Chapter 12.3, Problem 12.107P

To determine

(a)

The speed of the spacecraft at $A$.

Answer to Problem 12.107P

The speed of the spacecraft at $A$ ,

$v_A=15,965 m{s}^{-1}$

Explanation of Solution

Given information:

$r_A=202\times {10}^{6}mi$

$r_B=92\times {10}^{6}mi$

$r_{A^{\text{'}}}=164.5\times {10}^{6}mi$

$r_{B^{\text{'}}}=85.5\times {10}^{6}mi$

$M_{Sun}=332.8\times {10}^3\times M_{Earth}$

If the trajectory is an elliptical orbit,

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

Angular momentum of a unit mass,

$h=r.v$

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,

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

Since, $G{M}_{Earth}=g{R}_{Earth}^2$ and $M_{Sun}=332.8\times {10}^3\times M_{Earth}$ ,

$\frac{1}{r_A}+\frac{1}{r_B}=\frac{2G\times 332.8\times {10}^3\times M_{Earth}}{h_1^2}=\frac{2\times 332.8\times {10}^3\times g{R}_{Earth}^2}{h_1^2}$

$\left[\frac{1}{202\times {10}^{6}mi}+\frac{1}{92\times {10}^{6}mi}\right]\times \frac{1}{1609.344 m/mi}=\frac{2\times 332.8\times {10}^3\times 9.81 m s^{-2}\times {(6.37\times {10}^{6}m )}^2}{h_1^2}$

$h_1=5.19\times {10}^{15}{m}^2{s}^{-1}$

Angular momentum of first elliptical orbit,

$h_1=r_A.v_A$

$5.19\times {10}^{15} m^2{s}^{-1}=202\times {10}^{6}mi\times 1609.344m/mi\times v_A$

$v_A=15,965 m{s}^{-1}$

Thus, the speed of the spacecraft at $A$ ,

$v_A=15,965 m{s}^{-1}$

To determine

(b)

The amounts by which the speed of the spacecraft should be reduced at $A$ and $B^{\text{'}}$ to insert it into the desired elliptic orbit.

Answer to Problem 12.107P

The speed reduction at point $A$ ,

$△v_A=400 m{s}^{-1}$

The speed reduction at point $B^{\text{'}}$ ,

$△v_{B^{\text{'}}}=1162 m{s}^{-1}$

Explanation of Solution

Given information:

$r_A=202\times {10}^{6}mi$

$r_B=92\times {10}^{6}mi$

$r_{A^{\text{'}}}=164.5\times {10}^{6}mi$

$r_{B^{\text{'}}}=85.5\times {10}^{6}mi$

$M_{Sun}=332.8\times {10}^3\times M_{Earth}$

If the trajectory is an elliptical orbit,

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

Angular momentum of a unit mass,

$h=r.v$

Calculation:

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

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

$1 mi=1609.344 m$

Consider $A{B}^{\text{'}}$ elliptical path,

$\frac{1}{r_A}+\frac{1}{r_{B^{\text{'}}}}=\frac{2G{M}_{Sun}}{h_2{}^2}$

Since, $G{M}_{Earth}=g{R}_{Earth}^2$ and $M_{Sun}=332.8\times {10}^3\times M_{Earth}$ ,

$\frac{1}{r_A}+\frac{1}{r_{B^{\text{'}}}}=\frac{2G\times 332.8\times {10}^3\times M_{Earth}}{h_2^2}=\frac{2\times 332.8\times {10}^3\times g{R}_{Earth}^2}{h_2^2}$

$\left[\frac{1}{202\times {10}^{6}mi}+\frac{1}{85.5\times {10}^{6}mi}\right]\times \frac{1}{1609.344 m/mi}=\frac{2\times 332.8\times {10}^3\times 9.81 m s^{-2}\times {(6.37\times {10}^{6}m )}^2}{h_2^2}$

$h_2=5.06\times {10}^{15}{m}^2{s}^{-1}$

Angular momentum of the $A{B}^{\text{'}}$ elliptical path,

$h_2=r_A.v_{A^{\text{'}}}$

$5.06\times {10}^{15} m^2{s}^{-1}=202\times {10}^{6}mi\times 1609.344m/mi\times v_{A^{\text{'}}}$

$v_A=15,565 m{s}^{-1}$

Therefore, the speed reduction at point $A$ ,

$\begin{array}{l}△v_A=v_A-v_{A^{\text{'}}}\\ △v_A=15,965-15,565\end{array}$

$\underset{\_}{\underset{\_}{△v_A=400 m{s}^{-1}}}$

The speed at point $B^{\text{'}}$ corresponding to $A{B}^{\text{'}}$ elliptical path,

$h_2=r_{B^{\text{'}}}.v_{B^{\text{'}}}$

$5.06\times {10}^{15} m^2{s}^{-1}=85.5\times {10}^{6}mi\times 1609.344m/mi\times v_{B^{\text{'}}}$

$v_{B^{\text{'}}}=36,773 m{s}^{-1}$

Consider the second elliptical orbit.

$\frac{1}{r_{A^{\text{'}}}}+\frac{1}{r_{B^{\text{'}}}}=\frac{2G{M}_{Sun}}{h_3{}^2}$

Since, $G{M}_{Earth}=g{R}_{Earth}^2$ and $M_{Sun}=332.8\times {10}^3\times M_{Earth}$ ,

$\frac{1}{r_{A^{\text{'}}}}+\frac{1}{r_{B^{\text{'}}}}=\frac{2G\times 332.8\times {10}^3\times M_{Earth}}{h_2^2}=\frac{2\times 332.8\times {10}^3\times g{R}_{Earth}^2}{h_2^2}$

$\left[\frac{1}{164.5\times {10}^{6}mi}+\frac{1}{85.5\times {10}^{6}mi}\right]\times \frac{1}{1609.344 m/mi}=\frac{2\times 332.8\times {10}^3\times 9.81 m s^{-2}\times {(6.37\times {10}^{6}m )}^2}{h_3^2}$

$h_3=4.9\times {10}^{15}{m}^2{s}^{-1}$

Angular momentum of the $A{B}^{\text{'}}$ elliptical path,

$h_3=r_{B^{\text{'}}}.v_{B^{\text{'}}}^{\text{'}}$

$4.9\times {10}^{15} m^2{s}^{-1}=85.5\times {10}^{6}mi\times 1609.344m/mi\times v_{{}_{B^{\text{'}}}}^{\text{'}}$

$v_{B^{\text{'}}}^{\text{'}}=35,611 m{s}^{-1}$

Therefore, the speed reduction at point $B^{\text{'}}$ ,

$\begin{array}{l}△v_{B^{\text{'}}}=v_{B^1}-v_{B^{\text{'}}}^{\text{'}}\\ △v_{B^{\text{'}}}=36,773-35,611\end{array}$

$\underset{\_}{\underset{\_}{△v_{B^{\text{'}}}=1162 m{s}^{-1}}}$