Chapter 9, Problem 34P

A rocket has total mass M_i = 360 kg, including M_fuel = 330 kg of fuel and oxidizer. In interstellar space, it starts from rest at the position x = 0, turns on its engine at time t = 0, and puts out exhaust with relative speed v_e = 1 500 m/s at the constant rate k = 2.50 kg/s. The fuel will last for a burn time of T_b = M_fuel/k = 330 kg/(2.5 kg/s) = 132 s. (a) Show that during the burn the velocity of the rocket as a function of time is given by

$v(t)=-v_e\ln \left(1-\frac{kt}{M_i}\right)$

(b) Make a graph of the velocity of the rocket as a function of time for times running from 0 to 132 s. (c) Show that the acceleration of the rocket is

$a(t)=\frac{k{v}_e}{M_i-kt}$

(d) Graph the acceleration as a function of time. (c) Show that the position of the rocket is

$x(t)=v_e\left(\frac{M_i}{k}-t\right)\ln \left(1-\frac{kt}{M_i}\right)+v_{e}t$

(f) Graph the position during the burn as a function of time.

To determine

The velocity of the rocket is $v\left(t\right)=-v_{\text{e}}\ln \left(1-\frac{kt}{M_{\text{i}}}\right)$.

Answer to Problem 34P

The velocity of the rocket is $v\left(t\right)=-v_{\text{e}}\ln \left(1-\frac{kt}{M_{\text{i}}}\right)$.

Explanation of Solution

The total mass of rocket is $M_{\text{i}}=360\text{kg}$ and mass of fuel is $330\text{kg}$. The relative speed of exhaust is $1500\text{m}/\text{s}$. The rate discharge of exhaust is $2.5\text{kg}/\text{s}$.

Rocket works on Newton’s third law, the thrust generate by the exhaust creates a reaction which causing the rocket to fly.

The basic expression of the rocket propulsion is,

  $v_{\text{f}}-v_{\text{i}}=v_{\text{e}}\ln \left(\frac{M_{\text{i}}}{M_{\text{f}}}\right)$        (I)

Here, $v_{\text{e}}$ is the exhaust speed, $v_{\text{f}}$ is the final velocity of rocket and $v_{\text{i}}$ is the initial velocity of rocket.

Formula to calculate the total mass of rocket is,

    $\begin{array}{c}{M}_{\text{i}}=M_{\text{f}}+kt\\ M_{\text{f}}=M_{\text{i}}-kt\end{array}$

Here, $M_{\text{f}}$ is the mass of fuel and $t$ is the time duration.

Conclusion:

Substitute $0$ for $v_{\text{i}}$ and $M_{\text{i}}-kt$ for $M_{\text{f}}$ in equation (I).

    $\begin{array}{c}{v}_{\text{f}}-0=v_{\text{e}}\ln \left(\frac{M_{\text{i}}}{M_{\text{i}}-kt}\right)\\ v_{\text{f}}=-v_{\text{e}}\ln \left(\frac{M_{\text{i}}-kt}{M_{\text{i}}}\right)\\ =-v_{\text{e}}\ln \left(1-\frac{kt}{M_{\text{i}}}\right)\end{array}$

Therefore, The velocity of the rocket is $v\left(t\right)=-v_{\text{e}}\ln \left(1-\frac{kt}{M_{\text{i}}}\right)$.

To determine

The graph of the velocity of the rocket as a function of time for times running from 0 to $132\text{s}$.

Answer to Problem 34P

The graph of the velocity of the rocket as a function of time for times running from 0 to $132\text{s}$ is shown in figure I.

Explanation of Solution

The velocity of the rocket is,

  $v\left(t\right)=-v_{\text{e}}\ln \left(1-\frac{kt}{M_{\text{i}}}\right)$.

Substitute $1500\text{m}/\text{s}$ for $v_{\text{e}}$, $2.5\text{kg}/\text{s}$ for $k$ and $360\text{kg}$ for $M_{\text{i}}$ in above equation.

  $\begin{array}{c}v\left(t\right)=-\left(1500\text{m}/\text{s}\right)\ln \left(1-\frac{\left(2.5\text{kg}/\text{s}\right)t}{360\text{kg}}\right)\\ =-\left(1500\text{m}/\text{s}\right)\ln \left(1-6.944\times {10}^{-3}t\right)\end{array}$

The value of $v\left(t\right)$ at the different value of $t$ is calculated and shown below.

$t(s)$	$v(t)\left(\text{m}/\text{s}\right)$
0	0
10	107.95
20	224.28
30	350.39
40	488.09
50	639.72
60	808.43
70	998.53
80	1216.27
90	1471.08
100	1778.21
110	2164.86
120	2687.16
130	3495.24
132	3726.3

Draw the graph between $v\left(t\right)$ and $t$ as shown below.

Figure I

Conclusion:

Therefore, the graph of the velocity of the rocket as a function of time for times running from 0 to $132\text{s}$ is shown in figure I.

To determine

The acceleration of the rocket is $a(t)=\frac{k{v}_{\text{e}}}{M_{\text{i}}-kt}$.

Answer to Problem 34P

The acceleration of the rocket is $a(t)=\frac{k{v}_{\text{e}}}{M_{\text{i}}-kt}$.

Explanation of Solution

The velocity of the rocket is,

    $v\left(t\right)=-v_{\text{e}}\ln \left(1-\frac{kt}{M_{\text{i}}}\right)$.

Formula to calculate the acceleration is,

    $a=\frac{dv(t)}{dt}$

Conclusion:

Substitute $-v_{\text{e}}\ln \left(1-\frac{kt}{M_{\text{i}}}\right)$ for $v(t)$ in above equation.

    $\begin{array}{c}a=\frac{d\left[-v_{\text{e}}\ln \left(1-\frac{kt}{M_{\text{i}}}\right)\right]}{dt}\\ =-v_{\text{e}}\frac{1}{1-\frac{kt}{M_{\text{i}}}}\left(-\frac{k}{M_{\text{i}}}\right)\\ =\frac{k{v}_{\text{e}}}{M_{\text{i}}-kt}\end{array}$

Therefore, the acceleration of the rocket is $a(t)=\frac{k{v}_{\text{e}}}{M_{\text{i}}-kt}$.

To determine

The graph of the acceleration of the rocket as a function of time for times running from 0 to $132\text{s}$.

Answer to Problem 34P

The graph of the acceleration of the rocket as a function of time for times running from 0 to $132\text{s}$ is given in figure II.

Explanation of Solution

The acceleration of the rocket is,

    $a(t)=\frac{k{v}_{\text{e}}}{M_{\text{i}}-kt}$.

Substitute $1500\text{m}/\text{s}$ for $v_{\text{e}}$, $2.5\text{kg}/\text{s}$ for $k$ and $360\text{kg}$ for $M_{\text{i}}$ in above equation.

    $\begin{array}{c}a(t)=\frac{\left(2.5\text{kg}/\text{s}\right)\left(1500\text{m}/\text{s}\right)}{360\text{kg}-\left(2.5\text{kg}/\text{s}\right)t}\\ =\frac{3750\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{360\text{kg}-\left(2.5\text{kg}/\text{s}\right)t}\end{array}$

The value of $a\left(t\right)$ at the different value of $t$ is calculated and shown below.

$t(s)$	$a(t)\left(\text{m}/{\text{s}}^2\right)$
0	10.42
10	11.12
20	12.09
30	13.15
40	14.42
50	15.95
60	17.85
70	20.27
80	23.43
90	27.78
100	34.09
110	44.12
120	62.5
130	107.14
132	125

Draw the graph between $v\left(t\right)$ and $t$ as shown below.

Figure II

Conclusion:

Therefore, the graph of the acceleration of the rocket as a function of time for times running from 0 to $132\text{s}$ is shown in figure II.

To determine

The position of the rocket is $x(t)=v_{\text{e}}\left(\frac{M_{\text{i}}}{k}-t\right)\ln \left(1-\frac{kt}{M_{\text{i}}}\right)+v_{\text{e}}t$.

Answer to Problem 34P

The position of the rocket is $x(t)=v_{\text{e}}\left(\frac{M_{\text{i}}}{k}-t\right)\ln \left(1-\frac{kt}{M_{\text{i}}}\right)+v_{\text{e}}t$.

Explanation of Solution

The velocity of the rocket is,

    $v\left(t\right)=-v_{\text{e}}\ln \left(1-\frac{kt}{M_{\text{i}}}\right)$.

Formula to calculate the position is,

    $x\left(t\right)={\displaystyle {\int }_0^{\text{t}}v}\left(t\right)dt$

Substitute $-v_{\text{e}}\ln \left(1-\frac{kt}{M_{\text{i}}}\right)$ for $v(t)$ in above equation.

    $\begin{array}{c}x\left(t\right)={\displaystyle {\int }_0^{\text{t}}-v_{\text{e}}\ln \left(1-\frac{kt}{M_{\text{i}}}\right)}dt\\ =v_{\text{e}}\left(\frac{M_{\text{i}}}{k}-t\right)\ln \left(1-\frac{kt}{M_{\text{i}}}\right)+v_{\text{e}}t\end{array}$

Conclusion:

Therefore, the position of the rocket is $x(t)=v_{\text{e}}\left(\frac{M_{\text{i}}}{k}-t\right)\ln \left(1-\frac{kt}{M_{\text{i}}}\right)+v_{\text{e}}t$.

To determine

The graph of the position of the rocket as a function of time for times running from 0 to $132\text{s}$.

Answer to Problem 34P

The graph of the position of the rocket as a function of time for times running from 0 to $132\text{s}$ is shown in figure III.

Explanation of Solution

Rocket works on Newton’s third law the thrust generate by the exhaust creates a reaction which causing the rocket to fly.

The position of the rocket is,

    $x(t)=v_{\text{e}}\left(\frac{M_{\text{i}}}{k}-t\right)\ln \left(1-\frac{kt}{M_{\text{i}}}\right)+v_{\text{e}}t$

Substitute $1500\text{m}/\text{s}$ for $v_{\text{e}}$, $2.5\text{kg}/\text{s}$ for $k$ and $360\text{kg}$ for $M_{\text{i}}$ in above equation.

    $\begin{array}{c}x(t)=v_{\text{e}}\left(\frac{M_{\text{i}}}{k}-t\right)\ln \left(1-\frac{kt}{M_{\text{i}}}\right)+v_{\text{e}}t\\ =1500\text{m}/\text{s}\left(\frac{360\text{kg}}{2.5\text{kg}/\text{s}}-t\right)\ln \left(1-\frac{2.5\text{kg}/\text{s}\times t}{360\text{kg}}\right)+1500\text{m}/\text{s}\times t\\ =1500\text{m}/\text{s}\left(144\text{s}-\text{t}\right)\ln \left(1-6.944\times {10}^{-3}\times t\right)+1500\text{m}/\text{s}\times t\end{array}$

The value of $x\left(t\right)$ at the different value of $t$ is calculated and shown below.

$t(s)$	$x(t)\text{m}$
0	0
10	534.2866
20	2189.017
30	5054.74
40	9237.945
50	14865.69
60	22092.2
70	31108.67
80	42158.38
90	55561.47
100	71758.44
110	91394.47
120	115508.2
130	146066.6
132	153284.3

Draw the graph between $v\left(t\right)$ and $t$ as shown below.

Figure III

Conclusion:

Therefore, the graph of the position of the rocket as a function of time for times running from 0 to $132\text{s}$ is shown in figure III.