Chapter 3, Problem 3.13P

To determine

To show the rocket’s height as a function of $t$ as $y\left(t\right)=v_{\text{ex}}t-\frac{1}{2}g{t}^2-\frac{m{v}_{\text{ex}}}{k}\ln \left(\frac{m}{m_0}\right)$, and to estimate the space shuttle’s height after two minutes.

Answer to Problem 3.13P

The rocket’s height as a function of $t$ is shown as $\underset{\_}{y\left(t\right)=v_{\text{ex}}t-\frac{1}{2}g{t}^2-\frac{m{v}_{\text{ex}}}{k}\ln \left(\frac{m}{m_0}\right)}$, and the space shuttle’s height after two minutes is approximately $\underset{\_}{4.0\times {10}^4\text{m}}$.

Explanation of Solution

From Problem 11, the equation of motion of rocket is given by

    $m\dot{v}=-\dot{m}{v}_{\text{ext}}-mg$

Here, $m$ is the mass, $\dot{v}$ is the acceleration, $\dot{m}$ is the variable mass, $v_{\text{ext}}$ is the exhaust speed, and $g$ is the acceleration due to gravity.

Rearrange the above equation.

    $\begin{array}{c}m\dot{v}+mg=-\dot{m}{v}_{\text{ext}}\\ \frac{dv}{dt}+g=\frac{-\dot{m}}{m}{v}_{\text{ext}}\end{array}$

Integrate the above equation.

    $v+gt=-{\displaystyle \int \frac{\dot{m}}{m}{v}_{\text{ext}}}dt$

Since ${\displaystyle \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}}dx=\ln f\left(x\right)+C$, the above equation is reduced as

    $\begin{array}{c}v+gt=-v_{\text{ext}}\left(\ln {\left(m\right)}_{m_0}^m-C\right)\\ v+gt=-v_{\text{ext}}\left(\ln \left(m\right)-\ln \left(m_0\right)\right)\\ v+gt=-v_{\text{ex}}\ln \left(\frac{m}{m_0}\right)\end{array}$        (I)

From problem 3.11, the value of $m$ is obtained as $m_0-kt$.

Use the above equation in equation (I).

    $\begin{array}{c}v+gt=-v_{\text{ex}}\ln \left(\frac{m_0-kt}{m_0}\right)\\ v=-v_{\text{ex}}\ln \left(\frac{m_0-kt}{m_0}\right)-gt\end{array}$        (II)

Use velocity $v$ as $\frac{dy}{dt}$ in the above equation.

    $\begin{array}{c}\frac{dy}{dt}=-v_{\text{ex}}\ln \left(\frac{m_0-kt}{m_0}\right)-gt\\ dy=\left(-v_{\text{ex}}\ln \left(\frac{m_0-kt}{m_0}\right)-gt\right)dt\end{array}$

Integrate the above equation, $dy$ from $0\text{to }y\left(t\right)$, and $dt$ from $0\text{to }t$.

    $\begin{array}{c}{\displaystyle {\int }_0^{y\left(t\right)}dy}={\displaystyle {\int }_0^t\left(-v_{\text{ex}}\ln \left(\frac{m_0-kt}{m_0}\right)-gt\right)}dt\\ y\left(t\right)=-v_{\text{ex}}{\displaystyle {\int }_0^t\left(\ln \left(\frac{m_0-kt}{m_0}\right)\right)}dt-{\displaystyle {\int }_0^{t}gt}dt\\ =-v_{\text{ex}}{\displaystyle {\int }_0^t\left(\ln \left(1-\frac{kt}{m_0}\right)dt\right)}-\frac{1}{2}g{t}^2\\ =-v_{ex}{\left[\left(1-\frac{kt}{m_0}\right)\ln \left(1-\frac{kt}{m_0}\right)-\left(1-\frac{kt}{m_0}\right)\right]}_0^t\left(\frac{1}{-\frac{k}{m_0}}\right)-\frac{1}{2}g{t}^2\end{array}$        (III)

Since ${\displaystyle \int \ln }(1-ax)dx=\frac{\left(1-ax\right)\ln \left(1-ax\right)-\left(1-ax\right)}{-a}$, where $a=\frac{k}{m_0}$.

Use the above condition in equation (III).

    $\begin{array}{c}y(t)=\frac{m_0{v}_{ex}}{k}{\left[\left(1-\frac{kt}{m_0}\right)\ln \left(1-\frac{kt}{m_0}\right)-\left(1-\frac{kt}{m_0}\right)\right]}_0^t-\frac{1}{2}g{t}^2\\ =\frac{m_0{v}_{ex}}{k}\left[\left\{\left(1-\frac{kt}{m_0}\right)\ln \left(1-\frac{kt}{m_0}\right)-\left(1-\frac{kt}{m_0}\right)\right\}-(-1)\right]-\frac{1}{2}g{t}^2\\ =\frac{m_0{v}_{ex}}{k}\left[\left\{\left(1-\frac{kt}{m_0}\right)\ln \left(1-\frac{kt}{m_0}\right)-\left(1-\frac{kt}{m_0}\right)\right\}+1\right]-\frac{1}{2}g{t}^2\end{array}$        (IV)

Rearrange the equation $m=m_0-kt$, to substitute for $\frac{m}{m_0}$

    $\begin{array}{c}m=m_0-kt\\ \frac{m}{m_0}=1-\frac{kt}{m_0}\end{array}$        (V)

Use equation (V) in (IV).

    $\begin{array}{c}y(t)=\frac{m_0{v}_{ex}}{k}\left[\frac{m}{m_0}\ln \left(\frac{m}{m_0}\right)-\left(\frac{m}{m_0}\right)+1\right]-\frac{1}{2}g{t}^2\\ =\frac{m_0{v}_{ex}}{k}\left[-\frac{m}{m_0}\ln \left(\frac{m_0}{m}\right)+1-\left(\frac{m}{m_0}\right)\right]-\frac{1}{2}g{t}^2\\ =\frac{m_0{v}_{ex}}{k}\left[-\frac{m}{m_0}\ln \left(\frac{m_0}{m}\right)+\left(\frac{m_0-m}{m_0}\right)\right]-\frac{1}{2}g{t}^2\end{array}$        (VI)

Rearrange the equation $m=m_0-kt$, for $kt$.

    $\begin{array}{c}m=m_0-kt\\ kt=m_0-m\end{array}$

Use the above equation in equation (VI).

    $\begin{array}{c}y(t)=\frac{m_0{v}_{ex}}{k}\left[-\frac{m}{m_0}\ln \left(\frac{m_0}{m}\right)+\left(\frac{kt}{m_0}\right)\right]-\frac{1}{2}g{t}^2\\ =v_{ex}t-\frac{m{v}_{ex}}{k}\ln \left(\frac{m_0}{m}\right)-\frac{1}{2}g{t}^2\\ =v_{ex}t-\frac{1}{2}g{t}^2-\frac{m{v}_{ex}}{k}\ln \left(\frac{m_0}{m}\right)\end{array}$        (VII)

Substitute $\frac{m_0-m}{t}$ for $k$ in the above equation.

    $\begin{array}{c}y(t)=v_{ex}t-\frac{1}{2}g{t}^2-\frac{m{v}_{ex}}{\left(\frac{m_0-m}{t}\right)}\ln \left(\frac{m_0}{m}\right)\\ =v_{ex}t-\frac{1}{2}g{t}^2-\frac{m{v}_{ex}t}{\left(m_0-m\right)}\ln \left(\frac{m_0}{m}\right)\end{array}$        (VIII)

Conclusion:

Substitute $2\text{min }$ for $t$, $3000\text{m}/\text{s}$ for $v_{ex}$, $9.8\text{m}/{\text{s}}^2$ for $g$, $2\times {10}^6\text{kg}$ for $m_0$, and $1\times {10}^6\text{kg}$ for $m$ in equation (VIII), to find $y\left(t\right)$.

$\begin{array}{c}y=\left(3000\text{m}/\text{s}\right)\left(2\text{min }\times \frac{60\text{s}}{1\text{min}}\right)-\frac{1}{2}\left(9.8\text{m}/{\text{s}}^2\right){\left(2\text{min }\times \frac{60\text{s}}{1\text{min}}\right)}^2-\frac{\left(1\times {10}^6\text{kg}\right)\left(3000\text{m}/\text{s}\right)\left(2\text{min }\times \frac{60\text{s}}{1\text{min}}\right)}{\left(\left(2\times {10}^6\text{kg}\right)-\left(1\times {10}^6\text{kg}\right)\right)}\ln \left(\frac{2\times {10}^6\text{kg}}{1\times {10}^6\text{kg}}\right)\\ =\left(3000\text{m}/\text{s}\right)\left(120\text{s}\right)-\frac{1}{2}\left(9.8\text{m}/{\text{s}}^2\right){\left(120\text{s}\right)}^2-\frac{\left(1\times {10}^6\text{kg}\right)\left(3000\text{m}/\text{s}\right)\left(120\text{s}\right)}{\left(\left(2\times {10}^6\text{kg}\right)-\left(1\times {10}^6\text{kg}\right)\right)}\ln \left(\frac{2\times {10}^6\text{kg}}{1\times {10}^6\text{kg}}\right)\\ =4.0\times {10}^4\text{m}\end{array}$

Therefore, The rocket’s height as a function of $t$ is shown as $\underset{\_}{y\left(t\right)=v_{\text{ex}}t-\frac{1}{2}g{t}^2-\frac{m{v}_{\text{ex}}}{k}\ln \left(\frac{m}{m_0}\right)}$, and the space shuttle’s height after two minutes is approximately $\underset{\_}{4.0\times {10}^4\text{m}}$.