of force where the particle's potential energy is U=kr² (k is a positive constant, r is the distance of the particle from the center O of the field). Find the mass of the particle if it's minimum distance from O equals r₁ and it's velocity at the point farthest from O equals v₂.

Homework Equations

$rv=r^2\dot \theta=\text{constant}\equiv c_a$
$0.5mv^2+kr^2=0.5m(\dot r^2+r^2\dot \theta^2)+kr^2=\text{constant}\equiv c_b$
$\dot r = \dot \theta \frac{dr}{d\theta}$
$\vec F=-\nabla U=-2k\vec r=m(\ddot r-r\dot \theta^2)\hat r$

The Attempt at a Solution

My attempt ignored the force equation (the last of my "relevant equations").
What I did was to eliminate the time dependency in the energy equation to get a differential equation between r and θ, like this:
$c_b=kr^2+\frac{mc_a^2}{2r^4}\big ( \frac{dr}{d\theta} \big )^2+\frac{mc_a^2}{2r^2}$
which is a separable equation:
$d\theta=\frac{c_adr}{r^2\sqrt{\frac{2}{m}(c_b-kr^2)-\frac{c_a^2}{r^2}}}$
I haven't tried, but I don't think I can solve that integral. At any rate, there must be a simpler way. What I am doing will lead to the arbitrary path of an object in this field (the constants c_a and c_b are defined by the initial condition). I don't think it is necessary to solve for the entire trajectory like this... but I don't know what else to do.

[edited to include that k is a positive constant]

 

[TSny]

$rv=r^2\dot \theta=\text{constant}\equiv c_a$
$0.5mv^2+kr^2=0.5m(\dot r^2+r^2\dot \theta^2)+kr^2=\text{constant}\equiv c_b$

Use each of these two equations to relate $r_1, r_2, v_1$ and $v_2$. Combine the equations and simplify.

 

[Nathanael]

Use each of these two equations to relate $r_1, r_2, v_1$ and $v_2$. Combine the equations and simplify.

That's 3 unknowns with only two equations. We want to find the mass. r₂ and v₁ are unknown.

 

[andrewkirk]

It seems to me that more information is needed. There are three unknowns: $r_2,\ v_1$ and $m$. But there are only two equations - one from conservation of angular momentum and one from conservation of energy. Can we get another, independent, equation, or is the problem under-specified?

Edit: I see that Nathaneal has pointed this out. For some reason it didn't appear on my screen.

 

[Nathanael]

It seems to me that more information is needed. There are three unknowns: $r_2,\ v_1$ and $m$. But there are only two equations - one from conservation of angular momentum and one from conservation of energy. Can we get another, independent, equation, or is the problem under-specified?

I was thinking the same, but I think the final constraint may come from the fact that it is a "closed trajectory," because surely not all initial conditions will lead to a closed trajectory (right?). I'm just not sure how to use this constraint. (This is why I attempted to solve for the entire trajectory r(θ) in the OP by eliminating time.)

 

[Nathanael]

I just checked and the answer is supposed to be $m=2k(\frac{r_1}{v_2})^2$

This is what you get if you assume the path is circular, but I see no reason why this should be the answer in general. Perhaps the problem creator made a mistake with this one? (The problem statement is word for word.)

 

[haruspex]

I just checked and the answer is supposed to be $m=2k(\frac{r_1}{v_2})^2$

This is what you get if you assume the path is circular, but I see no reason why this should be the answer in general. Perhaps the problem creator made a mistake with this one? (The problem statement is word for word.)

No, it does not depend on the shape of the trajectory, and you do not need to solve the differential equation.
Just follow TSny's lead to obtain a differential equation that does not involve theta. From this you can obtain expressions for the min and max radii. You can use your energy equation to relate the maximum radius to the velocity at that point.

 

[TSny]

That's 3 unknowns with only two equations. We want to find the mass. r₂ and v₁ are unknown.

Just use the two equations
$rv=\text{constant}$
$0.5mv^2+kr^2=\text{constant}$.

The first equation tells you that $r_1v_1 = r_2v_2$. Similarly, you can get a relation involving $r_1, v_1, r_2, v_2$, and $m$ from the second equation. Due to the form of the potential energy, you will be able to solve for $m$ in terms of just $r_1$ and $v_2$.

 

[Nathanael]

$0.5m(\frac{r_2v_2}{r_1})^2+kr_1^2=0.5mv_2^2+kr_2^2$
$m\big(\frac{v_2}{r_1}\big)^2(r_2^2-r_1^2)=2k(r_2^2-r_1^2)$
$m=2k\big (\frac{r_1}{v_2}\big )^2$

@TSny How did you know it would cancel like this?

 

[TSny]

I didn't! I just wanted to see what you could conclude from the two equations. Bingo, the answer fell out unexpectedly.