Chapter 12.3, Problem 12.105P

(a)

To determine

Find the reduced velocity of the probe at A.

Answer to Problem 12.105P

The reduced velocity of the probe at A is $\underset{\_}{509\text{ft/s}}$.

Explanation of Solution

Given information:

The mass of the Venus is 0.82 times the mass of the earth.

The radius of parabolic orbit at point A $\left(r_A\right)$ is $9.3\times {10}^3\text{mi}$.

The radius of parabolic orbit at point B $\left(r_B\right)$ is $190\times {10}^3\text{mi}$.

The radius of the earth (R) is 3,960 mi.

The radius of the Venus $\left(R_V\right)$ is 5,600 mi.

Calculation:

Find the equation of product $\left(G{M}_{\text{earth}}\right)$ of the constant of gravitation G and the mass M of the earth using the equation:

$G{M}_{\text{earth}}=g{R}^2$

Substitute $32.2{\text{m/s}}^2$ for g and 3,960 mi for R.

$\begin{array}{c}G{M}_{\text{earth}}=32.2\times {\left(3,960\text{mi}\times \frac{5,280\text{ft}}{1\text{mi}}\right)}^2\\ =14.077\times {10}^{15}{\text{m}}^3{\text{/s}}^2\end{array}$

Find the equation of product $\left(G{M}_{\text{venus}}\right)$ of the constant of gravitation G and the mass M of the Venus using the equation:

$G{M}_{\text{venus}}=0.82G{M}_{\text{earth}}$

Substitute $14.077\times {10}^{15}{\text{m}}^3{\text{/s}}^2$ for $G{M}_{\text{earth}}$.

$\begin{array}{c}G{M}_{\text{venus}}=0.82\left(14.077\times {10}^{15}\right)\\ =11.543\times {10}^{15}{\text{ft}}^{\text{3}}{\text{/s}}^{\text{2}}\end{array}$

Find the angular momentum per unit mass $\left(h_{AB}\right)$ of first transfer orbit AB using the equation.

$\begin{array}{l}\frac{1}{r_A}+\frac{1}{r_B}=\frac{2G{M}_{\text{venus}}}{h_{AB}^2}\\ \frac{r_A+r_B}{r_A{r}_B}=\frac{2G{M}_{\text{venus}}}{h_{AB}^2}\\ h_{AB}^2=\frac{2G{M}_{\text{venus}}{r}_A{r}_B}{r_A+r_B}\\ h_{AB}=\sqrt{\frac{2G{M}_{\text{venus}}{r}_A{r}_B}{r_A+r_B}}\end{array}$

Substitute $11.543\times {10}^{15}{\text{ft}}^{\text{3}}{\text{/s}}^{\text{2}}$ for $G{M}_{\text{venus}}$, $9.3\times {10}^3\text{mi}$ for $r_A$, and $190\times {10}^3\text{mi}$ for $r_B$.

$\begin{array}{c}{h}_{AB}=\sqrt{\frac{2\left(11.543\times {10}^{15}\right)\left(9.3\times {10}^3\times \frac{5,280\text{ft}}{1\text{mi}}\right)\left(190\times {10}^3\times \frac{5,280\text{ft}}{1\text{mi}}\right)}{\left(9.3\times {10}^3\times \frac{5,280\text{ft}}{1\text{mi}}\right)+\left(190\times {10}^3\times \frac{5,280\text{ft}}{1\text{mi}}\right)}}\\ =1.039575\times {10}^{12}{\text{ft}}^2\text{/s}\end{array}$

Find the escaping velocity at A $\left[{\left(v_A\right)}_1\right]$ for the first transfer orbit using the equation:

${\left(v_A\right)}_1=\sqrt{\frac{2G{M}_{\text{venus}}}{r_A}}$

Substitute $11.543\times {10}^{15}{\text{ft}}^{\text{3}}{\text{/s}}^{\text{2}}$ for $G{M}_{\text{venus}}$ and $9.3\times {10}^3\text{mi}$ for $r_A$

$\begin{array}{c}{\left(v_A\right)}_1=\sqrt{\frac{2\left(11.543\times {10}^{15}\right)}{\left(9.3\times {10}^3\times \frac{5,280\text{ft}}{1\text{mi}}\right)}}\\ =21.683\times {10}^3\text{ft/s}\end{array}$

Find the escaping velocity at A $\left[{\left(v_A\right)}_2\right]$ for the second transfer orbit using the equation:

${\left(v_A\right)}_2=\frac{h_{AB}}{r_A}$

Substitute $1.039575\times {10}^{12}{\text{ft}}^2\text{/s}$ for $h_{AB}$ and $9.3\times {10}^3\text{mi}$ for $r_A$.

$\begin{array}{c}{\left(v_A\right)}_2=\frac{1.039575\times {10}^{12}}{\left(9.3\times {10}^3\times \frac{5,280\text{ft}}{1\text{mi}}\right)}\\ =21.174\times {10}^3\text{ft/s}\end{array}$

Find the escaping velocity at B $\left[{\left(v_B\right)}_1\right]$ for the first transfer orbit using the equation:

${\left(v_B\right)}_1=\frac{h_{AB}}{r_B}$

Substitute $1.039575\times {10}^{12}{\text{ft}}^2\text{/s}$ for $h_{AB}$ and $190\times {10}^3\text{mi}$ for $r_B$.

$\begin{array}{c}{\left(v_B\right)}_1=\frac{1.039575\times {10}^{12}{\text{ft}}^2\text{/s}}{\left(190\times {10}^3\times \frac{5,280\text{ft}}{1\text{mi}}\right)}\\ =1.03626\times {10}^3\text{ft/s}\end{array}$

The radius of orbit $\left(r_C\right)$ at C is equal to the radius of the planet Venus. (5,600 mi).

Find the angular momentum per unit mass $\left(h_{BC}\right)$ of first transfer orbit BC using the equation.

$h_{BC}=\sqrt{\frac{2G{M}_{\text{venus}}{r}_B{r}_C}{r_B+r_C}}$

Substitute $11.543\times {10}^{15}{\text{ft}}^{\text{3}}{\text{/s}}^{\text{2}}$ for $G{M}_{\text{venus}}$, $190\times {10}^3\text{mi}$ for $r_B$, and $5,600\text{mi}$ for $r_C$.

$\begin{array}{c}{h}_{AB}=\sqrt{\frac{2\left(11.543\times {10}^{15}\right)\left(190\times {10}^3\times \frac{5,280\text{ft}}{1\text{mi}}\right)\left(5,600\times \frac{5,280\text{ft}}{1\text{mi}}\right)}{\left(190\times {10}^3\times \frac{5,280\text{ft}}{1\text{mi}}\right)+\left(5,600\times \frac{5,280\text{ft}}{1\text{mi}}\right)}}\\ =814.287\times {10}^9{\text{ft}}^2\text{/s}\end{array}$

Find the escaping velocity at B $\left[{\left(v_B\right)}_2\right]$ for the second transfer orbit using the equation:

${\left(v_B\right)}_2=\frac{h_{BC}}{r_B}$

Substitute $814.287\times {10}^9{\text{ft}}^2\text{/s}$ for $h_{BC}$ and $190\times {10}^3\text{mi}$ for $r_B$.

$\begin{array}{c}{\left(v_B\right)}_2=\frac{814.287\times {10}^9}{\left(190\times {10}^3\times \frac{5,280\text{ft}}{1\text{mi}}\right)}\\ =811.69\text{ft/s}\end{array}$

Find the escaping velocity at C $\left[{\left(v_C\right)}_1\right]$ for the second orbit using the equation:

${\left(v_C\right)}_1=\frac{h_{BC}}{r_C}$

Substitute $814.287\times {10}^9{\text{ft}}^2\text{/s}$ for $h_{BC}$ and $5,600\text{mi}$ for $r_C$.

$\begin{array}{c}{\left(v_C\right)}_1=\frac{814.287\times {10}^9}{\left(5,600\times \frac{5,280\text{ft}}{1\text{mi}}\right)}\\ =27.539\times {10}^3\text{ft/s}\end{array}$

Find the escaping velocity at C $\left[{\left(v_C\right)}_1\right]$ for the final circular orbit using the equation:

${\left(v_C\right)}_1=\sqrt{\frac{G{M}_{\text{venus}}}{r_C}}$

Substitute $11.543\times {10}^{15}{\text{ft}}^{\text{3}}{\text{/s}}^{\text{2}}$ for $G{M}_{\text{venus}}$ and $5,600\text{mi}$ for $r_C$.

$\begin{array}{c}{\left(v_C\right)}_2=\sqrt{\frac{\left(11.543\times {10}^{15}\right)}{\left(5,600\times \frac{5,280\text{ft}}{1\text{mi}}\right)}}\\ =19.758\times {10}^3\text{ft/s}\end{array}$

Find the reduced velocity $\left(\Delta v_A\right)$ of the probe at A using the equation:

$\Delta v_A={\left(v_A\right)}_1-{\left(v_A\right)}_2$

Substitute $21.683\times {10}^3\text{ft/s}$ for ${\left(v_A\right)}_1$ and $21.174\times {10}^3\text{ft/s}$ for ${\left(v_A\right)}_2$.

$\begin{array}{c}\Delta v_A=\left(21.683\times {10}^3\right)-\left(21.174\times {10}^3\right)\\ =509\text{ft/s}\end{array}$

Thus, the reduced velocity of the probe at A is $\underset{\_}{509\text{ft/s}}$.

(b)

To determine

Find the reduced velocity of the probe at B.

Answer to Problem 12.105P

The reduced velocity of the probe at B is $\underset{\_}{224\text{ft/s}}$.

Explanation of Solution

Calculation:

Find the reduced velocity $\left(\Delta v_B\right)$ of the probe at B using the equation:

$\Delta v_B={\left(v_B\right)}_1-{\left(v_B\right)}_2$

Substitute $1.036\times {10}^3\text{ft/s}$ for ${\left(v_A\right)}_1$ and $811.69\text{ft/s}$ for ${\left(v_A\right)}_2$.

$\begin{array}{c}\Delta v_A=\left(1.036\times {10}^3\right)-\left(811.69\right)\\ =224\text{ft/s}\end{array}$

Thus, the reduced velocity of the probe at B is $\underset{\_}{224\text{ft/s}}$.

(b)

To determine

Find the reduced velocity of the probe at C.

Answer to Problem 12.105P

The reduced velocity of the probe at C is $\underset{\_}{7.78\times {10}^3\text{ft/s}}$.

Explanation of Solution

Calculation:

Find the reduced velocity $\left(\Delta v_C\right)$ of the probe at C using the equation:

$\Delta v_C={\left(v_C\right)}_1-{\left(v_C\right)}_2$

Substitute $27.539\times {10}^3\text{ft/s}$ for ${\left(v_A\right)}_1$ and $19.578\times {10}^3\text{ft/s}$ for ${\left(v_A\right)}_2$.

$\begin{array}{c}\Delta v_A=\left(27.539\times {10}^3\right)-\left(19.758\times {10}^3\right)\\ =7.78\times {10}^3\text{ft/s}\end{array}$

Thus, the reduced velocity of the probe at C is $\underset{\_}{7.78\times {10}^3\text{ft/s}}$.