91Ó°ÊÓ

Rocket Flight. A model rocket having initial mass mo kg is launched vertically from the ground. The rocket expels gas at a constant rate of a kg/sec and at a constant velocity of b m/sec relative to the rocket. Assume that the magnitude of the gravitational force is proportional to the mass with proportionality constant g. Because the mass is not constant, Newton’s second law leads to the equation (mo - α³Ù) dv/dt - αβ = -g(m0 – α³Ù), where v = dx/dt is the velocity of the rocket, x is its height above the ground, and m0 - α³Ù is the mass of the rocket at t sec after launch. If the initial velocity is zero, solve the above equation to determine the velocity of the rocket and its height above ground for 0≤t<m0/α.

Step by step solution

01

Find the value of velocity

$\mathbf{(}{\mathbf{m}}_{\mathbf{o}}\mathbf{-}\mathrm{Î\pm }\mathbf{t}\mathbf{)}\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{-}\mathrm{Î\pm }\mathrm{β}\mathbf{=}\mathbf{-}\mathbf{g}\mathbf{(}{\mathbf{m}}_{\mathbf{o}}\mathbf{-}\mathrm{Î\pm }\mathbf{t}\mathbf{)}$

Since, the initial velocity is zero so v(0) = 0

$$\begin{gathered} \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{Î\pm }\mathrm{β}}{{\mathrm{m}}_{\mathrm{o}}-\mathrm{Î\pm }\mathrm{t}}-\mathrm{g} \\ ∫\mathrm{dv}=∫\frac{\mathrm{Î\pm }\mathrm{β}}{{\mathrm{m}}_{\mathrm{o}}-\mathrm{Î\pm }\mathrm{t}}-\mathrm{g}\mathrm{dt}\left(\text{Integrating on both sides}\right) \\ \mathrm{v}\left(\mathrm{t}\right)=-\mathrm{β}\ln \left|{\mathrm{m}}_{\mathrm{o}}-\mathrm{Î\pm ³Ù}\right|-\mathrm{gt}+\mathrm{C} \end{gathered}$$

Thus, $m_o-\mathrm{Î\pm }t>0,t<m_o/\mathrm{Î\pm }\text{and}v\left(0\right)=0thenC=-\mathrm{β}\ln \left|{\mathrm{m}}_{\mathrm{o}}\right|$

$$\begin{gathered} \mathrm{v}\left(\mathrm{t}\right)=-\mathrm{β}\ln \left|{\mathrm{m}}_{\mathrm{o}}-\mathrm{Î\pm }\mathrm{t}\right|-\mathrm{gt}+-\mathrm{β}\ln \left|{\mathrm{m}}_{\mathrm{o}}\right| \\ \mathrm{v}\left(\mathrm{t}\right)=-\mathrm{β}\ln \left|\frac{{\mathrm{m}}_{\mathrm{o}}-\mathrm{Î\pm }\mathrm{t}}{{\mathrm{m}}_{\mathrm{o}}}\right|-\mathrm{gt} \end{gathered}$$

Hence, the velocity is $\mathbf{v}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathrm{β}\mathbf{ln}\left|\frac{{\mathbf{m}}_{\mathbf{o}}\mathbf{-}\mathrm{Î\pm }\mathbf{t}}{{\mathbf{m}}_{\mathbf{o}}}\right|\mathbf{-}\mathbf{gt}$

02

Find the height of the rocket

$$\begin{gathered} \mathrm{x}\left(\mathrm{t}\right)=∫(-\mathrm{β}\ln \left|\frac{{\mathrm{m}}_{\mathrm{o}}-\mathrm{Î\pm }\mathrm{t}}{{\mathrm{m}}_{\mathrm{o}}}\right|-\mathrm{gt})\mathrm{dt} \\ \mathrm{x}\left(\mathrm{t}\right)=-\frac{\mathrm{β}({\mathrm{m}}_{\mathrm{o}}-\mathrm{Î\pm }\mathrm{t})}{{\mathrm{m}}_{\mathrm{o}}}\left[\ln \left|\frac{{\mathrm{m}}_{\mathrm{o}}-\mathrm{Î\pm }\mathrm{t}}{{\mathrm{m}}_{\mathrm{o}}}\right|-1\right]+\mathrm{C} \end{gathered}$$

When x(0) = 0 then C = -,then

$\mathrm{x}\left(\mathrm{t}\right)=\frac{\mathrm{β}(\mathrm{Î\pm }\mathrm{t}-{\mathrm{m}}_{\mathrm{o}})}{{\mathrm{m}}_{\mathrm{o}}}\left[\ln \left|\frac{{\mathrm{m}}_{\mathrm{o}}-\mathrm{Î\pm }\mathrm{t}}{{\mathrm{m}}_{\mathrm{o}}}\right|-1\right]-\mathrm{β}$

Hence, the height of the rocket is role="math" localid="1664218019166" $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathrm{β}\mathbf{(}\mathrm{Î\pm }\mathbf{t}\mathbf{-}{\mathbf{m}}_{\mathbf{o}}\mathbf{)}}{{\mathbf{m}}_{\mathbf{o}}}\left[\mathbf{ln}\left|\frac{{\mathbf{m}}_{\mathbf{o}}\mathbf{-}\mathrm{Î\pm }\mathbf{t}}{{\mathbf{m}}_{\mathbf{o}}}\right|\mathbf{-}\mathbf{1}\right]\mathbf{-}\mathrm{β}$.

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