SOLUTION:
Using polar coordinates. The position vector of the particle is\begin{equation} \vec{r}=r\hat{r}+r\theta \hat{\theta } \tag{1} \end{equation} We now find the Lagrangian\begin{align*} T & =\frac{1}{2}m\left ( \dot{r}^{2}+r^{2}\dot{\theta }^{2}\right ) \\ U & =V\left ( r\right ) \\ L & =\frac{1}{2}m\left ( \dot{r}^{2}+r^{2}\dot{\theta }^{2}\right ) -V\left ( r\right ) \end{align*}
Since we are asked about the angular momentum part, we will just find the equation of motion for the \(\theta \) generalized coordinates. \begin{align*} \frac{\partial L}{\partial \theta } & =0\\ \frac{\partial L}{\partial \dot{\theta }} & =mr^{2}\dot{\theta } \end{align*}
Hence the EQM is\[ \frac{d}{dt}\left ( mr^{2}\dot{\theta }\right ) =Q_{\theta }\] Where \(Q_{\theta }\) is the generalized force corresponding to generalized coordinate \(\theta \). From (1)\[ d\vec{r}=dr\hat{r}+rd\theta \hat{\theta }\] Hence\begin{align*} \frac{d\vec{r}}{dt} & =\frac{dr}{dt}\hat{r}+r\frac{d\theta }{dt}\hat{\theta }\\ & =\dot{r}\hat{r}+r\dot{\theta }\hat{\theta } \end{align*}
Therefore, the drag force can be written as \begin{align} \vec{F} & =-mk\frac{d\vec{r}}{dt}\nonumber \\ & =-mk\left ( \dot{r}\hat{r}+r\dot{\theta }\hat{\theta }\right ) \tag{2} \end{align}
Applying the definition of \(Q_{\theta }=\vec{F}\cdot \frac{\partial \vec{r}}{\partial \theta }\) gives\begin{align} Q_{\theta } & =-mk\left ( \dot{r}\hat{r}+r\dot{\theta }\hat{\theta }\right ) \cdot \frac{\partial }{\partial \theta }\left ( r\hat{r}+r\theta \hat{\theta }\right ) \nonumber \\ & =-mk\left ( \dot{r}\hat{r}+r\dot{\theta }\hat{\theta }\right ) \cdot \left ( r\hat{\theta }\right ) \nonumber \\ & =-mkr^{2}\dot{\theta } \tag{3} \end{align}
Now that we found \(Q_{\theta }\), the EQM is\[ \frac{d}{dt}\left ( mr^{2}\dot{\theta }\right ) =-mkr^{2}\dot{\theta }\] We notice the same term on both sides (but for a constant \(k\)). The above is the same as \[ \frac{d}{dt}\left ( Z\right ) =-kZ \] The solution must be exponential \(Z=e^{-kt}+C\) where \(C\) is some constant. This means\[ mr^{2}\dot{\theta }=e^{-kt}+C \] But \(mr^{2}\dot{\theta }\) is the angular momentum. Hence, for positive \(k\), the angular momentum decays exponentially with time.
SOLUTION:
From class notes, we found \[ \frac{v_{o}}{v_{c}}=\sqrt{\frac{2r_{1}}{r_{1}+r_{o}}}=\sqrt{\frac{2}{1+\frac{r_{o}}{r_{1}}}}\] Where \(v_{c}\) is the velocity in the circular orbit just before speed boost, and \(v_{o}\) is the speed at the perigee of the ellipse just after the speed boost, and \(r_{0}\) is the perigee distance and \(r_{1}\) is the apogee distance. We need to find \(\frac{\delta \left ( \frac{v_{o}}{v_{c}}\right ) }{\left ( \frac{v_{o}}{v_{c}}\right ) }\). To make the calculation easier, let \(\frac{v_{o}}{v_{c}}=z\). Then we have\[ z=\left ( \frac{2}{1+\frac{r_{o}}{r_{1}}}\right ) ^{\frac{1}{2}}\] Hence\[ \frac{\delta z}{\delta r_{1}}=\frac{1}{2}\frac{1}{\left ( \frac{2}{1+\frac{r_{o}}{r_{1}}}\right ) ^{\frac{1}{2}}}\frac{\delta }{\delta r_{1}}\left ( \frac{2}{1+\frac{r_{o}}{r_{1}}}\right ) \] But \(\left ( \frac{2}{1+\frac{r_{o}}{r_{1}}}\right ) ^{\frac{1}{2}}=z\) so the above becomes\begin{align*} \frac{\delta z}{\delta r_{1}} & =\frac{1}{2}\frac{1}{z}\frac{\delta }{\delta r_{1}}\left ( \frac{2}{1+\frac{r_{o}}{r_{1}}}\right ) \\ & =\frac{1}{2}\frac{1}{z}\left ( 2\frac{\delta }{\delta r_{1}}\left ( 1+\frac{r_{o}}{r_{1}}\right ) ^{-1}\right ) \\ & =\frac{1}{2}\frac{1}{z}\left ( 2\left ( -1\right ) \left ( 1+\frac{r_{o}}{r_{1}}\right ) ^{-2}\frac{\delta }{\delta r_{1}}\left ( \frac{r_{o}}{r_{1}}\right ) \right ) \\ & =\frac{1}{2}\frac{1}{z}\left ( 2\left ( -1\right ) \left ( 1+\frac{r_{o}}{r_{1}}\right ) ^{-2}\left ( -r_{o}\right ) r_{1}^{-2}\right ) \\ & =\frac{1}{2}\frac{1}{z}\left ( \frac{2}{\left ( 1+\frac{r_{o}}{r_{1}}\right ) ^{2}}\frac{r_{o}}{r_{1}^{2}}\right ) \end{align*}
Since \(\frac{2}{\left ( 1+\frac{r_{o}}{r_{1}}\right ) }=z^{2}\) the above simplifies to\begin{align*} \frac{\delta z}{\delta r_{1}} & =\frac{1}{2}\frac{1}{z}\left ( z^{2}\frac{1}{\left ( 1+\frac{r_{o}}{r_{1}}\right ) }\frac{r_{o}}{r_{1}^{2}}\right ) \\ & =\frac{1}{2}z\frac{r_{o}}{r_{1}^{2}\left ( 1+\frac{r_{o}}{r_{1}}\right ) }\\ & =\frac{1}{2}z\frac{r_{o}}{r_{1}\left ( r_{1}+r_{o}\right ) } \end{align*}
We want to find \(\frac{\delta z}{z}\), therefore the above can be written as\[ \frac{\delta z}{z}=\frac{\delta r_{1}}{r_{1}}\frac{1}{2}\frac{r_{o}}{\left ( r_{1}+r_{o}\right ) }\] Or in terms of \(\frac{\delta r_{1}}{r_{1}}\) the above becomes \[ \frac{\delta r_{1}}{r_{1}}=\frac{\delta z}{z}\left ( 2\frac{\left ( r_{1}+r_{o}\right ) }{r_{o}}\right ) \] Since \(z=\frac{v_{o}}{v_{c}}\), the reduces to\[ \fbox{$\frac{\delta r_1}{r_1}=\frac{\delta \left ( \frac{v_o}{v_c}\right ) }{\left ( \frac{v_o}{v_c}\right ) }\left ( 2\frac{\left ( r_1+r_o\right ) }{r_o}\right ) $}\]
For \(\frac{\delta \left ( \frac{v_{o}}{v_{c}}\right ) }{\left ( \frac{v_{o}}{v_{c}}\right ) }=0.01\) then\[ \frac{\delta r_{1}}{r_{1}}=0.01\left ( 2\frac{\left ( r_{1}+r_{o}\right ) }{r_{o}}\right ) \] Using \(r_{0}=\frac{1}{60}r_{1}\) in the above gives\begin{align*} \frac{\delta r_{1}}{r_{1}} & =0.01\left ( 2\frac{\left ( r_{1}+\frac{1}{60}r_{1}\right ) }{\frac{1}{60}r_{1}}\right ) \\ & =1.22 \end{align*}
This means that \(\delta r_{1}\) is \(22\%\) of \(r_{1}\). The spacecraft will miss the moon by \(22\%\) of \(r_{1}\). (This seems like a big miss for such small speed boost error)
SOLUTION:
One way to find \(U_{eff}\left ( r\right ) \) is to find the Largrangian \(L\) and pick the terms in it that have \(r\) without time derivative in them. \[ T=\frac{1}{2}m\dot{r}^{2}+\frac{1}{2}mr^{2}\dot{\theta }^{2}\] To find \(U\left ( r\right ) \), since we are given \(f\left ( r\right ) \) and since \(f\left ( r\right ) =-\frac{\partial U\left ( r\right ) }{\partial r}\), then \begin{align*} U\left ( r\right ) & =-\int f\left ( r\right ) dr\\ & =\int \frac{ce^{-rb}}{r^{2}}dr \end{align*}
Hence \begin{align*} L & =T-U\\ & =\frac{1}{2}m\dot{r}^{2}+\frac{1}{2}mr^{2}\dot{\theta }^{2}-\int \frac{ce^{-rb}}{r^{2}}dr \end{align*}
Hence \[ U_{eff}\left ( r\right ) =\frac{1}{2}mr^{2}\dot{\theta }^{2}-\int \frac{ce^{-rb}}{r^{2}}dr \] In terms of \(l=mr^{2}\dot{\theta }\), the above can be written as\[ U_{eff}\left ( r\right ) =\frac{1}{2}l\dot{\theta }-\int \frac{ce^{-rb}}{r^{2}}dr \] Or, it can also be written, as done in class notes, as \[ U_{eff}\left ( r\right ) =\frac{1}{2}\frac{l^{2}}{mr^{2}}-\int \frac{ce^{-rb}}{r^{2}}dr \]
\[ L=\frac{1}{2}m\dot{r}^{2}+\frac{1}{2}mr^{2}\dot{\theta }^{2}-\int \frac{ce^{-rb}}{r^{2}}dr \] Hence\begin{align*} \frac{\partial L}{\partial r} & =mr\dot{\theta }^{2}-\frac{ce^{-rb}}{r^{2}}\\ \frac{\partial L}{\partial \dot{r}} & =m\dot{r} \end{align*}
The equation of motion for \(r\) is\begin{align*} m\ddot{r}-\left ( mr\dot{\theta }^{2}-\frac{ce^{-rb}}{r^{2}}\right ) & =0\\ m\ddot{r}-mr\dot{\theta }^{2}+\frac{ce^{-rb}}{r^{2}} & =0\\ m\ddot{r}-mr\dot{\theta }^{2} & =F\left ( r\right ) \end{align*}
Written in terms of angular momentum, since \(\dot{\theta }=\frac{l}{mr^{2}}\) (integral of motion) where \(l\) is the angular momentum, the above becomes\begin{equation} m\ddot{r}-\frac{l^{2}}{mr^{3}}=F\left ( r\right ) \tag{1} \end{equation} For \(\theta \), \begin{align*} \frac{\partial L}{\partial \theta } & =0\\ \frac{\partial L}{\partial \dot{\theta }} & =mr^{2}\dot{\theta } \end{align*}
The equation of motion for \(\theta \) is\[ \frac{d}{dt}\left ( mr^{2}\dot{\theta }\right ) =C \] Where \(C\) is some constant. The full EQM for \(\theta \) is\begin{align*} m\left ( 2r\dot{r}\dot{\theta }+r^{2}\ddot{\theta }\right ) & =0\\ r^{2}\ddot{\theta }+2r\dot{r}\dot{\theta } & =0 \end{align*}
To check for stability, since this is circular orbit, the radius is constant, say \(a\). Then we perturb it by replacing \(a\) by \(x+a\) where \(x\ll a\) in the equation of motion \(m\ddot{r}-\frac{l^{2}}{mr^{3}}=F\left ( r\right ) \) and it becomes\begin{align*} m\ddot{x}-\frac{l^{2}}{m\left ( x+a\right ) ^{3}} & =F\left ( x+a\right ) \\ m\ddot{x} & =\frac{l^{2}\left ( x+a\right ) ^{-3}}{m}+F\left ( a+x\right ) \end{align*}
Since \(x\ll a\), we expand \(\left ( x+a\right ) ^{-3}\) in Binomial and obtain\begin{align*} m\ddot{x} & =\frac{l^{2}}{ma^{3}}\left ( 1+\frac{x}{a}\right ) ^{-3}+F\left ( a+x\right ) \\ & \approx \frac{l^{2}}{ma^{3}}\left ( 1-\frac{3x}{a}+\cdots \right ) +\overset{\text{Taylor expansion}}{\overbrace{F\left ( a\right ) +xF^{\prime }\left ( a\right ) +\cdots }} \end{align*}
Since circular orbit, then \(\ddot{r}=0\) and the EQM motion becomes \(-\frac{l^{2}}{ma^{3}}=F\left ( a\right ) \). Using this to replace \(\frac{l^{2}}{ma^{3}}\) with in the above expression we find\begin{align*} m\ddot{x} & \approx -F\left ( a\right ) \left ( 1-\frac{3x}{a}\right ) +F\left ( a\right ) +xF^{\prime }\left ( a\right ) \\ & =-F\left ( a\right ) +F\left ( a\right ) \frac{3x}{a}+F\left ( a\right ) +xF^{\prime }\left ( a\right ) \\ & =F\left ( a\right ) \frac{3x}{a}+xF^{\prime }\left ( a\right ) \end{align*}
Hence\begin{align*} m\ddot{x}+\left ( -F\left ( a\right ) \frac{3x}{a}-xF^{\prime }\left ( a\right ) \right ) & =0\\ m\ddot{x}+\left ( -\frac{3}{a}F\left ( a\right ) -F^{\prime }\left ( a\right ) \right ) x & =0 \end{align*}
This perturbation motion is stable if \(\left ( -\frac{3}{a}F\left ( a\right ) -F^{\prime }\left ( a\right ) \right ) >0.\) But \(F\left ( a\right ) =-\frac{ce^{-ba}}{a}\) and \(F^{\prime }\left ( a\right ) =\frac{ce^{-ab}}{a^{2}}+\frac{bce^{-ab}}{a}\), hence\begin{align*} \Delta & =-\frac{3}{a}F\left ( a\right ) -F^{\prime }\left ( a\right ) \\ & =-\frac{3}{a}\left ( -\frac{ce^{-ba}}{a}\right ) -\left ( \frac{ce^{-ab}}{a^{2}}+\frac{bce^{-ab}}{a}\right ) \end{align*}
We want the above to be positive for stability. Simplifying gives\begin{align*} \Delta & =\frac{3ce^{-ba}}{a^{2}}-\frac{ce^{-ab}}{a^{2}}-\frac{bce^{-ab}}{a}\\ & =\frac{2ce^{-ba}}{a^{2}}-\frac{bce^{-ab}}{a}\\ & =\frac{2ce^{-ba}-abce^{-ab}}{a^{2}}\\ & =\frac{ce^{-ba}}{a^{2}}\left ( 2-ab\right ) \end{align*}
Therefore, we want \(\left ( 2-ab\right ) >0\) or \(2>ab\) or \[ \fbox{$b<\frac{2}{a}$}\]
The angle \(\psi \) is found from\begin{equation} \psi =\frac{T_{p}}{2}\dot{\theta }\tag{1} \end{equation} Where \(T_{p}\) is the period of oscillation due to the perturbation from the exact circular orbit, and \(\dot{\theta }\) is the angular velocity on the circular orbit. But\begin{equation} \dot{\theta }\approx \frac{l}{ma^{2}}\tag{2} \end{equation} But from part(3) we found that\begin{align*} -\frac{l^{2}}{ma^{3}} & =F\left ( a\right ) \\ l & =\sqrt{-F\left ( a\right ) ma^{3}} \end{align*}
Therefore (2) becomes\begin{align*} \dot{\theta } & \approx \frac{1}{ma^{2}}\sqrt{-F\left ( a\right ) ma^{3}}\\ & =\sqrt{\frac{-F\left ( a\right ) }{ma}} \end{align*}
We now find \(T_{p}\). Since the perturbation equation of motion, from part (3) is \(m\ddot{x}+\left ( -\frac{3}{a}F\left ( a\right ) -F^{\prime }\left ( a\right ) \right ) x=0\), which is of the form \[ \ddot{x}+\overset{\omega _{0}^{2}}{\overbrace{\left ( \frac{-\frac{3}{a}F\left ( a\right ) -F^{\prime }\left ( a\right ) }{m}\right ) }}x=0 \] Then, the natural frequency is \(\omega =\sqrt{\frac{\left ( -\frac{3}{a}F\left ( a\right ) -F^{\prime }\left ( a\right ) \right ) }{m}}\), therefore\begin{align*} \frac{2\pi }{T_{p}} & =\sqrt{\frac{-\frac{3}{a}F\left ( a\right ) -F^{\prime }\left ( a\right ) }{m}}\\ T_{p} & =2\pi \sqrt{\frac{m}{-\frac{3}{a}F\left ( a\right ) -F^{\prime }\left ( a\right ) }} \end{align*}
Equation (1) now becomes\begin{align*} \psi & =\frac{T_{p}}{2}\dot{\theta }\\ & =\pi \sqrt{\frac{m}{-\frac{3}{a}F\left ( a\right ) -F^{\prime }\left ( a\right ) }}\sqrt{\frac{-F\left ( a\right ) }{ma}}\\ & =\pi \sqrt{\frac{-F\left ( a\right ) }{-3F\left ( a\right ) -aF^{\prime }\left ( a\right ) }}\\ & =\pi \sqrt{\frac{F\left ( a\right ) }{3F\left ( a\right ) +aF^{\prime }\left ( a\right ) }} \end{align*}
But \(F\left ( a\right ) =-\frac{ce^{-ba}}{a^{2}}\) and \(F^{\prime }\left ( a\right ) =\frac{ce^{-ab}}{a^{2}}+\frac{bce^{-ab}}{a}\) then the above becomes\begin{align*} \psi & =\pi \sqrt{\frac{-\frac{ce^{-ba}}{a^{2}}}{3F\left ( a\right ) +aF^{\prime }\left ( a\right ) }}\\ & =\pi \sqrt{\frac{-\frac{ce^{-ba}}{a^{2}}}{3\left ( -\frac{ce^{-ba}}{a^{2}}\right ) +a\left ( \frac{ce^{-ab}}{a^{2}}+\frac{bce^{-ab}}{a}\right ) }}\\ & =\pi \sqrt{\frac{-\frac{ce^{-ba}}{a^{2}}}{-3\frac{ce^{-ba}}{a^{2}}+\left ( \frac{ce^{-ab}+abce^{-ab}}{a}\right ) }}\\ & =\pi \sqrt{\frac{-ce^{-ba}}{-3ce^{-ba}+\left ( ace^{-ab}+a^{2}bce^{-ab}\right ) }}\\ & =\pi \sqrt{\frac{-1}{-3+a+a^{2}b}} \end{align*}
Hence \[ \fbox{$\psi =\pi \sqrt{\frac{1}{3-a\left ( 1+ab\right ) }}$}\]
SOLUTION:
The first time the ball falls from height \(h\) it will have speed of \(v_{1}=\sqrt{2gh}\) just before hitting the platform, which is found using\[ mgh=\frac{1}{2}mv_{1}^{2}\] On bouncing back, it will have speed of \(v_{1}^{\prime }=\varepsilon \sqrt{2gh}\). It will then travel up a distance of \(h_{1}=\varepsilon ^{2}h\) which is found by solving for \(h_{1}\)from\[ mgh_{1}=\frac{1}{2}m\left ( v_{1}^{\prime }\right ) ^{2}\] The second time it it falls back it will have speed of \(v_{2}=\varepsilon \sqrt{2gh_{1}}\). When it bounces back up, it will have speed \(v_{2}^{\prime }=\varepsilon ^{2}\sqrt{2gh_{1}}\) and now it will travel up a distance of \(h_{2}=\varepsilon ^{4}h\) which is found by solving for \(h_{2}\) from\[ mgh_{2}=\frac{1}{2}m\left ( v_{2}^{\prime }\right ) ^{2}\] This process will continue until the ball stops. We see that the distance travelled at each bouncing is\[ \Delta =\left \{ h,2\varepsilon ^{2}h,2\varepsilon ^{4}h,2\varepsilon ^{6}h,\cdots ,2\varepsilon ^{2n}h\right \} \] We added \(2\) to each bounce after the first one to count for going up and then coming down the same distance. The first time it will only have one \(h\). We now can calculate total distance travelled \(\Delta \) as\begin{align*} \Delta & =h+2\varepsilon ^{2}h+2\varepsilon ^{4}h+\cdots \\ & =h\left ( 1+2\varepsilon ^{2}+2\varepsilon ^{4}+\cdots \right ) \end{align*}
The above can be written as\begin{equation} \Delta =h\left ( 2+2\varepsilon ^{2}+2\varepsilon ^{4}+\cdots \right ) -h \tag{1} \end{equation} But since \(\varepsilon \leq 1\) the series sum is\[ 2+2\varepsilon ^{2}+2\varepsilon ^{4}+\cdots =2{\displaystyle \sum \limits _{n=0}^{\infty }} \varepsilon ^{2n}=2\frac{1}{1-\varepsilon ^{2}}\] Therefore (1) becomes\begin{align*} \Delta & =\frac{2h}{1-\varepsilon ^{2}}-h\\ & =\frac{2h-h\left ( 1-\varepsilon ^{2}\right ) }{1-\varepsilon ^{2}}\\ & =\frac{2h-h+h\varepsilon ^{2}}{1-\varepsilon ^{2}} \end{align*}
Hence total distance is\[ \fbox{$\frac{h\left ( 1+\varepsilon ^2\right ) }{1-\varepsilon ^2}$}\] To find the total time of all ball bounces, we need to find the time it takes to travel in each bounce. The time it takes to fall distance \(h\) is \(\sqrt{\frac{2h}{g}}\), using the information we found about each \(h_{i}\) from above, we now set up the sequence of times we we did for distances\[ \Delta _{time}=\left \{ \sqrt{\frac{2h}{g}},2\sqrt{\frac{2\varepsilon ^{2}h}{g}},2\sqrt{\frac{2\varepsilon ^{4}h}{g}},2\sqrt{\frac{2\varepsilon ^{6}h}{g}},\cdots \right \} \] Adding the times gives\begin{align*} \Delta & =\sqrt{\frac{2h}{g}}+2\sqrt{\frac{2\varepsilon ^{2}h}{g}}+2\sqrt{\frac{2\varepsilon ^{4}h}{g}}+2\sqrt{\frac{2\varepsilon ^{6}h}{g}}\\ & =\sqrt{\frac{2h}{g}}\left ( 1+2\varepsilon +2\varepsilon ^{2}+2\varepsilon ^{3}+2\varepsilon ^{4}\cdots \right ) \\ & =\sqrt{\frac{2h}{g}}\left ( 2+2\varepsilon +2\varepsilon ^{2}+2\varepsilon ^{3}+2\varepsilon ^{4}\cdots \right ) -\sqrt{\frac{2h}{g}}\\ & =\sqrt{\frac{2h}{g}}\sum _{n=0}^{\infty }2\varepsilon ^{n}-\sqrt{\frac{2h}{g}} \end{align*}
But \(2\sum _{n=0}^{\infty }\varepsilon ^{n}=2\frac{1}{1-\varepsilon }\), hence the above becomes\begin{align*} \Delta & =\sqrt{\frac{2h}{g}}\frac{2}{1-\varepsilon }-\sqrt{\frac{2h}{g}}\\ & =\sqrt{\frac{2h}{g}}\left ( \frac{2}{1-\varepsilon }-1\right ) \\ & =\sqrt{\frac{2h}{g}}\left ( \frac{2-\left ( 1-\varepsilon \right ) }{1-\varepsilon }\right ) \end{align*}
Hence total time is\[ \fbox{$\sqrt{\frac{2h}{g}}\left ( \frac{1+\varepsilon }{1-\varepsilon }\right ) $}\]
SOLUTION:
First we make a diagram showing the geometry involved
We resolve the incoming velocity into its \(x,y\,\ \)components and apply conservation of linear momentum to each part. The vertical component remain the same after collision since it is parallel to the wall. Hence\[ v_{y}^{\prime }=v_{y}=v\cos \theta \] While the \(x\) component will change to \[ v_{x}^{\prime }=\varepsilon v_{x}=\varepsilon v\sin \theta \] By definition of \(\varepsilon .\) Therefore we see that after collision\begin{align*} \tan \alpha & =\frac{\varepsilon v\sin \theta }{v\cos \theta }\\ & =\varepsilon \tan \theta \end{align*}
Hence \[ \fbox{$\alpha =\arctan \left ( \varepsilon \tan \theta \right ) $}\]
