Chapter 13.2, Problem 13.115P

To determine

(a)

Prove the given condition.

Answer to Problem 13.115P

We proved as $\sin {\varphi }_0=(1+\alpha )\sqrt{1-\frac{\alpha }{(1+\alpha )}{\left(\frac{v_{esc}}{v_0}\right)}^2}\text{ proved}$

Explanation of Solution

Given information:

Max altitude $\alpha R$

Initial velocity $v_0$

Radius of planet $\text{R}$

Launch angle ${\varphi }_0$

Escape velocity $v_{esc}$

Concept used:

Following formula will be used.

$T_A+V_A=T_B+V_B$

Calculation:

We know,

$T_A+V_A=T_B+V_B$

$\frac{1}{2}m{v}_0{}^2-\frac{g{R}^{2}m}{R}=\frac{1}{2}m{(v_B)}^2-\frac{g{R}^{2}m}{(1+\alpha )R}$

$\frac{1}{2}{v}_0{}^2-gR=\frac{1}{2}{(v_B)}^2-\frac{gR}{(1+\alpha )}$

Conservation of angular momentum,

$\begin{array}{l}(1+\alpha )Rm{v}_B=Rmv\sin {\varphi }_0\\ v_B=\frac{v_0\sin {\varphi }_0}{1+\alpha }\end{array}$

So,

$\begin{array}{l}$ \frac{1}{2}{v}_0{}^2-gR=\frac{1}{2}{(v_B)}^2-\frac{gR}{(1+\alpha )}$$ \frac{1}{2}{v}_0{}^2-gR=\frac{1}{2}{(\frac{v_0\sin {\varphi }_0}{1+\alpha })}^2-\frac{gR}{(1+\alpha )}$\\ \text{also,}\\ v_{esc}=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}\end{array}$

$\begin{array}{l}$ \frac{1}{2}{v}_0{}^2-\frac{v_{esc}{}^2}{2}=\frac{1}{2}{(\frac{v_0\sin {\varphi }_0}{1+\alpha })}^2-\frac{v_{esc}{}^2}{2(1+\alpha )}$$ \frac{1}{2}{v}_0{}^2-\frac{1}{2}{(\frac{v_0\sin {\varphi }_0}{1+\alpha })}^2=+\frac{v_{esc}{}^2}{2}-\frac{v_{esc}{}^2}{2(1+\alpha )}$v_0{}^2\left(1-\frac{{\sin }^2{\varphi }_0}{{(1+\alpha )}^2}\right)=v_{esc}{}^2\left(1-\frac{1}{(1+\alpha )}\right)\\ \sin {\varphi }_0=(1+\alpha )\sqrt{1-\frac{\alpha }{(1+\alpha )}{\left(\frac{v_{esc}}{v_0}\right)}^2}\end{array}$

Conclusion:

We proved as $\sin {\varphi }_0=(1+\alpha )\sqrt{1-\frac{\alpha }{(1+\alpha )}{\left(\frac{v_{esc}}{v_0}\right)}^2}\text{ proved}$

To determine

(b)

Allowable values of $v_0$.

Answer to Problem 13.115P

We got values as $v_{esc}\sqrt{\frac{\alpha }{1+\alpha }}<v_0<v_{esc}\sqrt{\frac{1+\alpha }{2+\alpha }}$

Explanation of Solution

Given information:

Max altitude $\alpha R$

Initial velocity $v_0$

Radius of planet $\text{R}$

Launch angle ${\varphi }_0$

Escape velocity $v_{esc}$

Concept used:

Following formula will be used.

$0\le \sin {\varphi }_0\le 1$

Calculation:

We know,

$T_A+V_A=T_B+V_B$

$\frac{1}{2}m{v}_0{}^2-\frac{g{R}^{2}m}{R}=\frac{1}{2}m{(v_B)}^2-\frac{g{R}^{2}m}{(1+\alpha )R}$

$\frac{1}{2}{v}_0{}^2-gR=\frac{1}{2}{(v_B)}^2-\frac{gR}{(1+\alpha )}$

Conservation of angular momentum,

$\begin{array}{l}(1+\alpha )Rm{v}_B=Rmv\sin {\varphi }_0\\ v_B=\frac{v_0\sin {\varphi }_0}{1+\alpha }\end{array}$

So,

$\begin{array}{l}$ \frac{1}{2}{v}_0{}^2-gR=\frac{1}{2}{(v_B)}^2-\frac{gR}{(1+\alpha )}$$ \frac{1}{2}{v}_0{}^2-gR=\frac{1}{2}{(\frac{v_0\sin {\varphi }_0}{1+\alpha })}^2-\frac{gR}{(1+\alpha )}$\\ \text{also,}\\ v_{esc}=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}\end{array}$

$\begin{array}{l}$ \frac{1}{2}{v}_0{}^2-\frac{v_{esc}{}^2}{2}=\frac{1}{2}{(\frac{v_0\sin {\varphi }_0}{1+\alpha })}^2-\frac{v_{esc}{}^2}{2(1+\alpha )}$$ \frac{1}{2}{v}_0{}^2-\frac{1}{2}{(\frac{v_0\sin {\varphi }_0}{1+\alpha })}^2=+\frac{v_{esc}{}^2}{2}-\frac{v_{esc}{}^2}{2(1+\alpha )}$v_0{}^2\left(1-\frac{{\sin }^2{\varphi }_0}{{(1+\alpha )}^2}\right)=v_{esc}{}^2\left(1-\frac{1}{(1+\alpha )}\right)\\ \sin {\varphi }_0=(1+\alpha )\sqrt{1-\frac{\alpha }{(1+\alpha )}{\left(\frac{v_{esc}}{v_0}\right)}^2}\end{array}$

If $\sin {\varphi }_0=0$ ,

$\begin{array}{l}\sin {\varphi }_0=(1+\alpha )\sqrt{1-\frac{\alpha }{(1+\alpha )}{\left(\frac{v_{esc}}{v_0}\right)}^2}\\ 0=(1+\alpha )\sqrt{1-\frac{\alpha }{(1+\alpha )}{\left(\frac{v_{esc}}{v_0}\right)}^2}\\ v_0=v_{esc}\sqrt{\frac{\alpha }{1+\alpha }}\end{array}$

If $\sin {\varphi }_0=1$ ,

$\begin{array}{l}\sin {\varphi }_0=(1+\alpha )\sqrt{1-\frac{\alpha }{(1+\alpha )}{\left(\frac{v_{esc}}{v_0}\right)}^2}\\ 1=(1+\alpha )\sqrt{1-\frac{\alpha }{(1+\alpha )}{\left(\frac{v_{esc}}{v_0}\right)}^2}\\ v_0=v_{esc}\sqrt{\frac{1+\alpha }{2+\alpha }}\end{array}$

Conclusion:

We got limit as $v_{esc}\sqrt{\frac{\alpha }{1+\alpha }}<v_0<v_{esc}\sqrt{\frac{1+\alpha }{2+\alpha }}$