Chapter 12.3, Problem 12.104P

(a)

To determine

Find the increase in speed required at point A for the satellite to achieve the escape velocity and enter a parabolic orbit.

Answer to Problem 12.104P

The increase in speed required at point A for the satellite to achieve the escape velocity and enter a parabolic orbit is $\underset{\_}{1.637\times {10}^3\text{m/s}}$.

Explanation of Solution

Given information:

The altitude of circular orbit of the satellite from the surface of the earth (r) is 19,110 km.

The radius of the earth (R) is 6,370 km.

Calculation:

Find the equation of product (GM) of the constant of gravitation G and the mass M of the earth using the equation:

$GM=g{R}^2$

Substitute $9.81{\text{m/s}}^2$ for g and 6,370 km for R.

$\begin{array}{c}GM=9.81\times {\left(6,370\text{km}\times \frac{1,000\text{m}}{1\text{km}}\right)}^2\\ =398.06\times {10}^{12}{\text{m}}^3{\text{/s}}^2\end{array}$

Find the altitude of circular orbit of the satellite $\left(r_A\right)$ from the center of the earth using the Equation:

$r_A=R+r$

Substitute 6,370 km for R and 19,110 km for r.

$\begin{array}{c}{r}_A=6,370+19,110\\ =25,480\text{km}\times \frac{1,000\text{m}}{1\text{km}}\\ =25.48\times {10}^6\text{m}\end{array}$

Find the velocity of satellite $\left(v_{\text{circ}}\right)$ in the circular orbit using the equation:

$v_{\text{circ}}=\sqrt{\frac{GM}{r_A}}$

Substitute $398.06\times {10}^{12}{\text{m}}^3{\text{/s}}^2$ for GM and $25.48\times {10}^6\text{m}$ for $r_A$.

$\begin{array}{c}{v}_{\text{circ}}=\sqrt{\frac{398.06\times {10}^{12}}{25.48\times {10}^6}}\\ =3.9525\times {10}^3\text{m/s}\end{array}$

Find the escape velocity of satellite $\left(v_{\text{circ}}\right)$ from the circular orbit using the equation:

$v_{\text{esc}}=\sqrt{\frac{2GM}{r_A}}$

Substitute $398.06\times {10}^{12}{\text{m}}^3{\text{/s}}^2$ for GM and $25.48\times {10}^6\text{m}$ for $r_A$.

$\begin{array}{c}{v}_{\text{esc}}=\sqrt{\frac{2\times 398.06\times {10}^{12}}{25.48\times {10}^6}}\\ =5.5897\times {10}^3\text{m/s}\end{array}$

Find the decrease in speed $\left(\Delta v\right)$ required at point A for the satellite to enter an elliptic orbit of minimum altitude 6370 km using the equation:

$\Delta v=v_{\text{esc}}-v_{\text{circ}}$

Substitute $5.5897\times {10}^3\text{m/s}$ for $v_{\text{esc}}$ and $3.9525\times {10}^3\text{m/s}$ for $v_{\text{circ}}$.

$\begin{array}{c}\Delta v=\left(5.5897\times {10}^3\right)-\left(3.9525\times {10}^3\right)\\ =1.637\times {10}^3\text{m/s}\end{array}$

Thus, the increase in speed required at point A for the satellite to achieve the escape velocity and enter a parabolic orbit is $\underset{\_}{1.637\times {10}^3\text{m/s}}$.

(b)

To determine

Find the decrease in speed required at point A for the satellite to enter an elliptic orbit of minimum altitude 6370 km.

Answer to Problem 12.104P

The decrease in speed required at point A for the satellite to enter an elliptic orbit of minimum altitude 6,370 km is $\underset{\_}{725\text{m/s}}$.

Explanation of Solution

Calculation:

Find the radius $\left(r_B\right)$ of the elliptical path.

$\begin{array}{c}{r}_B=6,370+6,370\\ =12,740\text{km}\times \frac{1,000\text{m}}{1\text{km}}\\ =12.74\times {10}^6\text{m}\end{array}$

Find the angular momentum per unit mass h using the equation.

$\begin{array}{l}\frac{1}{r_A}+\frac{1}{r_B}=\frac{2GM}{h^2}\\ \frac{r_A+r_B}{r_A{r}_B}=\frac{2GM}{h^2}\\ h^2=\frac{2GM{r}_A{r}_B}{r_A+r_B}\\ h=\sqrt{\frac{2GM{r}_A{r}_B}{r_A+r_B}}\end{array}$

Substitute $398.06\times {10}^{12}{\text{m}}^3{\text{/s}}^2$ for GM, $25.48\times {10}^6\text{m}$ for $r_A$, and $12.74\times {10}^6\text{m}$ for $r_B$.

$\begin{array}{c}h=\sqrt{\frac{2\left(398.06\times {10}^{12}\right)\left(25.48\times {10}^6\right)\left(12.74\times {10}^6\right)}{\left(25.48\times {10}^6\right)+\left(12.74\times {10}^6\text{m}\right)}}\\ =82.230\times {10}^9{\text{m}}^2\text{/s}\end{array}$

Find the velocity at A $\left(v_A\right)$ in parabolic orbit using the equation:

$\begin{array}{l}h=r_A{v}_A\\ v_A=\frac{h}{r_A}\end{array}$

Substitute $82.230\times {10}^9{\text{m}}^2\text{/s}$ for h and $25.48\times {10}^6\text{m}$ for $r_A$.

$\begin{array}{c}{v}_A=\frac{82.230\times {10}^9}{25.48\times {10}^6}\\ =3.2272\times {10}^3\text{m/s}\end{array}$

Find the decrease $\left(\Delta v\right)$ in speed required at point A for the satellite to enter an elliptic orbit of minimum altitude 6370 km using the equation:

$\Delta v=v_{\text{circ}}-v_A$

Substitute $3.9525\times {10}^3\text{m/s}$ for $v_{\text{circ}}$ and $3.2272\times {10}^3\text{m/s}$ for $v_A$.

$\begin{array}{c}\Delta v=\left(3.9525\times {10}^3\right)-\left(3.2272\times {10}^3\right)\\ =725\text{m/s}\end{array}$

(c)

To determine

Find the eccentricity of the elliptic orbit.

Answer to Problem 12.104P

The eccentricity of the elliptic orbit is $\underset{\_}{0.333}$.

Explanation of Solution

Calculation:

Write the equation of angle at B.

${\theta }_B={\theta }_A+180°$

Apply cosine on both sides.

$\begin{array}{l}\cos {\theta }_B=\cos \left({\theta }_A+180°\right)\\ \cos {\theta }_B=-\cos {\theta }_A\end{array}$

Find the constant C using the equation:

$\begin{array}{l}\frac{1}{r_B}-\frac{1}{r_A}=C\cos {\theta }_B-C\cos {\theta }_A\\ \frac{r_A-r_B}{r_A{r}_B}=C\left(\cos {\theta }_B-\cos {\theta }_A\right)\end{array}$

Substitute $-\cos {\theta }_A$ for $\cos {\theta }_B$.

$\begin{array}{l}\frac{r_A-r_B}{r_A{r}_B}=C\left(-\cos {\theta }_A-\cos {\theta }_A\right)\\ \frac{r_A-r_B}{r_A{r}_B}=-2C\cos {\theta }_A\end{array}$

Substitute $180°$ for ${\theta }_A$.

$\begin{array}{l}\frac{r_A-r_B}{r_A{r}_B}=-2C\cos \left(180°\right)\\ \frac{r_A-r_B}{r_A{r}_B}=-2C\left(-1\right)\\ \frac{r_A-r_B}{r_A{r}_B}=2C\\ C=\frac{r_A-r_B}{2r_A{r}_B}\end{array}$

Substitute $25.48\times {10}^6\text{m}$ for $r_A$ and $12.74\times {10}^6\text{m}$ for $r_B$.

$\begin{array}{c}C=\frac{\left(25.48\times {10}^6\right)-\left(12.74\times {10}^6\right)}{2\left(25.48\times {10}^6\right)\left(12.74\times {10}^6\right)}\\ =19.623\times {10}^{-9}{\text{m}}^{-1}\end{array}$

Find the eccentricity $\left(\epsilon \right)$ of the elliptic orbit using the equation:

$\epsilon =\frac{C{h}^2}{GM}$

Substitute $19.623\times {10}^{-9}{\text{m}}^{-1}$ for C, $82.230\times {10}^9{\text{m}}^2\text{/s}$ for h, and $398.06\times {10}^{12}{\text{m}}^3{\text{/s}}^2$ for GM.

$\begin{array}{c}\epsilon =\frac{\left(19.623\times {10}^{-9}\right){\left(82.230\times {10}^9\right)}^2}{398.06\times {10}^{12}}\\ =0.333\end{array}$

Thus, the eccentricity of the elliptic orbit is $\underset{\_}{0.333}$.