Let us derive the final velocity of a rocket facing straight up that is attempting to leave the earth's surface. Basically, we are deriving Tsiolkovsky's rocket equation.
We can see that initially, the rocket has some velocity v and a mass m.
After a small change in time, we denote $dt$. The rocket has expelled burnt fuel that has a mass $dm$ and a velocity $(v-v_{e})$ where $v_{e}$ is the velocity of the gas relative to the rocket. The rocket now has a mass of $(m-dm)$ and an increased velocity of $(v+dv)$
Since the net external force is not zero momentum is not conserved, however, we can do the following.
We can begin by showing that a small change in the momentum of the system is equal to a small change in the impulse. Since everything is happening in the same dimension we leave out the unit vector notation.
$$d\boldsymbol{\vec{p}} = d\boldsymbol{\vec{J}}$$
$$d\boldsymbol{\vec{p}} = p_{f} – p_{i}$$
$$d\boldsymbol{\vec{p}} = (m-dm)(v+dv)+dm(v-v_{e})-mv$$
After expanding and cancelling some terms ,
$$d\boldsymbol{\vec{p}} = m\hspace{0.2em}dv – v_{e}\hspace{0.2em}dm$$
Since the gravitational force is the net external force
on the system we can define the impulse to be ,
$$d\boldsymbol{\vec{J}} = F_{net} \hspace{0.2em}dt$$
$$ F_{net} = -mg $$
$$d\boldsymbol{\vec{J}} = -mg \hspace{0.2em}dt$$
After equating $d\boldsymbol{\vec{p}}$ and $d\boldsymbol{\vec{J}}$
$$m\hspace{0.2em}dv – v_{e}\hspace{0.2em}dm = -mg\hspace{0.2em}dt$$
Solving for dv ,
$$dv = \frac{-mg\hspace{0.2em}dt + v_{e} \hspace{0.2em}dm}{m}$$
$$dv = \frac{v_{e}}{m}\hspace{0.2em}dm – g \hspace{0.2em}dt$$
Integrating both sides we get,
$$\int_{v_0}^{v}dv = \int_{m_0}^{m}\frac{v_{e}}{m}\hspace{0.2em}dm – \int_{0}^{t}g \hspace{0.2em}dt$$
$$$$
After evaluating the integrals and doing some algebra manipulation we get,
$$\boxed{v = v_{0} + v_{e}\ln\left(\frac{m_0}{m}\right) – gt}$$
$$$$
After this derivation I would like to ask the following questions?
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We define the net external force on the system to be the graviational force , however , the rocket is expending fuel in the y direction which is pushing it up by newtons third law , why dont we include the reaction force $F_{reaction}$.
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I notice that we define the velocity of the expelled burnt fuel to be $(v-v_{e})$ I understand that this is because we want to describe the velocity of the gas relative to the observer on the ground since the velocity of the rocket is defined to be relative to the ground. But why do we define it to be $(v-v_{e})$ I understand why the velocities are subtracted since they are in opposite directions but I'm still left handing… What is a general way to define the velocity of an object relative to something?
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How realistic is this equation? , it does not include the force of fluid friction opposing the motion of the upward rocket. How would I redefine the equation to include for that? The impulse would change and therefore become
$$d\boldsymbol{\vec{J}} = -mg\hspace{0.2em}dt-\frac{1}{2}C\rho Av^2\hspace{0.2em}dt$$
equating this to the change in momentum ,
$$d\boldsymbol{\vec{p}} = d\boldsymbol{\vec{J}}$$
$$m\hspace{0.2em}dv – v_{e}\hspace{0.2em}dm = -mg\hspace{0.2em}dt-\frac{1}{2}C\rho Av^2\hspace{0.2em}dt$$
here we would have to solve for dv and then integrate but it looks a bit complicated. Additionally to add even more realism , we know that the air denisty $\rho$ changes as a function of the distance above earths surface. But I dont think this is necessarily needed since it might not play a large role in the end.
Best Answer
1. Look at formula for $d\overrightarrow{p}$. In the final momentum you have considered momentum of both rocket (m-dm) and fuel (dm). This means that system includes rocket and fuel. Force between rocket and fuel is internal and will not be considered in $F_{net}$
2 In general, relative velocity is given by $$ v_{AG} = v_{AF} + v_{FG} $$ Here $v_{AG}$ is velocity of "A" with respect to ground (G). $v_{AF}$ is velocity of A with respect to Frame (F). $v_{FG}$ is velocity of frame with respect to ground. (In reality A, F and G can be anything. However it is easier to visualize it this way)
In this example: $v_{FG}$ is +v (velocity of rocket (frame) with respect to ground); $v_{AF}$ is $-v_e$ (velocity of fuel (A) with respect to rocket (frame)).
This means $v_{AG} = -v_e + v$ (velocity of A with respect to ground)
3 You are right. This formula is approximate in many ways. Apart from air drag and variation in density, gravity also decreases as you move upwards. Variation in density and gravity will not matter for small toy rockets (used in firework display). But air drag is unlikely to be negligible.
So why are those factors ignored? I don's know for sure. But I have a theory. As soon as air drag is considered, obtaining a direct result for velocity becomes complicated (as you have noticed). So air drag is left out for pedagogical reasons.