# 9.7 Rocket propulsion  (Page 4/8)

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## Rocket in a gravitational field

Let’s now analyze the velocity change of the rocket during the launch phase, from the surface of Earth. To keep the math manageable, we’ll restrict our attention to distances for which the acceleration caused by gravity can be treated as a constant g .

The analysis is similar, except that now there is an external force of $\stackrel{\to }{F}=\text{−}mg\stackrel{^}{j}$ acting on our system. This force applies an impulse $d\stackrel{\to }{J}=\stackrel{\to }{F}dt=\text{−}mgdt\stackrel{^}{j}$ , which is equal to the change of momentum. This gives us

$\begin{array}{ccc}\hfill d\stackrel{\to }{p}& =\hfill & d\stackrel{\to }{J}\hfill \\ \hfill {\stackrel{\to }{p}}_{\text{f}}-{\stackrel{\to }{p}}_{\text{i}}& =\hfill & \text{−}mgdt\stackrel{^}{j}\hfill \\ \hfill \left[\left(m-d{m}_{g}\right)\left(v+dv\right)+d{m}_{g}\left(v-u\right)-mv\right]\stackrel{^}{j}& =\hfill & \text{−}mgdt\stackrel{^}{j}\hfill \end{array}$

and so

$mdv-d{m}_{g}u=\text{−}mgdt$

where we have again neglected the term $d{m}_{g}dv$ and dropped the vector notation. Next we replace $d{m}_{g}$ with $\text{−}dm$ :

$\begin{array}{ccc}\hfill mdv+dmu& =\hfill & \text{−}mgdt\hfill \\ \hfill mdv& =\hfill & \text{−}dmu-mgdt.\hfill \end{array}$

Dividing through by m gives

$dv=\text{−}u\frac{dm}{m}-gdt$

and integrating, we have

$\text{Δ}v=u\phantom{\rule{0.2em}{0ex}}\text{ln}\left(\frac{{m}_{\text{i}}}{m}\right)-g\text{Δ}t.$

Unsurprisingly, the rocket’s velocity is affected by the (constant) acceleration of gravity.

Remember that $\text{Δ}t$ is the burn time of the fuel. Now, in the absence of gravity, [link] implies that it makes no difference how much time it takes to burn the entire mass of fuel; the change of velocity does not depend on $\text{Δ}t$ . However, in the presence of gravity, it matters a lot. The − g $\text{Δ}t$ term in [link] tells us that the longer the burn time is, the smaller the rocket’s change of velocity will be. This is the reason that the launch of a rocket is so spectacular at the first moment of liftoff: It’s essential to burn the fuel as quickly as possible, to get as large a $\text{Δ}v$ as possible.

## Summary

• A rocket is an example of conservation of momentum where the mass of the system is not constant, since the rocket ejects fuel to provide thrust.
• The rocket equation gives us the change of velocity that the rocket obtains from burning a mass of fuel that decreases the total rocket mass.

## Key equations

 Definition of momentum $\stackrel{\to }{p}=m\stackrel{\to }{v}$ Impulse $\stackrel{\to }{J}\equiv {\int }_{{t}_{\text{i}}}^{{t}_{\text{f}}}\stackrel{\to }{F}\left(t\right)dt\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{J}={\stackrel{\to }{F}}_{\text{ave}}\Delta t$ Impulse-momentum theorem $\stackrel{\to }{J}=\Delta \stackrel{\to }{p}$ Average force from momentum $\stackrel{\to }{F}=\frac{\Delta \stackrel{\to }{p}}{\Delta t}$ Instantaneous force from momentum (Newton’s second law) $\stackrel{\to }{F}\left(t\right)=\frac{d\stackrel{\to }{p}}{dt}$ Conservation of momentum $\frac{d{\stackrel{\to }{p}}_{1}}{dt}+\frac{d{\stackrel{\to }{p}}_{2}}{dt}=0\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}{\stackrel{\to }{p}}_{1}+{\stackrel{\to }{p}}_{2}=\text{constant}$ Generalized conservation of momentum $\sum _{j=1}^{N}{\stackrel{\to }{p}}_{j}=\text{constant}$ Conservation of momentum in two dimensions $\begin{array}{c}{p}_{\text{f},x}={p}_{\text{1,i},x}+{p}_{\text{2,i},x}\hfill \\ {p}_{\text{f},y}={p}_{\text{1,i},y}+{p}_{\text{2,i},y}\hfill \end{array}$ External forces ${\stackrel{\to }{F}}_{\text{ext}}=\sum _{j=1}^{N}\frac{d{\stackrel{\to }{p}}_{j}}{dt}$ Newton’s second law for an extended object $\stackrel{\to }{F}=\frac{d{\stackrel{\to }{p}}_{\text{CM}}}{dt}$ Acceleration of the center of mass ${\stackrel{\to }{a}}_{\text{CM}}=\frac{{d}^{2}}{d{t}^{2}}\left(\frac{1}{M}\sum _{j=1}^{N}{m}_{j}{\stackrel{\to }{r}}_{j}\right)=\frac{1}{M}\sum _{j=1}^{N}{m}_{j}{\stackrel{\to }{a}}_{j}$ Position of the center of mass for a system of particles ${\stackrel{\to }{r}}_{\text{CM}}\equiv \frac{1}{M}\sum _{j=1}^{N}{m}_{j}{\stackrel{\to }{r}}_{j}$ Velocity of the center of mass ${\stackrel{\to }{v}}_{\text{CM}}=\frac{d}{dt}\left(\frac{1}{M}\sum _{j=1}^{N}{m}_{j}{\stackrel{\to }{r}}_{j}\right)=\frac{1}{M}\sum _{j=1}^{N}{m}_{j}{\stackrel{\to }{v}}_{j}$ Position of the center of mass of a continuous object ${\stackrel{\to }{r}}_{\text{CM}}\equiv \frac{1}{M}\int \stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}dm$ Rocket equation $\Delta v=u\phantom{\rule{0.2em}{0ex}}\text{ln}\left(\frac{{m}_{\text{i}}}{m}\right)$

## Conceptual questions

It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. When that is the case, the gas velocity and gas momentum are in the same direction as that of the rocket. How is the rocket still able to obtain thrust by ejecting the gases?

Yes, the rocket speed can exceed the exhaust speed of the gases it ejects. The thrust of the rocket does not depend on the relative speeds of the gases and rocket, it simply depends on conservation of momentum.

## Problems

(a) A 5.00-kg squid initially at rest ejects 0.250 kg of fluid with a velocity of 10.0 m/s. What is the recoil velocity of the squid if the ejection is done in 0.100 s and there is a 5.00- N frictional force opposing the squid’s movement?

(b) How much energy is lost to work done against friction?

(a) 0.413 m/s, (b) about 0.2 J

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