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4.7 HW 7

This HW in one PDF

4.7.1 Problem 1

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.

4.7.2 Problem 2

SOLUTION:

4.7.2.1 Part (1)

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 $$\fbox{$\frac{\delta r_1}{r_1}=\frac{\delta \left ( \frac{v_o}{v_c}\right ) }{\left ( \frac{v_o}{v_c}\right ) }\left ( 2\frac{\left ( r_1+r_o\right ) }{r_o}\right ) $}$$

4.7.2.2 Part (2)

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)

4.7.3 Problem 3

SOLUTION:

4.7.3.1 Part (1)

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 $$U_{eff}\left ( r\right ) =\frac{1}{2}\frac{l^{2}}{mr^{2}}-\int \frac{ce^{-rb}}{r^{2}}dr$$

4.7.3.2 Part (2)

$$L=\frac{1}{2}m\dot{r}^{2}+\frac{1}{2}mr^{2}\dot{\theta }^{2}-\int \frac{ce^{-rb}}{r^{2}}dr$$ 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 $$\begin{align*} m\left ( 2r\dot{r}\dot{\theta }+r^{2}\ddot{\theta }\right ) & =0\\ r^{2}\ddot{\theta }+2r\dot{r}\dot{\theta } & =0 \end{align*}$$

4.7.3.3 Part (3)

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

$$\fbox{$b<\frac{2}{a}$}$$

4.7.3.4 Part (4)

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

$$\fbox{$\psi =\pi \sqrt{\frac{1}{3-a\left ( 1+ab\right ) }}$}$$

4.7.4 Problem 4

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

$$\fbox{$\sqrt{\frac{2h}{g}}\left ( \frac{1+\varepsilon }{1-\varepsilon }\right ) $}$$

4.7.5 Problem 5

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

$$\fbox{$\alpha =\arctan \left ( \varepsilon \tan \theta \right ) $}$$

4.7.6 HW 7 key solution