Chapter 14.3, Problem 14.102P

(a)

To determine

Find the altitude at which stage A of the rocket is released.

Answer to Problem 14.102P

The altitude at which stage A of the rocket is released is $\underset{\_}{31.2\text{km}}$.

Explanation of Solution

Given information:

Consider the initial mass of the rocket is denoted by $m_0$.

The rate of consumption of the fuel is denoted by q.

The velocity of the rocket after time t is denoted by v.

The acceleration due to gravity is denoted by g.

Refer Problem 14.96.

The mass of the rocket is $540\text{kg}$.

The mass of the stage A and B is $9.5\text{Mg}$.

Show the unit conversion of the mass as follows:

$\begin{array}{c}{m}_s=9.5\text{Mg}\\ =\text{9500}\text{kg}\end{array}$

The rate of fuel consumption is $225\text{kg/s}$.

The relative velocity of the rocket is $3600\text{m/s}$.

Calculation:

Show the thrust force (P) of the rocket as follows:

$\begin{array}{c}P=u\frac{dm}{dt}\\ =uq\end{array}$ (1)

Show the combined mass (m) of the rocket and the unspent fuel as follows:

$m=m_0-qt$ (2)

Show the weight force (W) as follows:

$W=mg$

Show the acceleration (a) of the rocket as follows:

$\begin{array}{c}F=ma\\ a=\frac{F}{m}\\ =\frac{P-W}{m}\\ =\frac{P-mg}{m}\end{array}$

$a=\frac{P}{m}-g$ (3)

Modify Equation (3) using (2) and (1).

$a=\frac{uq}{m_0-qt}-g$

Show the velocity of the rocket as follows:

Integrate above Equation with respect to time.

$\begin{array}{c}v=v_0+{\displaystyle {\int }_0^{t}adt}\\ =v_0+{\displaystyle {\int }_0^t\left(\frac{uq}{m_0-qt}-g\right)dt}\\ =v_0-u\ln \frac{m_0-qt}{m_0}-gt\end{array}$

Show the displacement of the rocket as follows:

Integrate above Equation with respect to time.

$s=s_0+v_{0}t+u{\displaystyle {\int }_0^t\ln \frac{m_0-qt}{m_0}dt}-\frac{1}{2}g{t}^2$ (4)

Consider the value of z as follows:

$z=\frac{m_0-qt}{m_0}$ (5)

Differentiate the above Equation with respect to time t.

$\begin{array}{l}\frac{dz}{dt}=\frac{d}{dt}\left(\frac{m_0-qt}{m_0}\right)\\ \frac{dz}{dt}=-\frac{q}{m_0}\\ dz=-\frac{q}{m_0}dt\\ dt=-\frac{m_0}{q}dz\end{array}$ (6)

Consider the value of the $z=z_0$ at time $t=0$.

Substitute $z_0$ for $z$ and $0$ for $t$ in Equation (5).

$\begin{array}{c}{z}_0=\frac{m_0+q\times 0}{m_0}\\ =1\end{array}$

Modify Equation (4) using Equation (5) and (6).

Take the lower and upper limit of the integral as $z_0$ and z.

$\begin{array}{c}s=s_0+v_{0}t-u\frac{m_0}{q}{\displaystyle {\int }_{z_0}^z\ln zdz}-\frac{1}{2}g{t}^2\\ =s_0+v_{0}t-u\frac{m_0}{q}\left[z{\left(\ln z+z\right)}_{z_0}^z\right]-\frac{1}{2}g{t}^2\\ =s_0+v_{0}t-u\frac{m_0}{q}\left[z\left(\ln z+z\right)-z_0\left(\ln z_0+z_0\right)\right]-\frac{1}{2}g{t}^2\end{array}$

Substitute 1 for $z_0$ and $\frac{m_0-qt}{m_0}$ for z.

$\begin{array}{c}s=s_0+v_{0}t-u\frac{m_0}{q}\left[\frac{m_0-qt}{m_0}\left(\ln \frac{m_0-qt}{m_0}+\frac{m_0-qt}{m_0}\right)-1\left(\ln 1+1\right)\right]-\frac{1}{2}g{t}^2\\ =s_0+v_{0}t-u\frac{m_0}{q}\left[\frac{m_0-qt}{m_0}\left(\ln \frac{m_0-qt}{m_0}+\frac{m_0-qt}{m_0}\right)-1\right]-\frac{1}{2}g{t}^2\\ =s_0+v_{0}t-u\left[t+\left(\frac{m_0}{q}-t\right)\ln \frac{m_0-qt}{m_0}\right]-\frac{1}{2}g{t}^2\end{array}$ (7)

The mass of the fuel is $m=8900\text{kg}$.

The velocity of the rocket is $u=3600\text{m/s}$.

The rate of the fuel consumption is $q=225\text{kg/s}$.

Calculate the time taken (t) as follows:

$\begin{array}{c}t=\frac{m}{q}\\ =\frac{8900}{225}\\ =39.556\text{s}\end{array}$

Consider the first stage:

The initial velocity and initial distance covered are $v_0=0$ and $s_0=0$.

Calculate the initial mass of the rocket $m_0$ as follows:

$\begin{array}{c}{m}_0=540\text{kg}+2\times 9500\text{kg}\\ \text{=19,540}\text{kg}\end{array}$

Calculate the final velocity of the rocket using the relation:

$v_1=v_0-u\ln \frac{m_0-qt}{m_0}-gt$

Substitute 0 for $v_0$, $\text{19,540}\text{kg}$ for $m_0$, $225\text{kg/s}$ for q, $3600\text{m/s}$ for $u$, $39.556\text{s}$ for t, and $9.81\text{m/s}$ for g.

$\begin{array}{c}{v}_1=0-3600\ln \frac{19540-225\times 39.556}{m_0}-9.81\times 39.556\\ =1800.1\text{m/s}\end{array}$

Calculate the altitude $\left(s_1\right)$ at which stage A of the rocket is released using the relation:

Substitute 0 for $s_0$, 0 for $v_0$, $\text{19,540}\text{kg}$ for $m_0$, $225\text{kg/s}$ for q, $3600\text{m/s}$ for $u$, $39.556\text{s}$ for t, and $9.81\text{m/s}$ for g in Equation (7).

$\begin{array}{c}s=\left\{\begin{array}{l}0+0\times 39.556\\ -3600\left[39.556+\left(\frac{19,540}{225}-39.556\right)\ln \frac{19540-225\times 39.556}{19540}\right]\\ -\frac{1}{2}\times 9.81\times {39.556}^2\end{array}\right\}\\ =31249\text{m}\times \frac{1\text{km}}{1000\text{m}}\\ =31.2\text{km}\end{array}$

Thus, the altitude $\left(s_1\right)$ at which stage A of the rocket is released is $\underset{\_}{31.2\text{km}}$.

(b)

To determine

Find the altitude at which the fuel of both stages are consumed.

Answer to Problem 14.102P

The altitude at which the fuel of both stages are consumed is $\underset{\_}{197.5\text{km}}$.

Explanation of Solution

Given information:

Calculation:

Consider the fuel of both the stages are consumed.

Calculate the final mass $\left(m_1\right)$ of the rocket as follows:

$\begin{array}{c}{m}_1=540\text{kg}+9500\text{kg}\\ \text{=10,040}\text{kg}\end{array}$

Calculate the altitude $\left(s_2\right)$ at which stage A of the rocket is released using the relation:

$s_2=s_1+v_{1}t-u\left[t+\left(\frac{m_1}{q}-t\right)\ln \frac{m_1-qt}{m_1}\right]-\frac{1}{2}g{t}^2$

Substitute $31249\text{m}$ for $s_1$, $1800.1\text{m/s}$ for $v_1$, $\text{10,040}\text{kg}$ for $m_1$, $225\text{kg/s}$ for q, $3600\text{m/s}$ for $u$, $39.556\text{s}$ for t, and $9.81\text{m/s}$ for g in Equation (7).

$\begin{array}{c}s=\left\{\begin{array}{l}31249+1800.1\times 39.556\\ -3600\left[39.556+\left(\frac{19,540}{225}-39.556\right)\ln \frac{10040-225\times 39.556}{10040}\right]\\ -\frac{1}{2}\times 9.81\times {39.556}^2\end{array}\right\}\\ =197,502\text{m}\times \frac{1\text{km}}{1000\text{m}}\\ =197.5\text{km}\end{array}$

Thus, the altitude $\left(s_2\right)$ at which both stage is consumed is $\underset{\_}{197.5\text{km}}$.