Chapter 13.2, Problem 13.105P

(a)

To determine

Find the required increased in speed at A $\left(\Delta v_A\right)$ and at B $\left(\Delta v_B\right)$.

Answer to Problem 13.105P

The required increased in speed at point A $\left(\Delta v_A\right)$ and at point B $\left(\Delta v_B\right)$ are $\underset{\_}{147\text{m}/\text{s}}$ and $\underset{\_}{117\text{m}/\text{s}}$ respectively.

Explanation of Solution

Given information:

The altitude between the earth to point A $\left(h_1\right)$ is $200\text{mi}$.

The altitude between the earth to point B $\left(h_2\right)$ is $500\text{mi}$.

The radius of the earth (R) is $6370\text{km}$.

The acceleration due to gravity (g) is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Write the expression for the geocentric force acting on the spacecraft when it is orbiting around the earth $\left(F_0\right)$ as follows:

$F_0=\frac{GMm}{R^2}$

Here, G is the universal gravitational constant, M is the mass of the earth, and m is the mass of the space vehicle.

Write the expression for the force acting on the space vehicle on the surface of the earth due to gravity $\left(F_0\right)$ as follows:

$F_0=mg$

Substitute $mg$ for $F$.

$\begin{array}{l}F=\frac{GMm}{R^2}\\ mg=\frac{GMm}{R^2}\\ GM=g{R}^2\end{array}$

Substitute 9.81 m/s² for g and $\text{6}\text{.37}\times {\text{10}}^6\text{m}$ for R.

$\begin{array}{c}GM=\left(9.81{\text{ m/s}}^{\text{2}}\right){\left(6.37\times {10}^6\text{m}\right)}^2\\ =\left(9.81\right)\left(40.58\times {10}^{12}\right)\\ =398.06\times {10}^{12}{\text{m}}^3/{\text{s}}^2\end{array}$

Write the expression for the centripetal force (F) acting on the spacecraft rotating around the earth at the given altitude as follows:

$F=\frac{m{v}^2}{r}$ (1)

Here, v is the velocity of the spacecraft describing a circular path around the earth.

Write the expression for the geocentric force (F) acting on the spacecraft rotating at the given altitude around the earth as follows:

$F=\frac{GMm}{r^2}$ (2)

Equate equations (1) and (2).

$\begin{array}{l}\frac{m{v}^2}{r}=\frac{GMm}{r^2}\\ v^2=\frac{GM}{r}\\ v=\sqrt{\frac{GM}{r}}\end{array}$

Calculate the radius of the circular orbit described by the space vehicle around the earth at height $h_1$ $\left(r_A\right)$ using the relation:

$r_A=R+h_1$

Substitute $6370\text{km}$ for R and 200 mi for $h_1$.

$\begin{array}{c}{r}_A=6,370\text{km}+200\text{mi}\\ =6,370\text{km}+200\times \left(\frac{1.61\text{km}}{1\text{mi}}\right)\\ =6,370\text{km}+320\text{km}\end{array}$

$\begin{array}{c}=6,690\text{km}\\ =6,690\times \left(\frac{1,000\text{m}}{1\text{km}}\right)\\ =6.69\times {10}^6\text{m}\end{array}$

Calculate the velocity of the space vehicle at point A $\left[{\left(v_A\right)}_{\text{circ}}\right]$ using the formula:

${\left(v_A\right)}_{\text{circ}}=\sqrt{\frac{GM}{r_A}}$

Substitute $398.06\times {10}^{12}{\text{m}}^3/{\text{s}}^2$ for GM and $6.69\times {10}^6\text{m}$ for $r_A$.

$\begin{array}{c}{\left(v_A\right)}_{\text{circ}}=\sqrt{\frac{398.06\times {10}^{12}{\text{m}}^3/{\text{s}}^2}{6.69\times {10}^6\text{m}}}\\ =\sqrt{59.5\times {\text{10}}^6}\\ =7714\text{m}/\text{s}\end{array}$

Calculate the radius of the circular orbit described by the space vehicle around the earth at height $h_2$ $\left(r_B\right)$ using the relation:

$r_B=R+h_2$

Substitute $6370\text{km}$ for $R$ and 500 mi for $h_2$.

$\begin{array}{c}{r}_B=6,370\text{km}+500\text{mi}\\ =6,370\text{km}+500\times \left(\frac{1.61\text{km}}{1\text{mi}}\right)\\ =6,370\text{km}+800\text{km}\end{array}$

$\begin{array}{c}=7,170\text{km}\\ =7,170\times \left(\frac{1,000}{1\text{km}}\right)\\ =7.170\times {10}^6\text{m}\end{array}$

Calculate the velocity of space vehicle at point B $\left[{\left(v_B\right)}_{\text{circ}}\right]$ just after of the space vehicle is placed on the orbit at B using the formula:

${\left(v_B\right)}_{\text{circ}}=\sqrt{\frac{GM}{r_B}}$

Substitute $398.06\times {10}^{12}{\text{m}}^3/{\text{s}}^2$ for GM and $7.170\times {10}^6\text{m}$ for $r_B$.

$\begin{array}{c}{\left(v_B\right)}_{\text{circ}}=\sqrt{\frac{398.06\times {10}^{12}{\text{m}}^3/{\text{s}}^2}{7.17\times {10}^6\text{m}}}\\ =\sqrt{55.5\times {\text{10}}^6}\\ =7451\text{m}/\text{s}\end{array}$

Use the principle of conservation of angular momentum states that, in the absence of external torque acting on the body, the angular momentum remains constant and no change of the momentum occurs during the entire process.

Write the expression for the principle of conservation of angular momentum as follows:

$\begin{array}{l}m{r}_A{v}_A=m{r}_B{v}_B\\ v_A=\frac{r_B{v}_B}{r_A}\end{array}$

Here, $v_B$ is the velocity of the space vehicle at position B and $v_A$ is the velocity of the space shuttle at position A.

Substitute $6.69\times {10}^6\text{m}$ for $r_A$ and $7.170\times {10}^6\text{m}$ for $r_B$.

$\begin{array}{c}{v}_A=\frac{\left(7.170\times {10}^6\text{m}\text{m}\right)v_B}{\left(6.69\times {10}^6\text{m}\right)}\\ =1.0717v_B\end{array}$ (1)

Write the expression for the kinetic energy of the space vehicle at point A $\left(T_A\right)$ in the path AB as follows:

$T_A=\frac{1}{2}m{v}_A^2$

Write the expression for the kinetic energy of the space vehicle at point B $\left(T_B\right)$ in the path AB as follows;

$T_B=\frac{1}{2}m{v}_B^2$

Write the expression for the gravitational potential energy of the space vehicle at position A in the path AB $\left(V_A\right)$ as follows:

$V_A=-\frac{GMm}{r_A}$

Write the expression for the gravitational potential energy of the space vehicle at position B in the path AB $\left(V_B\right)$ as follows:

$V_B=-\frac{GMm}{r_B}$

Use the principle of conservation of energy states that sum of the kinetic and potential energy of a particle remains constant.

Calculate the speed of the space vehicle at positions A $\left(v_A\right)$ and at position B $\left(v_B\right)$ when the space shuttle is following the path AB using the relation:

$T_A+V_A=T_B+V_B$

Substitute $\frac{1}{2}m{v}_A^2$ for $T_A$, $-\frac{GMm}{r_A}$ for $V_A$, $\frac{1}{2}m{v}_B^2$ for $T_B$,and $-\frac{GMm}{r_B}$ for $V_B$.

$\begin{array}{l}\frac{1}{2}m{v}_A^2-\frac{GMm}{r_A}=\frac{1}{2}m{v}_B^2-\frac{GMm}{r_B}\\ \frac{1}{2}\left(v_A^2-v_B^2\right)=\frac{GM}{r_A}-\frac{GM}{r_B}\\ v_A^2-v_B^2=2GM\left(\frac{1}{r_A}-\frac{1}{r_B}\right)\\ v_B^2=v_A^2-2GM\left(\frac{1}{r_A}-\frac{1}{r_B}\right)\end{array}$

Substitute $1.0717v_B$ for $v_A$, $398.060\times {10}^{12}{\text{m}}^3/{\text{s}}^2$ for $GM$, $6.690\times {10}^6\text{m}$ for $r_A$ and $7.170\times {10}^6\text{m}$ for $r_B$.

$\begin{array}{l}{v}_B^2={\left(1.0717v_B\right)}^2-2\left(398.060\times {10}^{12}{\text{m}}^3/{\text{s}}^2\right)\left(\frac{1}{6.690\times {10}^6\text{m}}-\frac{1}{7.170\times {10}^6\text{m}}\right)\\ v_B^2=1.149v_B^2-\left(796.12\times {10}^{12}\right)\left(0.01\times {\text{10}}^{-6}\right)\\ 1.149v_B^2-v_B^2=\left(7.9612\times {10}^6\right)\end{array}$      

$\begin{array}{l}0.149v_B^2=7.9612\times {10}^6\\ v_B^2=\frac{7.9612\times {10}^6}{0.149}\\ v_B=\sqrt{53.43\times {10}^6}\\ v_B=7334\text{m}/\text{s}\end{array}$

Consider the equation (1).

Calculate the velocity of space shuttle at point A $\left(v_A\right)$:

$v_A=1.0717v_B$

Substitute $7334\text{m}/\text{s}$ for $v_B$.

$\begin{array}{c}{v}_A=1.0717\left(7334\text{m}/\text{s}\right)\\ =7861\text{m}/\text{s}\end{array}$

Calculate the increase in velocity at point A $\left(\Delta v_A\right)$ using the relation:

$\Delta v_A=v_A-{\left(v_A\right)}_{\text{circ}}$

Substitute $7,861\text{m}/\text{s}$ for $v_A$ and $7,714\text{m}/\text{s}$ for ${\left(v_A\right)}_{\text{circ}}$.

$\begin{array}{c}\Delta v_A=7,861\text{m}/\text{s}-7,714\text{m}/\text{s}\\ =147\text{m}/\text{s}\end{array}$

Calculate the increase in the velocity required at B $\left(\Delta v_B\right)$ using the relation:

$\Delta v_B={\left(v_B\right)}_{\text{circ}}-v_B$

Substitute $7,451\text{m}/\text{s}$ for ${\left(v_B\right)}_{\text{circ}}$ and $7,334\text{m}/\text{s}$ for $v_B$.

$\begin{array}{c}\Delta v_B=\left(7,459\text{m}/\text{s}\right)-\left(7,334\text{m}/\text{s}\right)\\ =117\text{m}/\text{s}\end{array}$

Therefore, the required increased in speed at A $\left(\Delta v_A\right)$ and at B $\left(\Delta v_B\right)$ are $\underset{\_}{147\text{m}/\text{s}}$ and $\underset{\_}{117\text{m}/\text{s}}$ respectively.

(b)

To determine

Find the total energy per unit mass $\left(\frac{E}{m}\right)$ required to execute the transfer.

Answer to Problem 13.105P

The total energy per unit mass $\left(\frac{E}{m}\right)$ required to execute the transfer is $\underset{\_}{2.01\times {10}^6\text{J}/\text{kg}}$.

Explanation of Solution

Given information:

The altitude between the earth to point A $\left(h_1\right)$ is $200\text{mi}$.

The altitude between the earth to point B $\left(h_2\right)$ is $500\text{mi}$.

The radius of the earth (R) is $6370\text{km}$.

The acceleration due to gravity (g) is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the total energy per unit mass of the vehicle to execute the transfer of space vehicle $\left(\frac{E}{m}\right)$ using the relation:

$\begin{array}{l}E=\frac{1}{2}m\left[v_A^2-{\left(v_A\right)}_{\text{circ}}^2+{\left(v_B\right)}_{\text{circ}}^2-v_B^2\right]\\ \frac{E}{m}=\frac{1}{2}\left[v_A^2-{\left(v_A\right)}_{\text{circ}}^2+{\left(v_B\right)}_{\text{circ}}^2-v_B^2\right]\end{array}$

Here, E is the total energy to execute the transfer.

Substitute $7861\text{m}/\text{s}$ for $v_A$, $7714\text{m}/\text{s}$ for ${\left(v_A\right)}_{\text{circ}}$, $7451\text{m}/\text{s}$ for ${\left(v_B\right)}_{\text{circ}}$ and $7334\text{m}/\text{s}$ for $v_B$.

$\begin{array}{c}\frac{E}{m}=\frac{1}{2}\left[{\left(7861\text{m}/\text{s}\right)}^2-{\left(7714\text{m}/\text{s}\right)}^2+{\left(7451\text{m}/\text{s}\right)}^2-{\left(7334\text{m}/\text{s}\right)}^2\right]\\ =\frac{1}{2}\left(4.01\times {10}^6\right)\\ =2.01\times {10}^6\text{J}/\text{kg}\end{array}$

Therefore, the total energy per unit mass $\left(\frac{E}{m}\right)$ required to execute the transfer is $\underset{\_}{2.01\times {10}^6\text{J}/\text{kg}}$.