# Thread: Central Force Motion Problem For the Final~~~

1. ## Central Force Motion Problem

Hi all, this is a final review question, there are so many parts from (a) to (j), but i think it includes all the details, so any advice and help would be awesome~ thanks a lot~~

A particle of mass m moves under the influence of a central force whose magnitude is given by f(r) = kr, where k > 0. (The particle is thus repelled by the force center.) At some instant $t_{0}$, the particle is observed to be at ( $r=r_{0}$ $\theta=\theta_{0}$) and moving with a speed $v_{0}$ in a direction that makes an angle $\gamma$ with the polar axis (i.e., the positive x-axis). Assume $\gamma$ is not equal to { $0,\pi,2\pi$} and $v_{0}^2$ is not equal to $\frac{k}{m}r_{0}^2$.
(a) Show that the polar equation for the orbit of the particle has the form $\frac{\alpha}{r^2}=1+\varepsilon\sin[{2(\theta-\theta_{0})+\Omega}]$
(b) Determine the values of the constants $\alpha$ and $\varepsilon$ in terms of the parameters m, k, E, and L.
(c) Determine the values of the constants $\alpha$ and $\varepsilon$ in terms of the parameters m, k, $r_{0},v_{0}$, and $\gamma$.
(d) Show that the orbit equation, in Cartesian coordinates, is a quadratic form A $x^2$ + Bxy + C $y^2$ + Dx + Fy + G = 0 and determine explicitly the constants A, B, C, D, F, and G in
terms of $\alpha,\varepsilon,\theta_{0}$
(e) The orbit of the particle is thus a conic section! Determine the type of conic section.
(f) Show that the conic section (whose precise nature you determined in part (e)) is one whose center is at the origin and whose axes are rotated by an angle $\beta$ (with respect to the polar axis)
where the value of $\beta$ is given by the equation $\tan(2\beta)=-\cot(\theta_{0})$
Thus, the orientation of the orbit (w.r.t. a set of fixed coordinate axes) depends only on $\theta_{0}$!
(g) Prove that the shape of the orbit is independent of $\theta_{0}$(i.e., $\theta_{0}$only affects the orientation).
(h) Find the distance of closest approach to the center of force (i.e. $r_{min}$).
(i) Determine the eccentricity $e$ of the orbit.
(j) How do parts (a)-(i) change if the restriction ( $v_{0}^2$ is not equal to $\frac{k}{m}r_{0}^2$) is removed?

NOTE: In parts (h) and (i), consider the cases E < 0 and E > 0 separately, and express your answers in terms of $\alpha$ and $\varepsilon$

2. Originally Posted by zorop
A particle of mass m moves under the influence of a central force whose magnitude is given by f(r) = kr, where k > 0. (The particle is thus repelled by the force center.) At some instant $t_{0}$, the particle is observed to be at ( $r=r_{0}$ $\theta=\theta_{0}$) and moving with a speed $v_{0}$ in a direction that makes an angle $\gamma$ with the polar axis (i.e., the positive x-axis). Assume $\gamma$ is not equal to { $0,\pi,2\pi$} and $v_{0}^2$ is not equal to $\frac{k}{m}r_{0}^2$.
(a) Show that the polar equation for the orbit of the particle has the form $\frac{\alpha}{r^2}=1+\varepsilon\sin[{2(\theta-\theta_{0})}]$

The radial and transverse equations of motion are
$f(r) = m(\ddot{r} - r\dot{\theta}^2),\ \ r^2\dot{\theta} = H$ (H is a constant). Solving these is not easy. I'm guessing that you're supposed to know enough about orbital motion to know that these equations imply that $\frac{d^2s}{d\theta^2} + s = -\frac{f(1/s)}{H^2s^2m}$, where s=1/r. (You can find a derivation of that equation here—see Section 2 on that page.)

You are given that f(r) = kr, so the equation becomes $\frac{d^2s}{d\theta^2} + s = -\frac{k}{H^2s^3m}$. I don't know how to solve that differential equation. But looking at part (a) of the question, you can see that the solution is supposed to be of the form $s^2 = \alpha^{-1}\bigl(1+\varepsilon\sin[{2(\theta-\theta_{0})}]\bigr)$. -----(*)

So it makes sense to introduce a new variable $u=s^2$. Then $s=u^{1/2}$, with derivatives $s' = \tfrac12u^{-1/2}u'$ and $s'' = -\tfrac14u^{-3/2}(u')^2 + \tfrac12u^{-1/2}u''$ (where dashes indicate differentiation with respect to theta). The differential equation can then be written in terms of u as $2uu'' - (u')^2 + 4u^2 + \tfrac{k}{mH^2} = 0$. I don't know of a constructive way to solve that differential equation either. But by substituting the formula (*) for u, you can verify that it does indeed satisfy the equation, subject to some condition like $4(1-\varepsilon^2) + \frac{\alpha^2k}{mH^2} = 0$. -----(**)

Originally Posted by zorop
(b) Determine the values of the constants $\alpha$ and $\varepsilon$ in terms of the parameters m, k, E, and L.
(c) Determine the values of the constants $\alpha$ and $\varepsilon$ in terms of the parameters m, k, $r_{0},v_{0}$, and $\gamma$.

You haven't said what the constants E and L are. But the equation (**) above, together with the initial conditions of the problem, ought to give the answers here.

Originally Posted by zorop
(d) Show that the orbit equation, in Cartesian coordinates, is a quadratic form A $x^2$ + Bxy + C $y^2$ + Dx + Fy + G = 0 and determine explicitly the constants A, B, C, D, F, and G in
terms of $\alpha,\varepsilon,\theta_{0}$

The equation
$\frac{\alpha}{r^2}=1+\varepsilon\sin[{2(\theta-\theta_{0})}]$ can be written $\alpha= r^2 + \varepsilon2r\sin(\theta-\theta_{0})r\cos$ $(\theta-\theta_{0})$, which you can put into Cartesian form in the usual way ( $r^2=x^2+y^2$, $x=r\cos\theta$, $y=r\sin\theta$).

Originally Posted by zorop
(e) The orbit of the particle is thus a conic section! Determine the type of conic section.

Intuition suggests a hyperbola, since the force is repulsive.

3. Thanks a lot, I just received some messages from the professor on the corrections.

There is an additional phase factor in (a); call this additional phase Omega. Now note the following:

(1) Omega depends on all parameters EXCEPT $\theta_{0}$;

(2) in part (d), the constants A, B, etc... will then also depend on Omega;

(3) Part(f) is rendered incorrect by the addition of Omega in part (a) replace $\theta_{0}$ by $(\theta_{0}-\Omega/2)$.

Thanks very much~

4. For part (b) and (c), E is the total mechanical energy and L is the angular momentum. For the formulas,
E = $\frac{1}{2}mv^2 + V(r)$
L = $mr^2d\theta/dt$ = $mr^2v_{\theta}$

5. I think Kepler's equaitons should help~ $\frac{\alpha}{r}=1+\varepsilon\cos\theta$

6. Originally Posted by zorop
I think Kepler's equaitons should help~ $\frac{\alpha}{r}=1+\varepsilon\cos\theta$
No, Kepler's equations apply where there is an inverse square law of attraction, but what you have here is a linear law of repulsion (f(r)=kr).

7. yeah, i think i get it, thanks~ the result is something like

$\alpha=\frac{L^2}{mE}$
$\varepsilon=-\sqrt{1+\frac{kL^2}{mE^2}}$

8. do you mean that B=D=F=0? Because there is no xy, x and y terms in the equations.