Chapter 13.2, Problem 13.101P

While describing a circular orbit, 185 mi above the surface of the earth, a space shuttle ejects at point A an inertial upper stage (IUS) carrying a communications satellite to be placed in a geosynchronous orbit (see Prob. 13.87) at an altitude of 22,230 mi above the surface of the earth. Determine (a) the velocity of the IUS relative to the shuttle after its engine has been fired at A, (b) the increase in velocity required at B to place the satellite in its final orbit.

Fig. P13.101

To determine

Find the velocity of the IUS relative to the shuttle after its engine has been fired at A $\left(\Delta v_A\right)$.

Answer to Problem 13.101P

The velocity of the IUS relative to the shuttle after its engine has been fired at A $\left(\Delta v_A\right)$ is $\underset{\_}{7960\text{ft}/\text{s}}$.

Explanation of Solution

Given information:

The distance from above the earth surface to the circular orbit (x) is $185\text{mi}$.

The altitude of the geosynchronous orbit (A) is $22230\text{mi}$.

The radius of the earth (R) is $3960\text{mi}$.

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

Calculation:

Convert the unit of radius of earth (R) from miles to feet using the relation:

$R=R_E\times \left(\frac{5280\text{ft}}{1\text{mi}}\right)$

Here, $R_E$ is the radius of the earth.

Substitute $3960\text{mi}$ for $R_E$.

$\begin{array}{c}R=3960\text{mi}\times \left(\frac{5280\text{ft}}{1\text{mi}}\right)\\ =\text{20}\text{.909}\times {\text{10}}^6\text{ft}\end{array}$

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

$F_0=mg$

Write the expression for calculating the geocentric force acting on the spacecraft when it is on the surface of 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 $F_0$ is the force acting on the missile on the surface of the earth.

Substitute $mg$ for $F_0$.

$\begin{array}{l}{F}_0=\frac{GMm}{R^2}\\ mg=\frac{GMm}{R^2}\\ GM=g{R}^2\end{array}$

Substitute $32.2\text{ft}/{\text{s}}^2$ for g and $\text{20}\text{.909}\times {\text{10}}^6\text{ft}$ for R.

$\begin{array}{c}GM=\left(32.2\text{ft}/{\text{s}}^2\right){\left(\text{20}\text{.909}\times {\text{10}}^6\text{ft}\right)}^2\\ =\left(9.81\right)\left(437.19\times {10}^{12}\right)\\ =14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2\end{array}$

Write the expression for the centripetal force acting on the space shuttle carrying satellite rotating around the earth at the given altitude as follows:

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

Here, m is the mass of the space shuttle, $v_C$ is the velocity of the space shuttle describing a circular path around the earth and r is the distance or radius of the satellite from the center of the earth.

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

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

Equate the 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 altitude of the space shuttle from center of earth at position A$\left(r_A\right)$ using the relation:

$r_A=R+x$

Substitute $\text{20}\text{.909}\times {\text{10}}^6\text{ft}$ for R and $185\text{mi}$ for x.

$\begin{array}{c}{r}_A\text{=20}\text{.909}\times {\text{10}}^6\text{ft}+185\times \left(\frac{5280\text{ft}}{1\text{mi}}\right)\\ =\text{20}\text{.909}\times {\text{10}}^6+0.9768\times {\text{10}}^6\\ =\text{21}\text{.886}\times {\text{10}}^6\text{ft}\end{array}$

Calculate the velocity of space shuttle at point A inertial upper stage ${\left(v_A\right)}_{\text{circ}}$ using the formula:

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

Substitute $14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2$ for GM and $\text{21}\text{.886}\times {\text{10}}^6\text{ft}$ for $r_A$.

$\begin{array}{c}{\left(v_A\right)}_{\text{circ}}=\sqrt{\frac{14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2}{\text{21}\text{.886}\times {\text{10}}^6\text{ft}}}\\ =\sqrt{643.23\times {\text{10}}^6}\\ =25.362\times {\text{10}}^3\text{ft}/\text{s}\end{array}$

Calculate the altitude of the space shuttle from center of earth at position A$\left(r_B\right)$ using the relation:

$r_B=R+A$

Substitute $\text{20}\text{.909}\times {\text{10}}^6\text{ft}$ for R and $22230\text{mi}$ for A.

$\begin{array}{c}{r}_B\text{=20}\text{.909}\times {\text{10}}^6\text{ft}+22230\times \left(\frac{5280\text{ft}}{1\text{mi}}\right)\\ =\text{20}\text{.909}\times {\text{10}}^6+117.374\times {\text{10}}^6\\ =\text{138}\text{.283}\times {\text{10}}^6\text{ft}\end{array}$

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

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

Substitute $14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2$ for $GM$ and $\text{138}\text{.283}\times {\text{10}}^6\text{ft}$ for $r_B$.

$\begin{array}{c}{\left(v_B\right)}_{\text{circ}}=\sqrt{\frac{14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2}{\text{138}\text{.383}\times {\text{10}}^6\text{ft}}}\\ =\sqrt{101.787\times {\text{10}}^6}\\ =10.089\times {\text{10}}^3\text{ft}/\text{s}\end{array}$

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.

Find the velocity at B:

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

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

Substitute $\text{21}\text{.886}\times {\text{10}}^6\text{ft}$ for $r_A$ and $\text{138}\text{.283}\times {\text{10}}^6\text{ft}$ for $r_B$.

$\begin{array}{c}{v}_B=\frac{\left(\text{21}\text{.886}\times {\text{10}}^6\text{ft}\right)v_A}{\left(\text{138}\text{.283}\times {\text{10}}^6\text{ft}\right)}\\ =0.1581v_A\end{array}$ (1)

Write the expression for the kinetic energy of the space shuttle 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 shuttle 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 shuttle 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 shuttle 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 shuttle at position A $\left(v_A\right)$ and at position B $\left(v_B\right)$ when the space shuttle is following the path AB by using principle of conservation of energy.

Write the expression for the conservation of energy as follows:

$T_A+V_A=T_B+V_B$

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 $V_B$.

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

Find the velocity at A:

Substitute $0.1581v_A$ for $v_B$, $14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2$ for GM, $\text{21}\text{.886}\times {\text{10}}^6\text{ft}$ for $r_A$ and $\text{138}\text{.283}\times {\text{10}}^6\text{ft}$ for $r_B$.

$\begin{array}{l}{v}_B^2=v_A^2-2GM\left(\frac{1}{r_A}-\frac{1}{r_B}\right)\\ {\left(0.1581v_A\right)}^2=v_A^2-2\left(14.077\times {10}^{15}{\text{ft}}^3/{\text{s}}^2\right)\left(\frac{1}{\text{21}\text{.886}\times {\text{10}}^6\text{ft}}-\frac{1}{\text{138}\text{.283}\times {\text{10}}^6\text{ft}}\right)\\ 0.025v_A^2=v_A^2-\left(28.154\times {10}^{15}\right)\left(0.03846\times {\text{10}}^{-6}\right)\end{array}$

$\begin{array}{l}{v}_A^2-0.025v_A^2=\left(28.154\times {10}^{15}\right)\left(0.03846\times {\text{10}}^{-6}\right)\\ 0.975v_A^2=1.0828\times {10}^9\\ v_A^2=1.1106\times {10}^9\end{array}$

$\begin{array}{c}{v}_A=\sqrt{1.1106\times {10}^9}\\ v_A=33.325\times {10}^3\text{ft}/\text{s}\end{array}$

Consider the equation (1).

Find the velocity at B:

Substitute $33.325\times {10}^3\text{ft}/\text{s}$ for $v_A$.

$\begin{array}{c}{v}_B=0.1581\left(33.325\times {10}^3\text{ft}/\text{s}\right)\\ =7.963\times {10}^3\text{ft}/\text{s}\end{array}$

Calculate the velocity of the IUS relative to the shuttle after the engine has been fired at point A $\left(\Delta v_A\right)$ using the relation:

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

Substitute $33.325\times {10}^3\text{ft}/\text{s}$ for $v_A$ and $25.362\times {\text{10}}^3\text{ft}/\text{s}$ for ${\left(v_A\right)}_{\text{circ}}$.

$\begin{array}{c}\Delta v_A=33.325\times {10}^3\text{ft}/\text{s}-25.362\times {\text{10}}^3\text{ft}/\text{s}\\ =7.960\times {10}^3\text{ft}/\text{s}\\ =7960\text{ft}/\text{s}\end{array}$

Therefore, the velocity of the IUS relative to the shuttle after its engine has been fired at A $\left(\Delta v_A\right)$ is $\underset{\_}{7960\text{ft}/\text{s}}$.

To determine

Find the increase in velocity required at B $\left(\Delta v_B\right)$ to place the satellite in its final orbit.

Answer to Problem 13.101P

The increase in velocity required at B $\left(\Delta v_B\right)$ to place the satellite in its final orbit is $\underset{\_}{4820\text{ft}/\text{s}}$.

Explanation of Solution

Given information:

The distance from above the earth surface to the circular orbit (x) is $185\text{mi}$.

The altitude of the geosynchronous orbit (A) is $22230\text{mi}$.

The radius of the earth (R) is $3960\text{mi}$.

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

Calculation:

Calculate the increase in the velocity required at B to place the satellite in its final orbit $\left(\Delta v_B\right)$ using the relation:

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

Substitute $10.089\times {\text{10}}^3\text{ft}/\text{s}$ for ${\left(v_B\right)}_{\text{circ}}$ and $7.963\times {10}^3\text{ft}/\text{s}$ for $v_B$.

$\begin{array}{c}\Delta v_B=\left(10.089\times {\text{10}}^3\text{ft}/\text{s}\right)-\left(7.963\times {10}^3\text{ft}/\text{s}\right)\\ =4.820\times {\text{10}}^3\text{ft}/\text{s}\\ =4820\text{ft}/\text{s}\end{array}$

Therefore, the increase in velocity required at B $\left(\Delta v_B\right)$ to place the satellite in its final orbit is $\underset{\_}{4820\text{ft}/\text{s}}$.