Chapter 13.2, Problem 13.111P

To determine

Values of ${\varphi }_{b}and{v}_b$.

Answer to Problem 13.111P

Values of ${\varphi }_{b}and{v}_b$ are 58.9⁰ and 30.83x10³ ft/s respectively.

Explanation of Solution

Given information:

The altitude of the space vehicle at point A, h_a = 225mi.

The altitude of the space vehicle at point B, h_b = 40mi.

Radius of earth, R = 3960mi.

Calculation:

The radius of the orbit at point A,

$\begin{array}{l}{r}_a=h_a+R\\ r_a=225+3960\\ r_a=4185mi\times 5280ft\\ r_a=22097\times {10}^{3}ft\end{array}$

The radius of the orbit at point B,

$\begin{array}{l}{r}_b=h_b+R\\ r_b=40+3960\\ r_b=4000mi\times 5280ft\\ r_b=21120\times {10}^{3}ft\end{array}$

Velocity in circular orbit at point A,

$\begin{array}{l}{\left(v_a\right)}_{circular}=\sqrt{\frac{GM}{r_a}}\\ {\left(v_a\right)}_{circular}=\sqrt{\frac{g{R}^2}{r_a}}\\ {\left(v_a\right)}_{circular}=\sqrt{\frac{32.2\times {\left(20909\times {10}^3\right)}^2}{22097\times {10}^3}}\\ {\left(v_a\right)}_{circular}=25.25\times {10}^{3}ft/s\end{array}$

The kinetic energy at point A,

$E_a=\frac{1}{2}m{v}_a{}^2$

Potential energy at point A,

$U_a=-\frac{GMm}{r_a}$

The kinetic energy at point B,

$E_b=\frac{1}{2}m{v}_b{}^2$

Potential energy at point B,

$U_b=-\frac{GMm}{r_b}$

Law of conservation of energy,

$\begin{array}{l}{E}_a+U_a=E_b+U_b\\ \frac{1}{2}m{v}_a{}^2-\frac{GMm}{r_a}=\frac{1}{2}m{v}_b{}^2-\frac{GMm}{r_b}\\ v_a{}^2-v_b{}^2=2GM\left(\frac{1}{r_b}-\frac{1}{r_a}\right)\\ v_a{}^2-v_b{}^2=2g{R}^2\left(\frac{1}{r_b}-\frac{1}{r_a}\right)\\ v_a{}^2-v_b{}^2=2\times 32.2\times {\left(20909\times {10}^3\right)}^2\left(\frac{1}{22097\times {10}^3}-\frac{1}{21120\times {10}^3}\right)\\ v_a{}^2-v_b{}^2=-58.94\times {10}^6(1)\end{array}$

Conservation of angular momentum between A and B,

$\begin{array}{l}{r}_{a}m{v}_a=r_{b}m{v}_b\sin {\varphi }_b\\ v_b=\frac{r_a{v}_a}{r_b\sin {\varphi }_b}\\ v_b=\frac{22097\times {10}^3{v}_a}{21120\times {10}^3\sin 60}\\ v_b=1.2v_a(2)\end{array}$

Now from equation 1, putting the value of v_b.

$\begin{array}{l}{v}_a{}^2-v_b{}^2=-58.94\times {10}^6\\ v_a{}^2-{\left(1.2v_a\right)}^2=-58.94\times {10}^6\\ v_a=11.32\times {10}^{3}ft/s\end{array}$

From equation 2

$\begin{array}{l}{v}_b=1.2v_a\\ v_b=1.2\times 11.32\times {10}^3\\ v_b=13.58\times {10}^{3}ft/s\end{array}$

The energy expenditure,

$\begin{array}{l}\Delta E=\frac{1}{2}m\left[{\left(v_a\right)}^2{}_{circular}-v_a{}^2\right]\\ \Delta E=\frac{1}{2}m\left[{\left(25.25\times {10}^3\right)}^2-{\left(13.58\times {10}^3\right)}^2\right]\\ \Delta E=254.5\times {10}^{6}m\end{array}$

The energy expenditure at point A,

$\Delta E_a=0.5\Delta E$

Additional energy at point A,,

$\begin{array}{l}\Delta E_a=\frac{1}{2}m{\left(\Delta v_a\right)}^2\\ 0.5\times 254.5\times {10}^{6}m=0.5m{\left(\Delta v_a\right)}^2\\ {\left(\Delta v_a\right)}^2=254.5\times {10}^6\end{array}$

Kinetic energy at point A,

$E_a=\frac{1}{2}m\left[{\left(v_a\right)}^2{}_{circular}-{\left(\Delta v_a\right)}^2\right]$

Potential energy at point A,

$U_a=-\frac{GMm}{r_a}$

The kinetic energy at point B,

$E_b=\frac{1}{2}m{v}_b{}^2$

Potential energy at point B,

$U_b=-\frac{GMm}{r_b}$

Law of conservation of energy,

$\begin{array}{l}{E}_a+U_a=E_b+U_b\\ \frac{1}{2}m\left(v_a{{}^2}_{circular}+\Delta v_a{}^2\right)-\frac{GMm}{r_a}=\frac{1}{2}m{v}_b{}^2-\frac{GMm}{r_b}\\ v_a{{}^2}_{circular}+\Delta v_a{}^2-v_b{}^2=2GM\left(\frac{1}{r_b}-\frac{1}{r_a}\right)\\ v_a{{}^2}_{circular}+\Delta v_a{}^2-v_b{}^2=2g{R}^2\left(\frac{1}{r_b}-\frac{1}{r_a}\right)\\ {\left(25.25\times {10}^3\right)}^2+254.5\times {10}^6-v_b{}^2=2\times 32.2\times {\left(20909\times {10}^3\right)}^2\left(\frac{1}{22097\times {10}^3}-\frac{1}{21120\times {10}^3}\right)\\ 637.56\times {10}^6+254.5\times {10}^6-v_b{}^2=-58.94\times {10}^6\\ v_b{}^2=892.06\times {10}^6+58.94\times {10}^6\\ v_b{}^2=950\times {10}^6\\ v_b=30.83\times {10}^{3}ft/s\end{array}$

Conservation of angular momentum between A and B,

$\begin{array}{l}{r}_{a}m{\left(v_a\right)}_{circular}=r_{b}m{v}_b\sin {\varphi }_b\\ \sin {\varphi }_b=\frac{r_a{v}_a}{r_b{v}_b}\\ \sin {\varphi }_b=\frac{22097\times {10}^3\times 25.25\times {10}^3}{21120\times {10}^3\times 30.83\times {10}^3}\\ \sin {\varphi }_b=0.856\\ {\varphi }_b={58.9}^0\end{array}$