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3.2 second midterm

3.2.1 practice exam

3.2.1.1 questions

3.2.1.2 my solution to practice exam

Problem 1 SOLUTION:

Conservation of Linear momentum gives

$$\begin{equation} m_{1}v_{1}=m_{1}v_{1}^{\prime }+m_{2}v_{2}^{\prime }\tag{1} \end{equation}$$ Conservation of energy gives $$\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{1}\left ( v_{1}^{\prime }\right ) ^{2}+\frac{1}{2}m_{2}\left ( v_{2}^{\prime }\right ) ^{2}+Q$$ But since this is elastic collision, then $Q=0$. Hence the above becomes $$\begin{equation} m_{1}v_{1}^{2}=m_{1}\left ( v_{1}^{\prime }\right ) ^{2}+m_{2}\left ( v_{2}^{\prime }\right ) ^{2}\tag{2} \end{equation}$$ The goal now is to eliminate $v_{2}$ from (1) and (2) and solve for $v_{1}^{\prime }$ in terms of $v_{1}$ to be able to answer the question. Let $\frac{m_{2}}{m_{1}}=\gamma$, then (1,2) can be written as $$\begin{align} v_{1} & =v_{1}^{\prime }+\gamma v_{2}^{\prime }\tag{A1}\\ v_{1}^{2} & =\left ( v_{1}^{\prime }\right ) ^{2}+\gamma \left ( v_{2}^{\prime }\right ) ^{2}\tag{A2} \end{align}$$

We now move the $m_{1}$ terms to one side,

$$\begin{align} v_{1}-v_{1}^{\prime } & =\gamma v_{2}^{\prime }\tag{C1}\\ v_{1}^{2}-\left ( v_{1}^{\prime }\right ) ^{2} & =\gamma \left ( v_{2}^{\prime }\right ) ^{2}\tag{C2} \end{align}$$

Dividing (2) by (1), using long division (this step is tricky, must be careful), gives

$$\begin{align} \frac{v_{1}^{2}-\left ( v_{1}^{\prime }\right ) ^{2}}{v_{1}-v_{1}^{\prime }} & =v_{2}^{\prime }\nonumber \\ v_{1}+v_{1}^{\prime } & =v_{2}^{\prime }\tag{3} \end{align}$$

We now replace $v_{2}^{\prime }$ in (C1) with what (3) giving

$$\begin{align} v_{1}-v_{1}^{\prime } & =\gamma \left ( v_{1}+v_{1}^{\prime }\right ) \nonumber \\ v_{1}-\gamma v_{1} & =\gamma v_{1}^{\prime }+v_{1}^{\prime }\nonumber \\ v_{1}\left ( 1-\gamma \right ) & =v_{1}^{\prime }\left ( 1+\gamma \right ) \nonumber \\ v_{1}^{\prime } & =v_{1}\frac{\left ( 1-\gamma \right ) }{\left ( 1+\gamma \right ) }\tag{4} \end{align}$$

We achieved our goal of finding $v_{1}^{\prime }$ in terms of $v_{1}$. Now to answer the question. The question is asking to find

$$\begin{equation} \Delta =\frac{T_{1}-T_{1}^{\prime }}{T_{1}}\tag{5} \end{equation}$$ Which is the fraction of kinetic energy loss of $m_{1}$. So now we calculate the above, and see if it gives the answer we are asked to show. $$\begin{align*} \Delta & =\frac{\frac{1}{2}m_{1}v_{1}^{2}-\frac{1}{2}m_{1}\left ( v_{1}^{\prime }\right ) ^{2}}{\frac{1}{2}m_{1}v_{1}^{2}}\\ & =\frac{v_{1}^{2}-\left ( v_{1}^{\prime }\right ) ^{2}}{v_{1}^{2}} \end{align*}$$

Using (4) into the above gives

$$\Delta =\frac{v_{1}^{2}-\left ( v_{1}\frac{\left ( 1-\gamma \right ) }{\left ( 1+\gamma \right ) }\right ) ^{2}}{v_{1}^{2}}$$ But $\gamma =\frac{m_{2}}{m_{1}}$, expanding the above gives $$\begin{align*} \Delta & =\frac{v_{1}^{2}-\left ( v_{1}\frac{\left ( 1-\frac{m_{2}}{m_{1}}\right ) }{\left ( 1+\frac{m_{2}}{m_{1}}\right ) }\right ) ^{2}}{v_{1}^{2}}\\ & =\frac{v_{1}^{2}-v_{1}^{2}\frac{\left ( 1-\frac{m_{2}}{m_{1}}\right ) ^{2}}{\left ( 1+\frac{m_{2}}{m_{1}}\right ) ^{2}}}{v_{1}^{2}}\\ & =1-\frac{\left ( 1-\frac{m_{2}}{m_{1}}\right ) ^{2}}{\left ( 1+\frac{m_{2}}{m_{1}}\right ) ^{2}}\\ & =1-\frac{\left ( m_{1}-m_{2}\right ) ^{2}}{\left ( m_{1}+m_{2}\right ) ^{2}}\\ & =\frac{\left ( m_{1}+m_{2}\right ) ^{2}-\left ( m_{1}-m_{2}\right ) ^{2}}{\left ( m_{1}+m_{2}\right ) ^{2}}\\ & =\frac{\left ( m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\right ) -\left ( m_{1}^{2}+m_{2}^{2}-2m_{1}m_{2}\right ) }{\left ( m_{1}+m_{2}\right ) ^{2}} \end{align*}$$

Simplifying

$$\begin{align*} \Delta & =\frac{4m_{1}m_{2}}{\left ( m_{1}+m_{2}\right ) ^{2}}\\ & =4\frac{m_{1}m_{2}}{\left ( m_{1}+m_{2}\right ) }\frac{1}{\left ( m_{1}+m_{2}\right ) } \end{align*}$$

But $m=\frac{m_{1}m_{2}}{m_{1}+m_{2}}$ which is the reduced mass, and $M=m_{1}+m_{2}$. So the above becomes

$$\Delta =\frac{4m}{M}$$ Which is the result we are asked to show.

Problem 2 SOLUTION:

Part(a)

Let $v_{1}$ be the speed in the lower circular orbit. Let $v_{2}$ be the speed at the perigee just after speed boost. Let $GM\equiv \mu$. Since

$$\begin{align*} v_{1} & =\sqrt{\frac{\mu }{r_{A}}}\\ v_{2} & =\sqrt{\mu \left ( \frac{2}{r_{A}}-\frac{1}{a}\right ) } \end{align*}$$

Where $a=\frac{r_{A}+r_{B}}{2}$, then $\frac{v_{2}}{v_{2}}$ can now be evaluated

$$\begin{align*} \frac{v_{2}}{v_{1}} & =\frac{\sqrt{\mu \left ( \frac{2}{r_{A}}-\frac{1}{a}\right ) }}{\sqrt{\frac{\mu }{r_{A}}}}\\ & =\sqrt{\frac{\mu \left ( \frac{2}{r_{A}}-\frac{1}{\frac{r_{A}+r_{B}}{2}}\right ) }{\frac{\mu }{r_{A}}}}\\ & =\sqrt{r_{A}\left ( \frac{2}{r_{A}}-\frac{2}{r_{A}+r_{B}}\right ) }\\ & =\sqrt{2-\frac{2r_{A}}{r_{A}+r_{B}}}\\ & =\sqrt{\frac{2\left ( r_{A}+r_{B}\right ) -2r_{A}}{r_{A}+r_{B}}}\\ & =\sqrt{\frac{2r_{B}}{r_{A}+r_{B}}} \end{align*}$$

Part(b)

Using the period for an ellipse given in the formulas and dividing this by half, since we are looking for half the period, then

$$\begin{align*} T_{p} & =\pi \sqrt{\frac{a^{3}}{G\left ( m_{1}+M_{earth}\right ) }}\\ & =\pi \sqrt{\frac{\left ( \frac{r_{A}+r_{B}}{2}\right ) ^{3}}{G\left ( m_{1}+M_{e}\right ) }} \end{align*}$$

Assuming the mass of the satellite ($m_{1}$) is much smaller than $M_{earth}$, then the above becomes

$$T_{p}=\frac{\pi }{\sqrt{GM_{e}}}\sqrt{\left ( \frac{r_{A}+r_{B}}{2}\right ) ^{3}}$$

Part(c)

The time it takes $B$ to travel one circle $\left ( 2\pi \right )$ is

$$T_{c}=2\pi \sqrt{\frac{r_{B}^{3}}{GM}}$$ Therefore, the angle $B$ travels during $T_{p}$ is found by the equating the ratios $$\frac{2\pi }{\alpha }\Leftrightarrow \frac{2\pi \sqrt{\frac{r_{B}^{3}}{GM}}}{T_{p}}$$ But $\theta _{0}=\pi -\alpha$ (assuming the diagram given, where $\alpha$ is the angle between $B$ and the apogee, while $\theta$ is the angle between $B$ and the perigee). Therefore we use the above to solve for $\theta _{0}$ $$\begin{align*} \frac{2\pi }{\pi -\theta _{0}} & =\frac{2\pi \sqrt{\frac{r_{B}^{3}}{GM}}}{T_{p}}\\ \frac{T_{p}}{\pi -\theta _{0}} & =\sqrt{\frac{r_{B}^{3}}{GM}}\\ \pi -\theta _{0} & =T_{p}\sqrt{\frac{GM}{r_{B}^{3}}}\\ \theta _{0} & =\pi -T_{p}\sqrt{\frac{GM}{r_{B}^{3}}}\\ & =\pi -\frac{\pi }{\sqrt{GM_{e}}}\sqrt{\left ( \frac{r_{A}+r_{B}}{2}\right ) ^{3}}\sqrt{\frac{GM}{r_{B}^{3}}} \end{align*}$$

Therefore

$$\theta _{0}=\pi \left ( 1-\sqrt{\left ( \frac{r_{A}+r_{B}}{2r_{B}}\right ) ^{3}}\right )$$

3.2.1.3 key solution to practice exam

3.2.2 Review Problems by TA

3.2.2.1 questions

3.2.2.2 key solution to review problems

3.2.3 Exam, Nov 16, 2015

3.2.3.1 questions

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