Chapter 13.2, Problem 13.111P

To determine

Find the resulting values of angle $\left({\varphi }_B\right)$ and speed $\left(v_B\right)$ at the point B.

Answer to Problem 13.111P

The resulting values of angle $\left({\varphi }_B\right)$ and speed $\left(v_B\right)$ at the point B are $\underset{\_}{58.8°}$ and $\underset{\_}{30.9\times {10}^3\text{ft}/\text{s}}$ respectively.

Explanation of Solution

Given information:

The altitude of the space vehicle in a circular orbit $\left(h_A\right)$ is $225\text{mi}$.

The altitude of the surface of the earth from the center of the earth $\left(h_B\right)$ is $40\text{mi}$.

The radius of the earth (R) is $3960\text{mi}$.

The velocity at B forms an angle $\left({\varphi }_0\right)$ is $60°$.

The acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Assume the energy is used with only 50 percent of the energy expenditure used in Problem 110.

Calculation:

Convert the radius of the earth (R) from kilometer to meter:

$R=R_E\times 5280\frac{\text{ft}}{\text{mi}}$

Here, $R_E$ is the radius of the earth in miles.

Substitute $3960\text{mi}$ for $R_E$.

$\begin{array}{c}R=3960\text{mi}\times \times 5280\frac{\text{ft}}{\text{mi}}\\ =20.909\times {10}^6\text{ft}\end{array}$

The expression for the geocentric force acting on the spacecraft when it is on the surface of earth $\left(F_0\right)$ as follows:

$F_0=\frac{GMm}{R^2}$

Here, G is the universal gravitational constant, M is the mass of the earth and m is the mass of the space vehicle.

The expression for the force acting on the spacecraft on the surface of the earth due to gravity $\left(F_0\right)$ as follows:

$F_0=mg$

Substitute $mg$ for $F$.

$\begin{array}{l}F=\frac{GMm}{R^2}\\ mg=\frac{GMm}{R^2}\\ GM=g{R}^2\end{array}$

Substitute $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g and $20.909\times {10}^6\text{ft}$ for R.

$\begin{array}{c}GM=\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right){\left(20.909\times {10}^6\text{ft}\right)}^2\\ =14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2\end{array}$

Calculate the altitude of the point A from the center of the earth $\left(r_A\right)$ using the relation:

$r_A=R+h_A$

Substitute $20.909\times {10}^6\text{ft}$ for R and $225\text{mi}$ for $h_A$.

$\begin{array}{c}{r}_A=20.909\times {10}^6\text{ft}+\left(225\text{mi}\times 5280\frac{\text{ft}}{\text{mi}}\right)\\ =22097\times {\text{10}}^3\text{ ft}\end{array}$

Calculate the altitude of the point B from the center of the earth $\left(r_B\right)$ using the relation:

$r_B=R+h_B$

Substitute $20.909\times {10}^6\text{ft}$ for R and $40\text{mi}$ for $h_B$.

$\begin{array}{c}{r}_B=20.909\times {10}^6\text{ft}+\left(40\text{mi}\times \text{5280}\frac{\text{ft}}{\text{mi}}\right)\\ =21120\times {\text{10}}^3\text{ft}\end{array}$

The expression for the normal acceleration $\left(a_n\right)$ as follows:

$a_n=\frac{{\left(v_A\right)}_{circ}^2}{r_A}$

The expression for the geocentric force acting on the space vehicle when it is on the surface of earth (F) as follows:

$F=\frac{GMm}{r_A^2}$

Here, G is the universal gravitational constant, M is the mass of the earth and m is the mass of the space vehicle.

Calculate the velocity in circular orbit ${\left(v_A\right)}_{circ}$ by considering the force equilibrium using the relation:

$F=m{a}_n$

Substitute $\frac{GMm}{r_A^2}$ for F and $\frac{{\left(v_A\right)}_{circ}^2}{r_A}$ for $a_n$.

$\begin{array}{l}\frac{GMm}{r_A^2}=m\frac{{\left(v_A\right)}_{circ}^2}{r_A}\\ \frac{GM}{r_A}={\left(v_A\right)}_{circ}^2\\ {\left(v_A\right)}_{circ}=\sqrt{\frac{GM}{r_A}}\end{array}$

Substitute $14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2$ for GM and $22097\times {\text{10}}^3\text{ ft}$ for $r_A$.

$\begin{array}{c}{\left(v_A\right)}_{circ}=\sqrt{\frac{14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2}{22097\times {\text{10}}^3\text{ ft}}}\\ =25.24\times {10}^3\text{ft}/\text{s}\end{array}$

The expression for the kinetic energy at point A $\left(T_A\right)$ as follows:

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

Here, m is the mass of the satellite.

Calculate the gravitational potential energy at point A $\left(V_A\right)$ using the formula:

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

Substitute $14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2$ for GM and $22097\times {\text{10}}^3\text{ ft}$ for $r_A$.

$\begin{array}{c}{V}_A=-\frac{\left(14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2\right)m}{22097\times {\text{10}}^3\text{ ft}}\\ =-637.05\times {10}^{6}m\end{array}$

The expression for the kinetic energy of the satellite at point B $\left(T_B\right)$ as follows:

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

Calculate the gravitational potential energy at point B$\left(V_B\right)$ using the formula:

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

Substitute $14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2$ for GM and $21120\times {\text{10}}^3\text{ft}$ for $r_B$.

$\begin{array}{c}{V}_B=-\frac{\left(14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2\right)m}{21120\times {\text{10}}^3\text{ft}}\\ =-666.5\times {10}^{6}m\end{array}$

The expression for the principle of conservation of energy at the point A to point B as follows:

$T_A+V_A=T_B+V_B$

Substitute $\frac{1}{2}m{v}_A^2$ for $T_A$, $-637.05\times {10}^{6}m$ for $V_A$, $\frac{1}{2}m{v}_B^2$ for $T_B$, and $-666.5\times {10}^{6}m$ for $V_B$.

$\begin{array}{l}\frac{1}{2}m{v}_A^2-637.05\times {10}^{6}m=\frac{1}{2}m{v}_B^2-666.5\times {10}^{6}m\\ \frac{v_A^2}{2}-637.05\times {10}^6=\frac{v_B^2}{2}-666.5\times {10}^6\\ \frac{v_A^2}{2}-\frac{v_B^2}{2}=637.05\times {10}^6-666.5\times {10}^6\end{array}$

$\begin{array}{l}\frac{v_A^2}{2}-\frac{v_B^2}{2}=-29.45\times {10}^6\\ v_A^2-v_B^2=-58.9\times {10}^6\\ v_A^2=v_B^2-58.9\times {10}^6\end{array}$

Substitute $\left(\frac{r_A}{r_B\sin {\varphi }_B}\right)v_A$ for $v_B$.

$\begin{array}{c}{v}_A^2={\left(\left(\frac{r_A}{r_B\sin {\varphi }_B}\right)v_A\right)}^2-58.9\times {10}^6\\ ={\left(\frac{r_A}{r_B\sin {\varphi }_B}\right)}^2{v}_A^2-58.9\times {10}^6\end{array}$

Substitute $60°$ for ${\varphi }_B$, $21120\times {\text{10}}^3\text{ft}$ for $r_B$ and $22097\times {\text{10}}^3\text{ ft}$ for $r_A$.

$\begin{array}{l}{v}_A^2={\left(\frac{22097\times {10}^3}{21120\times {10}^3\sin \left(60\right)}\right)}^2{v}_A^2-58.9\times {10}^6\\ v_A^2=1.45955v_A^2-58.9\times {10}^6\\ v_A^2-1.45955v_A^2+58.9\times {10}^6=0\end{array}$

$\begin{array}{l}-0.45955v_A^2+58.9\times {10}^6=0\\ v_A=\sqrt{\frac{58.9\times {10}^6}{0.45955}}\\ v_A=11.32\times {10}^3\text{ft}/\text{s}\end{array}$

Calculate the energy expenditure $\left(\Delta E_{\exp }\right)$ using the formula:

$\Delta E_{\exp }=\frac{1}{2}m{\left(v_A\right)}_{circ}^2-\frac{1}{2}m{v}_A^2$

Substitute $11.32\times {10}^3\text{ft}/\text{s}$ for $v_A$ and $25.24\times {10}^3\text{ft}/\text{s}$ for ${\left(v_A\right)}_{circ}$.

$\begin{array}{c}\Delta E_{\exp }=\frac{1}{2}m{\left(25.24\times {10}^3\text{ft}/\text{s}\right)}^2-\frac{1}{2}m{\left(11.32\times {10}^3\text{ft}/\text{s}\right)}^2\\ =318.5288\times {10}^{6}m-64.0712\times {10}^{6}m\\ =254.46\times {10}^{6}m\end{array}$

Calculate the energy used $\left(\Delta E\right)$ using the formula:

$\left(\Delta E\right)=50\%\left({\left(\Delta E\right)}_{\exp }\right)$

Substitute $254.46\times {10}^{6}m$ for $\Delta E_{\exp }$.

$\begin{array}{c}\left(\Delta E\right)=\left(\frac{50}{100}\right)254.46\times {10}^{6}m\\ =127.23\times {10}^{6}m\end{array}$

Consider the additional kinetic energy at the point A:

$\Delta E=\frac{1}{2}m{\left(\Delta v_A\right)}^2$

Substitute $127.23\times {10}^{6}m$ for $\Delta E$.

$127.23\times {10}^{6}m=\frac{1}{2}m{\left(\Delta v_A\right)}^2$

The expression for the kinetic energy at point A $\left(T_A\right)$ as follows:

$T_A=\frac{1}{2}m\left[{\left(v_A\right)}_{circ}^2+{\left(\Delta v_A\right)}^2\right]$

The expression for the principle of conservation of energy at the point A to point B as follows:

$T_A+V_A=T_B+V_B$

Substitute $\frac{1}{2}m\left[{\left(v_A\right)}_{circ}^2+{\left(\Delta v_A\right)}^2\right]$ for $T_A$, $-637.05\times {10}^{6}m$ for $V_A$, $\frac{1}{2}m{v}_B^2$ for $T_B$, and $-666.5\times {10}^{6}m$ for $V_B$.

$\begin{array}{l}\frac{1}{2}m\left[{\left(v_A\right)}_{circ}^2+{\left(\Delta v_A\right)}^2\right]-637.05\times {10}^{6}m=\frac{1}{2}m{v}_B^2-666.5\times {10}^{6}m\\ \frac{1}{2}m{\left(v_A\right)}_{circ}^2+\frac{1}{2}m{\left(\Delta v_A\right)}^2-637.05\times {10}^{6}m=\frac{1}{2}m{v}_B^2-666.5\times {10}^{6}m\end{array}$

Substitute $127.23\times {10}^{6}m$ for $\frac{1}{2}m{\left(\Delta v_A\right)}^2$ and $25.24\times {10}^3\text{ft}/\text{s}$ for ${\left(v_A\right)}_{circ}$.

$\begin{array}{l}\frac{1}{2}m{\left(25.24\times {10}^3\text{ft}/\text{s}\right)}^2+127.23\times {10}^{6}m-637.05\times {10}^{6}m=\frac{1}{2}m{v}_B^2-666.5\times {10}^{6}m\\ 318.5288\times {10}^6+127.23\times {10}^6-637.05\times {10}^6=\frac{v_B^2}{2}-666.5\times {10}^6\\ v_B^2=950.4176\times {10}^6\\ v_B=\sqrt{950.4176\times {10}^6}\\ v_B=30.9\times {10}^3\text{ft}/\text{s}\end{array}$

The expression or the principle of conservation of angular momentum at point A to the point B as follows:

$\begin{array}{l}m{r}_A{\left(v_A\right)}_{circ}=m{r}_B{v}_B\sin {\varphi }_B\\ \sin {\varphi }_B=\frac{r_A{\left(v_A\right)}_{circ}}{r_B{v}_B}\end{array}$

Substitute $25.24\times {10}^3\text{ft}/\text{s}$ for ${\left(v_A\right)}_{circ}$, $30.9\times {10}^3\text{ft}/\text{s}$ for $v_B$, $21120\times {\text{10}}^3\text{ft}$ for $r_B$ and $22097\times {\text{10}}^3\text{ ft}$ for $r_A$.

$\begin{array}{l}\sin {\varphi }_B=\frac{\left(22097\times {\text{10}}^3\text{ ft}\right)\left(25.24\times {10}^3\text{ft}/\text{s}\right)}{\left(30.9\times {10}^3\text{ft}/\text{s}\right)\left(21120\times {\text{10}}^3\text{ft}\right)}\\ \sin {\varphi }_B=0.855\\ {\varphi }_B={\sin }^{-1}\left(0.855\right)\\ {\varphi }_B=58.8°\end{array}$

Therefore, the resulting values of angle $\left({\varphi }_B\right)$ and speed $\left(v_B\right)$ at the point B are $\underset{\_}{58.8°}$ and $\underset{\_}{30.9\times {10}^3\text{ft}/\text{s}}$ respectively.