Chapter 14.3, Problem 14.100P

A rocket weighs 2600 lb, including 2200 lb of fuel, which is consumed at the rate of 25 lb/s and ejected with a relative velocity of 13,000 ft/s. Knowing that the rocket is fired vertically from the ground, determine (a) its acceleration as it is fired, (b) its acceleration as the last particle of fuel is being consumed, (c) the altitude at which all the fuel has been consumed, (d) the velocity of the rocket at that time.

To determine

The acceleration of the rocket as it is fired.

Answer to Problem 14.100P

The acceleration of the rocket as it is fired is $\underset{\_}{a=92.8\text{ft}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

Given information:

The fuel consumed rate is $\frac{dm}{dt}=25\text{lb}/\text{s}$.

The relative velocity is $u=13,000\text{ft}/\text{s}$.

The gross weight of the rocket is $W_0=2,600\text{lb}$.

The weight of the fuel is $W_{\text{fuel}}=2,200\text{lb}$.

Calculation:

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

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

$P=u\frac{dm}{dt}$

Substitute $13,000\text{ft}/\text{s}$ for u and $25\text{lb}/\text{s}$ for $\frac{dm}{dt}$.

$\begin{array}{c}P=13,000\text{ft}/\text{s}\times \frac{25\text{lb}/\text{s}}{32.2\text{ft}/{\text{s}}^2}\\ =10,093\text{lb}\end{array}$

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

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

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

$\begin{array}{c}m=\frac{2,600}{32.2}\\ =80.745\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\times \frac{1\text{slugs}}{1\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}}\\ =80.745\text{slugs}\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F}=ma\\ P-mg=ma\end{array}$ (1)

Substitute $10,093\text{lb}$ for P, $80.745\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m$, and $32.2\text{ft}/{\text{s}}^2$ for g in Equation (1).

$\begin{array}{l}10,093-80.745\times 32.2=80.745a\\ 7,493=80.745a\\ a=92.8\text{ft}/{\text{s}}^2\end{array}$

Hence, acceleration of the rocket as it is fired is $\underset{\_}{a=92.8\text{ft}/{\text{s}}^{\text{2}}}$.

To determine

The acceleration of the rocket as the last particle of fuel is being consumed.

Answer to Problem 14.100P

The acceleration of the rocket as the last particle of the fuel is being consumed is $\underset{\_}{a=780\text{ft}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

Given information:

The fuel consumed rate is $\frac{dm}{dt}=25\text{lb}/\text{s}$.

The relative velocity is $u=13,000\text{ft}/\text{s}$.

The gross weight of the rocket is $W_0=2,600\text{lb}$.

The weight of the fuel is $W_{\text{fuel}}=2,200\text{lb}$.

Calculation:

Refer to part (a).

The thrust force $P=10,093\text{lb}$.

Calculate the weight of the rocket $\left(W\right)$ after the fuel is being consumed as shown below.

$W=W_0-W_{\text{fuel}}$

Substitute $2,600\text{lb}$ for $W_0$ and $2,200\text{lb}$ for $W_{\text{fuel}}$.

$\begin{array}{c}W=2,600-2,200\\ =400\text{lb}\end{array}$

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

$m=\frac{W}{g}$

Substitute $400\text{lb}$ for $W$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}m=\frac{400}{32.2}\\ =12.422\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\times \frac{1\text{slugs}}{1\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}}\\ =12.422\text{slugs}\end{array}$

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

Substitute $10,093\text{lb}$ for P, $12.422\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m$, and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{l}10,093-12.422\times 32.2=12.422a\\ 9,693=12.422a\\ a=780\text{ft}/{\text{s}}^2\end{array}$

Hence, acceleration of the rocket as the last particle of the fuel is being consumed is $\underset{\_}{a=780\text{ft}/{\text{s}}^{\text{2}}}$.

To determine

The altitude at which all the fuel has been consumed

Answer to Problem 14.100P

The altitude of the rocket is $\underset{\_}{h=119.3\text{mi}}$.

Explanation of Solution

Given information:

The fuel consumed rate is $q=\frac{dm}{dt}=25\text{lb}/\text{s}$.

The relative velocity is $u=13,000\text{ft}/\text{s}$.

The gross weight of the rocket is $W_0=2,600\text{lb}$.

The weight of the fuel is $W_{\text{fuel}}=2,200\text{lb}$.

Calculation:

Consider the weight of the fuel as $W_{\text{fuel}}=m=2,200\text{lb}$.

Calculate the time $\left(t\right)$ as shown below.

$\begin{array}{l}\frac{dm}{dt}=25\\ dt=\frac{1}{25}dm\end{array}$

Integrate both sides of the Equation.

$\begin{array}{l}{\displaystyle {\int }_0^{t}dt}=\frac{1}{25}{\displaystyle {\int }_0^{m}dm}\\ t=\frac{1}{25}m\end{array}$

Substitute $2,200\text{lb}$ for $m$.

$\begin{array}{c}t=\frac{2,200}{25}\\ =88\text{s}\end{array}$

Refer to sample problem 14.8 in the Text book,

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

$\begin{array}{c}v=u\ln \frac{m_0}{m_0-qt}-gt\\ =-u\ln \frac{m_0-qt}{m_0}-gt\end{array}$ (2)

Consider $\frac{dy}{dt}=v$.

Substitute $\frac{dy}{dt}$ for v in Equation (2).

$\frac{dy}{dt}=-u\ln \frac{m_0-qt}{m_0}-gt$

Calculate the altitude $\left(h\right)$ as shown below.

Integrate both sides of the Equation with respect to t.

$\begin{array}{c}h={\displaystyle {\int }_0^{h}dy}\\ ={\displaystyle {\int }_0^{t}vdt}\end{array}$

Substitute $-u\ln \frac{m_0-qt}{m_0}-gt$ for v.

$\begin{array}{c}h={\displaystyle {\int }_0^t\left(-u\ln \frac{m_0-qt}{m_0}-gt\right)dt}\\ =-u{\displaystyle {\int }_0^t\ln \frac{m_0-qt}{m_0}dt}-{\displaystyle {\int }_0^{t}gtdt}\\ =-u{\displaystyle {\int }_0^t\ln \frac{m_0-qt}{m_0}dt}-\frac{1}{2}g{t}^2\end{array}$ (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 $dt$ and apply the limits in Equation (3).

$\begin{array}{c}h=-u{\displaystyle {\int }_0^t\ln z\left(-\frac{m_0}{q}dz\right)}-\frac{1}{2}g{t}^2\\ =\frac{m_{0}u}{q}{\displaystyle {\int }_{z_0}^z\ln zdz}-\frac{1}{2}g{t}^2\\ =\frac{m_{0}u}{q}{\left[\left(z\ln z-z\right)\right]}_{z_0}^z-\frac{1}{2}g{t}^2\end{array}$

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

$\begin{array}{c}h=\frac{m_{0}u}{q}{\left[\left(\frac{m_0-qt}{m_0}\ln \frac{m_0-qt}{m_0}-\frac{m_0-qt}{m_0}\right)\right]}_0^t-\frac{1}{2}g{t}^2\\ =\frac{m_{0}u}{q}\left[\left(\frac{m_0-qt}{m_0}\ln \frac{m_0-qt}{m_0}-\frac{m_0-qt}{m_0}\right)-\left(\frac{m_0}{m_0}\ln \frac{m_0}{m_0}-\frac{m_0}{m_0}\right)\right]-\frac{1}{2}g{t}^2\\ =\frac{m_{0}u}{q}\left[\left(\frac{m_0-qt}{m_0}\left(\ln \frac{m_0-qt}{m_0}-1\right)\right)+1\right]-\frac{1}{2}g{t}^2\\ =\frac{m_{0}u}{q}\left[\left(1-\frac{qt}{m_0}\right)\left(\ln \frac{m_0-qt}{m_0}-1\right)+1\right]-\frac{1}{2}g{t}^2\end{array}$

$\begin{array}{c}=\frac{m_{0}u}{q}\left(\ln \frac{m_0-qt}{m_0}-1+1\right)-ut\left(\ln \frac{m_0-qt}{m_0}-1\right)-\frac{1}{2}g{t}^2\\ =u\left(t-\left(\frac{m_0}{q}-t\right)\ln \frac{m_0}{m_0-qt}\right)-\frac{1}{2}g{t}^2\end{array}$ (5)

Calculate the altitude $\left(h\right)$ as shown below.

Substitute $13,000\text{ft}/\text{s}$ for $u$, $2,600\text{lb}$ for $m_0$, $25\text{lb}/\text{s}$ for $q$, $88\text{s}$ for t, and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ in Equation (5).

$\begin{array}{c}h=13,000\left(88-\left(\frac{2,600}{25}-88\right)\ln \frac{2,600}{2,600-25\times 88}\right)-\frac{1}{2}\times 32.2\times {88}^2\\ =13,000\left(88-16\ln 6.5\right)-124,678.4\\ =629,987\text{ft}\times \frac{1\text{mi}}{5,280\text{ft}}\\ =119.3\text{mi}\end{array}$

Hence, the altitude of the rocket is $\underset{\_}{h=119.3\text{mi}}$.

To determine

The velocity of the rocket.

Answer to Problem 14.100P

The velocity of the rocket is $\underset{\_}{v=14,660\text{mi}/\text{h}}$.

Explanation of Solution

Given information:

The fuel consumed rate is $q=\frac{dm}{dt}=25\text{lb}/\text{s}$.

The relative velocity is $u=13,000\text{ft}/\text{s}$.

The gross weight of the rocket is $W_0=2,600\text{lb}$.

The weight of the fuel is $W_{\text{fuel}}=2,200\text{lb}$.

Calculation:

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

$\begin{array}{c}v=u\ln \frac{m_0}{m_0-qt}-gt\\ =-u\ln \frac{m_0-qt}{m_0}-gt\end{array}$ (2)

Substitute $13,000\text{ft}/\text{s}$ for $u$, $2,600\text{lb}$ for $m_0$, $25\text{lb}/\text{s}$ for $q$, $88\text{s}$ for t, and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ in Equation (2).

$\begin{array}{c}v=-13,000\ln \frac{2,600-25\times 88}{2,600}-32.2\times 88\\ =13,000\ln \frac{2,600}{2,600-25\times 88}-2,833.6\\ =21,500\text{ft}/\text{s}\times \frac{1\text{mi}}{5,280\text{ft}}\times \frac{3,600\text{s}}{1\text{h}}\\ =14,659\text{mi}/\text{h}\end{array}$

$\simeq 14,660\text{mi}/\text{h}$

Therefore, the velocity of the rocket is $\underset{\_}{v=14,660\text{mi}/\text{h}}$.