Chapter 14.3, Problem 14.99P

Determine the distance traveled by the spacecraft of Prob. 14.97 during the rocket engine firing, knowing that its initial speed was 7500 ft/s and the duration of the firing was 60 s.

14.97 The weight of a spacecraft, including fuel, is 11,600 lb when the rocket engines are fired to increase its velocity by 360 ft/s. Knowing that 1000 lb of fuel is consumed, determine the relative velocity of the fuel ejected.

Fig. P14.97 and P14.98

To determine

Find the distance travelled by the spacecraft.

Answer to Problem 14.99P

The distance travelled by the spacecraft is $\underset{\_}{s=87.2\text{mi}}$.

Explanation of Solution

Given information:

The initial speed of the spacecraft is $v_0=7,500\text{ft}/\text{s}$.

The duration of firing is $t=60\text{s}$.

The change in velocity is $v-v_0=360\text{ft}/\text{s}$.

The gross weight of the spacecraft is $W_0=11,600\text{lb}$.

The fuel consumed rate is $W=1,000\text{lb}$.

Calculation:

Calculate the thrust force $\left(P\right)$ acting to the aircraft as shown below.

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

Calculate the mass $\left(m\right)$ as shown below.

$m=m_0-qt$ (1)

Calculate the acceleration $\left(a\right)$ as shown below.

$a=\frac{P}{m}$

Substitute $uq$ for F.

$a=\frac{uq}{m}$

Substitute $m_0-qt$ for m.

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

Calculate the velocity $\left(v\right)$ using the relation as shown below.

$v=v_0+{\displaystyle {\int }_0^{t}adt}$

Substitute $\frac{uq}{m_0-qt}$ for a.

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

$=v_0-u\ln \frac{m_0-qt}{m_0}$ (2)

Calculate the displacement $\left(s\right)$ as shown below.

Integrate both sides of the Equation (2) as shown below.

$s=s_0+v_{0}t-u{\displaystyle {\int }_0^t\left(\ln \frac{m_0-qt}{m_0}\right)dt}$ (3)

Consider $z=\frac{m_0-qt}{m_0}$ (4)

Differentiate both sides of the Equation (4) as shown below.

$\begin{array}{l}dz=-\frac{q}{m_0}dt\\ dt=-\frac{m_0}{q}dz\end{array}$

Substitute z for $\frac{m_0-qt}{m_0}$ and $-\frac{m_0}{q}dz$ for $dz$ and apply the limits in Equation (3).

$\begin{array}{c}s=s_0+v_{0}t-u{\displaystyle {\int }_0^t\ln z\left(-\frac{m_0}{q}dz\right)}\\ =s_0+v_{0}t+\frac{u{m}_0}{q}{\displaystyle {\int }_{z_0}^{z_1}\ln zdz}\\ =s_0+v_{0}t+\frac{u{m}_0}{q}\times {\left(z\ln z-z\right)}_{z_0}^{z_1}\end{array}$

Substitute $\frac{m_0-qt}{m_0}$ for z and apply the limits as shown below.

$\begin{array}{c}s=s_0+v_{0}t+\frac{u{m}_0}{q}\times {\left(\frac{m_0-qt}{m_0}\ln \frac{m_0-qt}{m_0}-\frac{m_0-qt}{m_0}\right)}_0^t\\ =s_0+v_{0}t+\frac{u{m}_0}{q}\times \left(\frac{m_0-qt}{m_0}\ln \frac{m_0-qt}{m_0}-\frac{m_0-qt}{m_0}-\frac{m_0}{m_0}\ln \frac{m_0}{m_0}+\frac{m_0}{m_0}\right)\\ =s_0+v_{0}t+\frac{u{m}_0}{q}\left(\frac{m_0-qt}{m_0}\ln \frac{m_0-qt}{m_0}-\frac{m_0-qt}{m_0}+1\right)\\ =s_0+v_{0}t+\frac{u{m}_0}{q}\left(\frac{m_0-qt}{m_0}\left(\ln \frac{m_0-qt}{m_0}-1\right)+1\right)\end{array}$

$\begin{array}{c}=s_0+v_{0}t+\frac{u{m}_0}{q}\left(\left(1-\frac{qt}{m_0}\right)\left(\ln \frac{m_0-qt}{m_0}-1\right)+1\right)\\ =s_0+v_{0}t+\frac{u{m}_0}{q}\left(\ln \frac{m_0-qt}{m_0}-\frac{qt}{m_0}\ln \frac{m_0-qt}{m_0}-1+\frac{qt}{m_0}+1\right)\\ =s_0+v_{0}t+\frac{u{m}_0}{q}\ln \frac{m_0-qt}{m_0}-ut\ln \frac{m_0-qt}{m_0}+ut\\ =s_0+v_{0}t+\left(ut+\left(\frac{u{m}_0}{q}-ut\right)\ln \frac{m_0-qt}{m_0}\right)\end{array}$

$=s_0+v_{0}t+u\left(t-\left(\frac{m_0}{q}-t\right)\ln \frac{m_0}{m_0-qt}\right)$ (5)

Calculate the velocity $\left(v\right)$ as shown below.

$v-v_0=360$

Substitute $7,500\text{ft}/\text{s}$ for $v_0$.

$\begin{array}{l}v-7,500=360\\ v=7,860\text{ft}/\text{s}\end{array}$

Consider the acceleration due to gravity $g=32.2\text{ft}/{\text{s}}^2$.

Calculate the gross mass of the spacecraft $\left(m_0\right)$ as shown below.

$m_0=\frac{W_0}{g}$

Substitute $11,600\text{lb}$ for $W_0$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_0=\frac{11,600}{32.2}\\ =360.25\text{lb}\end{array}$

Consider that the mass of the fuel $m_{\text{fuel}}=\frac{W_{\text{fuel}}}{g}=qt$.

Substitute $1,000\text{lb}$ for $W_{\text{fuel}}$, $32.2\text{ft}/{\text{s}}^2$ for g and $60\text{s}$ for t.

$\begin{array}{l}q\times 60=\frac{1,000}{32.2}\\ q=0.5176\text{lb}/\text{s}\end{array}$

Calculate the mass $\left(m\right)$ as shown below.

Substitute $360.25\text{lb}$ for $m_0$, $0.5176\text{lb}/\text{s}$ for q and $60\text{s}$ for t in Equation (1).

$\begin{array}{c}m=360.25-0.5176\times 60\\ =329.19\text{lb}\end{array}$

Consider the initial displacement $s_0=0$.

Calculate the relative velocity $\left(u\right)$ as shown below.

Substitute $7,860\text{ft}/\text{s}$ for $v$, $7,860\text{ft}/\text{s}$ for $v_0$, $329.19\text{lb}$ for $m_0-qt$, and $360.25\text{lb}$ for $m_0$ in Equation (2).

$\begin{array}{l}7,860=7,500-u\ln \frac{329.19}{360.25}\\ 360=u\ln \frac{360.25}{329.19}\\ u=3,992.7\\ u\approx 3,993\text{ft}/\text{s}\end{array}$

Calculate distance travelled by the spacecraft $\left(s\right)$ as shown below.

Substitute $0$ for $s_0$, $7,500\text{ft}/\text{s}$ for $v_0$, $3,993\text{ft}/\text{s}$ for $u$, $329.19\text{lb}$ for $m_0-qt$, $0.5176\text{lb}/\text{s}$ for $q$, $60\text{s}$ for $t$, and $360.25\text{lb}$ for $m_0$ in Equation (2).

$\begin{array}{c}s=0+7,500\times 60+3,993\times \left(60-\left(\frac{360.25}{0.5176}-60\right)\ln \frac{360.25}{329.19}\right)\\ =450\times {10}^3+3,993\times \left(60-57.344\right)\\ =450\times {10}^3+10.6\times {10}^3\\ =460.6\times {10}^3\text{ft}\times \frac{1\text{mi}}{5,280\text{ft}}\end{array}$

$=87.2\text{mi}$

Therefore, the distance travelled by the spacecraft is $\underset{\_}{s=87.2\text{mi}}$.