Chapter 12.3, Problem 12.105P

A space probe is to be placed in a circular orbit of 5600-mi radius about the planet Venus in a specified plane. As the probe reaches A, the point of its original trajectory closest to Venus, it is inserted in a first elliptic transfer orbit by reducing its speed by $\Delta v_A$ . This orbit brings it to point B with a much reduced velocity. There the probe is inserted in a second transfer orbit located in the specified plane by changing the direction of its velocity and further reducing its speed by $\Delta v_B$ . Finally, as the probe reaches point C, it is inserted in the desired circular orbit by reducing its speed by $\Delta v_C$ . Knowing that the mass of Venus is 0.82 times the mass of the earth, that $r_A=9.3\times {10}^3$ mi and $r_B=190\times {10}^3$ mi, and that the probe approaches A on a parabolic trajectory, determine by how much the velocity of the probe should be reduced (a) at A, (b) at B, (c) at C.

  

To determine

(a)

The velocity of the probe should be reduced at point $A$.

Answer to Problem 12.105P

The velocity reduction at point $A$ ,

$△v_A=156 m{s}^{-1}$

Explanation of Solution

Given information:

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

$r_B=190\times {10}^{3}mi$

$r_C=5600mi$

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

Initial velocity corresponding to a parabolic orbit,

$v_0=\sqrt{\frac{2GM}{r_0}}$

Angular momentum of a unit mass,

$h=r.v$

If the trajectory is an elliptical orbit,

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

Calculation:

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

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

$1 mi=1609.344 m$

Initial velocity corresponding to a parabolic orbit,

$v_0=\sqrt{\frac{2GM}{r_0}}$

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

$v_A=\sqrt{\frac{2\times 0.82\times g{R}_{Earth}^2}{r_A}}$

$\begin{array}{l}{v}_A=\sqrt{\frac{2\times 0.82\times 9.81 m s^{-2}\times {(6.37\times {10}^{6}m)}^2}{9.3\times {10}^{3}mi\times 1609.344m/mi}}\\ \\ v_A=6604 m{s}^{-1}\end{array}$

Then consider the first transfer elliptical orbit,

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

$\left[\frac{1}{9.3\times {10}^{3}mi}+\frac{1}{190\times {10}^{3}mi}\right]\times \frac{1}{1609.344 m/mi}=\frac{2\times 9.81 m s^{-2}\times {(6.37\times {10}^{6}m )}^2}{h_1^2}$

$h_1=9.65\times {10}^{10}{m}^2{s}^{-1}$

Applying angular momentum equation,

$h_1=r_A.v_A^{\text{'}}$

$9.65\times {10}^{10}{m}^2{s}^{-1}=9.3\times {10}^{3}mi\times 1609.344m/mi\times v_A^{\text{'}}$

$v_A^{\text{'}}=6448 m{s}^{-1}$

Therefore the velocity reduction at point $A$ ,

$\begin{array}{l}△v_A=v_A-v_A^{\text{'}}\\ △v_A=6604-6448\\ △v_A=156 m{s}^{-1}\end{array}$

To determine

(b)

The velocity of the probe should be reduced at $B$

Answer to Problem 12.105P

The velocity reduction at point $B$ ,

$△v_B=68m{s}^{-1}$

Explanation of Solution

Given information:

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

$r_B=190\times {10}^{3}mi$

$r_C=5600mi$

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

Angular momentum of a unit mass,

$h=r.v$

If the trajectory is an elliptical orbit,

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

Calculation:

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

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

$1 mi=1609.344 m$

Consider the first transfer elliptical orbit,

$h_1=r_B.v_B$

$9.65\times {10}^{10}{m}^2{s}^{-1}=190\times {10}^{3}mi\times 1609.344m/mi\times v_B$

$v_B=315 m{s}^{-1}$

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

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

$\left[\frac{1}{190\times {10}^{3}mi}+\frac{1}{5600mi}+\right]\times \frac{1}{1609.344 m/mi}=\frac{2\times 0.82\times 9.81 m s^{-2}\times {(6.37\times {10}^{6}m )}^2}{h_2^2}$

$h_2=7.56\times {10}^{10}{m}^2{s}^{-1}$

Applying angular momentum equation,

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

$7.56\times {10}^{10}{m}^2{s}^{-1}=190\times {10}^{3}mi\times 1609.344m/mi\times v_B^{\text{'}}$

$v_B^{\text{'}}=247 m{s}^{-1}$

Therefore, the velocity reduction at point $B$ ,

$\begin{array}{l}△v_{\text{}B}=v_B-v_B^{\text{'}}\\ △v_B=315-247\\ △v_B=68 m{s}^{-1}\end{array}$

To determine

(b)

The velocity of the probe should be reduced at $C$

Answer to Problem 12.105P

The velocity reduction at point $B$ ,

$△v_c=2371 m{s}^{-1}$

Explanation of Solution

Given information:

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

$r_B=190\times {10}^{3}mi$

$r_C=5600mi$

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

Angular momentum of a unit mass,

$h=r.v$

Initial velocity corresponding to a circular orbit,

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

Calculation:

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

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

$1 mi=1609.344 m$

Consider the second transfer elliptical orbit,

$h_2=r_C.v_C$

$7.56\times {10}^{10}{m}^2{s}^{-1}=5600mi\times 1609.344m/mi\times v_C$

$v_C=8389 m{s}^{-1}$

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 $B$ ,

$\begin{array}{l}△v_{\text{}C}=v_C-v_C^{\text{'}}\\ △v_C=8389-6018 m{s}^{-1}\\ △v_C= 2371 m{s}^{-1}\end{array}$