Question

In Problem 21, the mass m(t) is the sum of three different masses: $\mathit{m}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{m}}_{\mathbf{p}}\mathbf{+}{\mathit{m}}_{\mathbf{v}}\mathbf{+}{\mathit{m}}_{\mathbf{f}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$where${\mathit{m}}_{\mathbf{p}}$is the constant mass of the payload,${\mathit{m}}_{\mathbf{v}}$is the constant mass of the vehicle, and${\mathit{m}}_{\mathbf{f}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$is the variable amount of fuel.

(a) Show that the rate at which the total mass m(t) of the rocket changes is the same as the rate at which the mass${\mathit{m}}_{\mathbf{f}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$of the fuel changes.

(b) If the rocket consumes its fuel at a constant rate , find m(t). Then rewrite the differential equation in Problem 21 in terms of $\mathit{\lambda }$and the initial total mass $\mathit{m}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{m}}_{\mathbf{0}}$.

(c) Under the assumption in part (b), show that the burnout time ${\mathit{t}}_{\mathbf{b}}\mathbf{>}\mathbf{0}$of the rocket, or the time at which all the fuel is consumed, is${\mathit{t}}_{\mathbf{b}}\mathbf{=}{\mathit{m}}_{\mathbf{f}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{/}\mathit{\lambda }$, where ${\mathit{m}}_{\mathbf{f}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$is the initial mass of the fuel.

Short answer

1. $m\text{'}\left(t\right)=m_f\text{'}\left(t\right)$is showed.
2. The equation is $(-\lambda t+m_0)\frac{dv}{dt}+v\frac{dm}{dt}+kv=-(-\lambda t+m_0)g+R$
   
3. $m_f\left(t_b\right)=0$ is showed.

Step-by-step solution

Definition of differential equation.

An equation containing the derivatives of one or more unknown functions (or dependent variables), with respect to one or more independent variables, is said to be a differential equation (DE).

Solution for part a).

Sincem_p and m_v are constants, their rate of change (that is, their derivative) is zero. We have,

$$\begin{gathered} m\text{'}\left(t\right)=m_p\text{'}+m_v\text{'}+m_f\text{'}\left(t\right) \\ m\text{'}\left(t\right)=m_f\text{'}\left(t\right) \end{gathered}$$

Hence, showed $m\text{'}\left(t\right)=m_f\text{'}\left(t\right)$.

Solution for part b).

We had the equation, $m\frac{dv}{dt}+v\frac{dm}{dt}+kv=-mg+R$

We are given that the rocket burns fuel at a constant rate of $\lambda$units of fuel per unit of time, meaning that the rate of change of the mass of the fuels decreases the rate of change has to be negative and we write,

role="math" $m_f\text{'}\left(t\right)=\lambda$

We can integrate this equation with respect to t,

$m_f\left(t\right)=-\lambda t+C$

And the equation of the mass becomes,

$m\left(t\right)=m_p+m_v-\lambda t+C$

Now from the initial condition we get,

$$\begin{gathered} m\left(0\right)=m_0 \\ {m}_p+m_v+C=m_0 \\ C=m_0-m_p-m_v \end{gathered}$$

So our equation is $m\left(t\right)=-\lambda t+m_0$

And the equation from problem 21, becomes:

$(-\lambda t+m_0)\frac{dv}{dt}+v\frac{dm}{dt}+kv=-(-\lambda t+m_0)g+R$

Solution for part c).

We had the equation, $m_f\left(t\right)=-\lambda t+C$

Since $m_f\left(0\right)=C$the equation becomes,

$m_f\left(t\right)=-\lambda t+m_f\left(0\right)$

Burnout time $t_b$indicates the time such that $m_f\left(t_b\right)=0$

Let’s put (as suggested) role="math" $t_b=\frac{m_f\left(0\right)}{\lambda }$

We have,

$\begin{array}{rcl}{m}_f\left(t_b\right)& =& m_f\left(\frac{m_f\left(0\right)}{\lambda }\right)\\ & =& -\left(\frac{m_f\left(0\right)}{\lambda }\right)\lambda +m_f\left(0\right)\\ m_f\left(t_b\right)& =& 0\end{array}$

Hence,$m_f\left(t_b\right)=0$is showed.