Chapter 13.2, Problem 13.118P

To determine

(a)

Angular momentum per unit mass and energy per unit mass.

Answer to Problem 13.118P

We got

Angular momentum per unit mass $h=r_{\min }{v}_{\max }$

Energy per unit mass $\frac{E}{m}=(\frac{1}{2}{v}_{\max }{}^2-\frac{GM}{r_{\min }})$

Explanation of Solution

Given information:

Radius $r_{\min }$

Speed $v_{\max }$

Planet mass $M$

Concept used:

Following formulae will be used.

$h=\frac{H}{m}\text{ and }\frac{E}{m}=\frac{1}{m}(T+V)$

Calculation:

Angular momentum per unit mass,

$\begin{array}{l}h=\frac{H}{m}\\ h=\frac{r_{\min }m{v}_{\max }}{m}\\ h=r_{\min }{v}_{\max }\end{array}$

Energy per unit mass,

$\begin{array}{l}\frac{E}{m}=\frac{1}{m}(T+V)\\ \frac{E}{m}=\frac{1}{m}(\frac{1}{2}m{v}_{\max }{}^2-\frac{GMm}{r_{\min }})\\ \frac{E}{m}=(\frac{1}{2}{v}_{\max }{}^2-\frac{GM}{r_{\min }})\end{array}$

Conclusion:

We got

Angular momentum per unit mass $h=r_{\min }{v}_{\max }$

Energy per unit mass $\frac{E}{m}=(\frac{1}{2}{v}_{\max }{}^2-\frac{GM}{r_{\min }})$

To determine

(b)

Formula derivation.

Answer to Problem 13.118P

We got required formula as $\frac{1}{r_{\min }}=\frac{GM}{h^2}\left[1+\sqrt{1+\frac{2E}{m}{\left(\frac{h}{GM}\right)}^2}\right]$

Explanation of Solution

Given information:

Radius $r_{\min }$

Speed $v_{\max }$

Planet mass $M$

Concept used:

Following formulae will be used.

$h=\frac{H}{m}\text{ and }\frac{E}{m}=\frac{1}{m}(T+V)$

Calculation:

Angular momentum per unit mass,

$\begin{array}{l}h=\frac{H}{m}\\ h=\frac{r_{\min }m{v}_{\max }}{m}\\ h=r_{\min }{v}_{\max }\\ \frac{h}{r_{\min }}=v_{\max }\end{array}$

Energy per unit mass,

$\begin{array}{l}\frac{E}{m}=\frac{1}{m}(T+V)\\ \frac{E}{m}=\frac{1}{m}(\frac{1}{2}m{v}_{\max }{}^2-\frac{GMm}{r_{\min }})\\ \frac{E}{m}=\left[\frac{1}{2}{\left(\frac{h}{r_{\min }}\right)}^2-\frac{GM}{r_{\min }}\right]\end{array}$

By solving of quadratic equation and rearranging,

$\frac{1}{r_{\min }}=\frac{GM}{h^2}\left[1+\sqrt{1+\frac{2E}{m}{\left(\frac{h}{GM}\right)}^2}\right]$

Conclusion:

We got required formula as $\frac{1}{r_{\min }}=\frac{GM}{h^2}\left[1+\sqrt{1+\frac{2E}{m}{\left(\frac{h}{GM}\right)}^2}\right]$

To determine

(c)

The eccentricity of the trajectory.

Answer to Problem 13.118P

We got eccentricity as $\epsilon =\sqrt{1+\frac{2E}{m}{\left(\frac{h}{GM}\right)}^2}$

Explanation of Solution

Given information:

Radius $r_{\min }$

Speed $v_{\max }$

Planet mass $M$

Concept used:

Following formulae will be used.

$\frac{1}{r}=\frac{GM}{h^2}(1+\epsilon \cos \theta )$

Calculation:

We know when,

$\begin{array}{l}\text{When }\theta \text{=0 cos}\theta \text{=1 and }r=r_{\min },so\\ \frac{1}{r_{\min }}=\frac{GM}{h^2}(1+\epsilon )\end{array}$

Also,

$\frac{1}{r_{\min }}=\frac{GM}{h^2}\left[1+\sqrt{1+\frac{2E}{m}{\left(\frac{h}{GM}\right)}^2}\right]$

By comparing,

$\begin{array}{l}(1+\epsilon )=\left[1+\sqrt{1+\frac{2E}{m}{\left(\frac{h}{GM}\right)}^2}\right]\\ \epsilon =\sqrt{1+\frac{2E}{m}{\left(\frac{h}{GM}\right)}^2}\end{array}$

Conclusion:

We got eccentricity as $\epsilon =\sqrt{1+\frac{2E}{m}{\left(\frac{h}{GM}\right)}^2}$

To determine

(d)

To show:

the trajectory shape depending on the energy type.

Answer to Problem 13.118P

Conditions have been shown below for different shape of trajectory at different eccentricity.

Explanation of Solution

Given information:

Radius $r_{\min }$

Speed $v_{\max }$

Planet mass $M$

Concept used:

Following formulae will be used.

$\epsilon =\sqrt{1+\frac{2E}{m}{\left(\frac{h}{GM}\right)}^2}$

Calculation:

We know when,

$\text{Hyperbola if }\epsilon \text{>1 that is only if E>0}$

$\text{Parabola if }\epsilon =\text{1 that is only if E=0}$

$\text{Ellipse if }\epsilon <\text{1 that is only if E<0}$

For circular orbit $\epsilon =0$ ,

$\begin{array}{l}0=\sqrt{1+\frac{2E}{m}{\left(\frac{h}{GM}\right)}^2}\\ 1+\frac{2E}{m}{\left(\frac{h}{GM}\right)}^2=0\\ E=-\frac{m}{2}{\left(\frac{GM}{h}\right)}^2\\ \text{but, }{v}^2=\frac{GM}{r}\text{ for circular orbit \& h=vr}\\ \text{so, }E=-\frac{1}{2}\left(\frac{GMm}{r}\right)\end{array}$

Conclusion:

Conditions have been shown above for different shape of trajectory at different eccentricity.