Chapter 12.3, Problem 12.106P

To determine

(a)

The distance from the center of Venus to point $B$

Answer to Problem 12.106P

The distance from the center of Venus to point $B$ ,

$r_B=5.35\times {10}^{4}mi$

Explanation of Solution

Given information:

$r_A=9.3\times {10}^{3}mi$

$v_A=20,000ft/s$

$r_C=5600mi$

$M_{Venus}=0.82M_{Earth}$

If the trajectory is an elliptical orbit,

$\frac{1}{r_0}+\frac{1}{r_1}=\frac{2GM}{h^2}$

Calculation:

$g=9.81 m s^{-2}$

$R=6.37\times {10}^{6}m$

$1 mi=1609.344 m$

$1ft/s=0.3048 m/s$

Consider the first transfer elliptical orbit,

$h_1=r_A.v_A$

$h_1=9.3\times {10}^{3}mi\times 1609.344m/mi\times 20,000ft{s}^{-1}\times 0.3048 m{s}^{-1}/ft{s}^{-1}$

$h_1=9.124\times {10}^{10} m^2{s}^{-1}$

Consider the first transfer elliptical orbit,

$\frac{1}{r_A}+\frac{1}{r_B}=\frac{2G{M}_{Venus}}{h_1{}^2}$

Since, $G{M}_{Earth}=g{R}_{Earth}^2$ and $M_{Venus}=0.82M_{Earth}$ ,

$\frac{1}{r_A}+\frac{1}{r_B}=\frac{2G\times 0.82M_{{}_{Earth}}}{h_1^2}=\frac{2\times 0.82\times g{R}_{Earth}^2}{h_1^2}$

$\left[\frac{1}{9.3\times {10}^{3}mi}+\frac{1}{r_B}\right]\times \frac{1}{1609.344 m/mi}=\frac{2\times 0.82\times 9.81 m s^{-2}\times {(6.37\times {10}^{6}m )}^2}{{(9.124\times {10}^{10} m{s}^{-1})}^2}$

$r_B=5.35\times {10}^{4}mi$

The distance from the center of Venus to point $B$ ,

$r_B=5.35\times {10}^{4}mi$

To determine

(b)

The amounts by which the velocity of the probe should be reduced at $B$ and $C$ respectively.

Answer to Problem 12.106P

The velocity reduction at point $B$ ,

$△v_B=212m{s}^{-1}$

The velocity reduction at point $C$ ,

$△v_C=2082 m{s}^{-1}$

Explanation of Solution

Given information:

$r_A=9.3\times {10}^{3}mi$

$v_A=20,000ft/s$

$M_{Venus}=0.82M_{Earth}$

$r_C=5600mi$

Angular momentum of a unit mass,

$h=r.v$

If the trajectory is an elliptical orbit,

$\frac{1}{r_0}+\frac{1}{r_1}=\frac{2GM}{h^2}$

Initial velocity corresponding to a circular orbit,

$v_{circ}=\sqrt{\frac{GM}{r_0}}$

Calculation:

$g=9.81 m s^{-2}$

$R=6.37\times {10}^{6}m$

$1 mi=1609.344 m$

From Part (a),

$h_1=9.124\times {10}^{10} m^2{s}^{-1}$

$r_B=5.35\times {10}^{4}mi$

Consider the first transfer elliptical orbit,

$h_1=r_B.v_B$

$9.124\times {10}^{10} m^2{s}^{-1}=5.35\times {10}^{4}mi\times 1609.344m/mi\times v_B$

$v_B=1060 m{s}^{-1}$

Then consider the second transfer elliptical orbit,

$\frac{1}{r_B}+\frac{1}{r_C}=\frac{2G{M}_{Venus}}{h_2{}^2}$

Since, $G{M}_{Earth}=g{R}_{Earth}^2$ and $M_{Venus}=0.82M_{Earth}$ ,

$\frac{1}{r_B}+\frac{1}{r_C}=\frac{2G{M}_{Venus}}{h_2{}^2}=\frac{2G\times 0.82M_{{}_{Earth}}}{h_2^2}=\frac{2\times 0.82\times g{R}_{Earth}^2}{h_2^2}$

$\left[\frac{1}{5.35\times {10}^{4}mi}+\frac{1}{5600mi}+\right]\times \frac{1}{1609.344 m/mi}=\frac{2\times 0.82\times 9.81 m s^{-2}\times {(6.37\times {10}^{6}m )}^2}{h_2^2}$

$h_2=7.3\times {10}^{10}{m}^2{s}^{-1}$

Applying angular momentum equation for second elliptic orbit,

$h_2=r_B.v_B^{\text{'}}$

$7.3\times {10}^{10}{m}^2{s}^{-1}=5.35\times {10}^{4}mi\times 1609.344m/mi\times v_B^{\text{'}}$

$v_B^{\text{'}}=848 m{s}^{-1}$

Therefore the velocity reduction at point $B$ ,

$\begin{array}{l}△v_{\text{}B}=v_B-v_B^{\text{'}}\\ △v_B=1060-848\end{array}$

$△v_B=212m{s}^{-1}$

Consider the second transfer elliptical orbit,

$h_2=r_C.v_C$

$7.3\times {10}^{10}{m}^2{s}^{-1}=5600mi\times 1609.344m/mi\times v_C$

$v_C=8100 m{s}^{-1}$

Then consider the circular orbit at $C$ ,

$v_C=v_{circ}=\sqrt{\frac{G{M}_{Venus}}{r_C}}$

Since, $G{M}_{Earth}=g{R}_{Earth}^2$ and $M_{Venus}=0.82M_{Earth}$ ,

$v_C^{,}=v_{circ}=\sqrt{\frac{G\times 0.82M_{{}_{Earth}}}{r_C}}=\sqrt{\frac{0.82g{R}_{{}_{Earth}}^2}{r_C}}$

$v_C^{\text{'}}=\sqrt{\frac{0.82\times 9.81 m s^{-2}\times {(6.37\times {10}^{6}m)}^2 }{5600mi\times 1609.344m/mi}}$

$v_C^{\text{'}}=6018 m{s}^{-1}$

Therefore, the velocity reduction at point $B$ ,

$\begin{array}{l}△v_{\text{}C}=v_C-v_C^{\text{'}}\\ △v_C=8100-6018 m{s}^{-1}\\ △v_C= 2082 m{s}^{-1}\end{array}$