Chapter 13.2, Problem 13.110P

To determine

(a)

The speed of the vehicle as it leaves its circular orbit at A at given condition.

Answer to Problem 13.110P

The required value of the speed is $11.32\times {10}^{3}ft/s.$

Explanation of Solution

Given information:

$\begin{array}{l}{h}_1=225mi\\ {\varphi }_B={60}^o\\ h_2=40mi\end{array}$

Formula used:

$GM=g{R}^2$

$T=\frac{1}{2}m{v}^2$

$V=-\frac{GMm}{r}$

$L=mvr$

Calculation:

As we have,

$r_A=3960mi+225mi=4185mi$

$r_A=4185mi\times 5280ft/mi=22097\times {10}^{3}ft$

And, $r_B=3960mi\times 40mi=4000mi$

$\Rightarrow r_B=4000\times 5280=21120\times {10}^{3}ft$

$\Rightarrow R=3960mi=20909\times {10}^{3}ft$

We know that:

$GM=g{R}^2$

Put the values in the above equation;

$\Rightarrow GM=\left(32.2ft/s^2\right){\left(20909\times {10}^{3}ft\right)}^2$

$\Rightarrow GM=14.077\times {10}^{15}f{t}^3/s^2$

According to conservation of energy;

$\Rightarrow T_A=\frac{1}{2}m{v}_A^2$

$\Rightarrow V_A=-\frac{GMm}{r_A}$

Put the values in the above equation;

$\Rightarrow -\frac{14.077\times {10}^{15}m}{22097\times {10}^3}$

$\Rightarrow -637.1\times {10}^{6}m$

$\Rightarrow T_B=\frac{1}{2}m{v}_B^2$

$\Rightarrow V_B=-\frac{GMm}{r_B}$

$\Rightarrow -\frac{14.077\times {10}^{15}m}{21120\times {10}^3}$

$\Rightarrow -666.5\times {10}^{6}m$

On applying conservation of energy:

$T_A+V_A=T_B+V_B$

$\Rightarrow \frac{1}{2}m{v}_A^2-637.1\times {10}^{6}m=\frac{1}{2}m{v}_B^2-666.5\times {10}^{6}m$

$\Rightarrow v_A^2=v_B^2-58.94\times {10}^6$   ....... (i)

According to conservation of angular momentum

$\Rightarrow r_{A}m{v}_A=r_{B}m{v}_B\sin {\varphi }_B$

Solve for $v_B$

$\Rightarrow v_B=\frac{\left(r_A\right)v_A}{\left(r_B\right)\left(\sin {\varphi }_B\right)}=\frac{4185}{4000}\left(\frac{1}{\sin {60}^o}\right)v_A$

$\Rightarrow v_B=1.208v_A$   ....... (ii)

Put the value of $v_B$ from equation (ii) in (i)

$\Rightarrow v_A^2={\left(1.208v_A\right)}^2-58.94\times {10}^6$

$\Rightarrow v_A^2\left[{\left(1.208\right)}^2-1\right]=58.94\times {10}^6$

$\Rightarrow v_A^2=128.27\times {10}^{6}f{t}^2/s^2$

$\Rightarrow v_A=11.32\times {10}^{3}ft/s.$

Conclusion:

The required value of the speed is $11.32\times {10}^{3}ft/s.$

To determine

(b)

The sped at point B at given condition.

Answer to Problem 13.110P

The required value of the speed is $13.68\times {10}^{3}ft/s.$

Explanation of Solution

Given information:

$\begin{array}{l}{h}_1=225mi\\ {\varphi }_B={60}^o\\ h_2=40mi\end{array}$

Formula used:

$v_B=1.208v_A$

Calculation:

We know that, $v_B=1.208v_A$   (from equation (ii))

Put the values in the above equation

$\Rightarrow 1.208\left(11.32\times {10}^6\right)$

$\Rightarrow 13.68\times {10}^{3}ft/s$

$\Rightarrow v_B=13.68\times {10}^{3}ft/s$