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Newton's 2nd law of motion can't be applied for mass-varying systems. Another force, known as Thrust must come to play. It can be measured using law of conservation of linear momentum. $$\text{Thrust} M.\dfrac{dv}{dt} = v_{rel} .\dfrac{dm}{dt}.$$ where $M$ is the mass of the system; $\frac{dm}{dt}$ is the rate at which the mass changes; & $v_{rel}$ is the relative velocity of the exhausted mass w.r.t to the main system.

This is what my book writes. Now, the left hand side of the eqn. represents force on mass $M$ whereas the right hand side represents the force due to variable mass. How can they be equal? During the application of thrust, $M$ does change and the equation had to be $$ (M - dm). \dfrac{dv}{dt} = v_{rel} .\dfrac{dm}{dt}$$ But that is not the book's equation. Why is mine wrong? What is the intuition?

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  • $\begingroup$ $\uparrow$ Which book? Which page? $\endgroup$ – Qmechanic Dec 6 '14 at 12:00
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Why is mine wrong? What is the intuition?

To see what happens, use finite differences to write

$$(M - \Delta m) \Delta v = v_{rel}\Delta m$$

which leads to

$$M \Delta v - \Delta m \Delta v = v_{rel} \Delta m$$

Dividing through by $\Delta t$ yields

$$M \frac{\Delta v}{\Delta t} - \Delta m\frac{ \Delta v}{\Delta t} = v_{rel} \frac{\Delta m}{\Delta t}$$

Taking the limit as $\Delta t \rightarrow 0$ yields

$$M \frac{dv}{dt} = v_{rel}\frac{dm}{dt}$$

since, in the limit, all $\Delta$ terms go to zero. See this Math Exchange answer regarding the product of two differentials.

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  • $\begingroup$ Sir, is thrust acting on the initial mass $M$ or the mass after the ejection ie. $M - dM$? I think the later is correct & so force at that time$(M - dM)\dfrac{dv}{dt} \approx M.\dfrac{dv}{dt}$ . Right? $\endgroup$ – user36790 Dec 6 '14 at 18:49
  • $\begingroup$ @user36790, I don't think that's correct; the right most term is exact as I showed in my answer and as is explained in the answer I linked to. Consider that $M(t + dt) = M(t) - dm$ $\endgroup$ – Alfred Centauri Dec 6 '14 at 19:11
  • $\begingroup$ @user36790, I do not understand what your argument is in the two preceding comments. Do you disagree with the mathematical argument in my answer and in the linked answer? If so, please point out the step that is in error. $\endgroup$ – Alfred Centauri Dec 7 '14 at 2:55
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    $\begingroup$ Oh! Enlightment! It's not negligence but rather it vanishes! So, at a time $t$ & mass being $M$, the thrust at that moment is what the eqn. gives,right? It's like the process of differentiation; first take a point ahead of the concerned one & then take limit when the further point approaches the concerned point! Thanks:) $\endgroup$ – user36790 Dec 7 '14 at 15:07
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    $\begingroup$ @user36790, because $V_e(t) = V(t) - v_e$ while $V_e(t + \Delta t) = V(t + \Delta t) - v_e$ $\endgroup$ – Alfred Centauri Dec 7 '14 at 15:58
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Newton's second law relates force to acceleration:

$$ F = Ma \tag{1} $$

but force can also be expressed as the rate of change of momentum:

$$ F = \frac{dp}{dt} \tag{2} $$

In the case of a rocket it's the rate of change of momentum of the exhaust gas has that produces the force. Momentum is given by $p = mv$ (note $M$ is the mass of the rocket and $m$ is the mass of the exhaust gas) so if the exhaust velocity is $v_e$ then:

$$ \frac{dp}{dt} = \frac{d(mv_e)}{dt} = v_e\frac{dm}{dt} \tag{3} $$

where $dm/dt$ is the rate of mass flow i.e. the mass of exhaust gas ejected per second. Equating (1) and (2), and substituting for $dp/dt$ from equation (3) gives:

$$ Ma = v_e\frac{dm}{dt} $$

which is the equation in your book.

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  • $\begingroup$ Sir, the mass of the rocket $M$ changes during the ejection of the gas. Then, how can the gas apply force to $M$ which will have changed due to the ejection?? $\endgroup$ – user36790 Dec 6 '14 at 11:05
  • $\begingroup$ @user36790: the equation gives the force, which by definition is simply the product of mass and acceleration, $Ma$. If you were calculating the motion of a rocket you would need to take into account the fact that $M$, and therefore $a$, would both be functions of time. This makes the equation of motion more complicated, but it's still just a differential equation. $\endgroup$ – John Rennie Dec 6 '14 at 11:47
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    $\begingroup$ Yeah,sir,you are right. $M$ is a function of time. So, I was confused as when the thrust was applied, during that time $M$ would have changed. But , instead of showing that mass, the eqn. has the same initial mass $M$ . $\endgroup$ – user36790 Dec 6 '14 at 11:54
  • $\begingroup$ @user36790: it's hard to comment without seeing exactly what your book says. It's certainly true that to calculate the trajetories of rockets you need to account for the changing mass. The Wikipedia article on the Tsiolkovsky equation is a good place to start looking at this. $\endgroup$ – John Rennie Dec 7 '14 at 7:20
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Newton's 2nd Law states that the 'the change in momentum (p) of a body is proportional to the force (F) exerted on the body'. Mathematically, we write:

$$F_{ext}=\frac{dp}{dt}$$

Where a body of mass $m$ and velocity $v$ has momentum $p=mv$. For a body of constant mass $m$, this becomes:

$$F_{ext}=\frac{dp}{dt}=m\frac{dv}{dt}=ma$$

Where the mass $m$ varies, we must apply the 'product rule' for derivatives, to get:

$$F_{ext}=\frac{dp}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}$$

EDIT:

Now, a rocket of mass $M$ travelling at velocity $v$ (with respect to an inertial frame of reference) has an initial momentum of $Mv$.

enter image description here

If it ejects a mass $\Delta m$ backwards at velocity $-v_e$ (relative to the rocket), that is, the momentum of the ejected mass with respect to the inertial frame is $\Delta m (v_e-v)$.

The rocket, now with mass $M-\Delta m$ and travelling at velocity $v+\Delta v$ has momentum $(M-\Delta m)(v+\Delta v)$.

enter image description here

Since there is no external force applied to the system (from the inertial frame of reference), applying conservation of momentum, we get:

$$p_i=Mv=p_f=(M-\Delta m)(v+\Delta v)-\Delta m(v_e-v)$$

ie:

$$M\Delta v - \Delta m v - \Delta m\Delta v - \Delta mv_e +\Delta mv=0$$

For small $\Delta m$ and small $\Delta v$, we can say $\Delta m\Delta v\approx0$

ie: $$M\Delta v=v_e \Delta m$$

Dividing both sides by $\Delta t$,

$$M\frac{\Delta v}{\Delta t}=v_e \frac{\Delta m}{\Delta t}$$

and taking $limit. \Delta t \rightarrow 0$, we get:

$$M\frac{dv}{dt}=v_e\frac{dm}{dt}$$

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  • $\begingroup$ Sir, can you show how the equation solved my problem? Also please see my comments to Rennie. Please help. $\endgroup$ – user36790 Dec 6 '14 at 12:47
  • $\begingroup$ See my edits for detail $\endgroup$ – theo Dec 6 '14 at 15:22
  • $\begingroup$ So, actually the exhausted gas exerted thrust on $(M - dM)$ & taking limit, it can be said that $(M - dM)\frac{dv}{dt} \approx M.\frac{dv}{dt}$ . So, we neglected the term, right?? $\endgroup$ – user36790 Dec 6 '14 at 16:01
  • $\begingroup$ @user36790 we sort of neglected it, right — we neglected infinitesimal compared to finite quantity. But to make it rigorous, you'd have to start from finite differences and then take the limit. In this case it's no longer necessary to view this as neglecting anything. AlfredCentauri's answer does exactly this. $\endgroup$ – Ruslan Dec 6 '14 at 16:22
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There is an indigenous group of people in South America, the Aimaras. When they mention something that happened in the past, they do not signal to their back, but to their front. To their perception, the past is in front of them, because they can see it. And the future is behind just for the opposite reason. At least in western countries, our perceptions is just the reverse. What do I mean with this? That you just took a diferent frame of reference. Your reference was the final point, instead of the initial one. But the result should be independent of the origin of reference. In your case, in your frame of reference a quantity appeared that you forgot to erase: Matematically, $dm<<M$ actually infinitesimally. The sum of one regular number and an infinitesimal is the regular number. Thus, M-dm=M+(-)dm=M. Now both results agree, as expected. In the same vein, when you are calculating something and get terms of the form $dx=Mdc+Rdcdz$ the term with $dc*dz$ is erased because it is infinitely small relative to $dc$

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  • $\begingroup$ +1 . Sir, why does the LHS of the eqn. contain $M$ , the initial mass?? The thrust doesn't act on the initial mass, but on the final changed mass after ejection of gas ie. $M - \Delta M$ . Please explain. $\endgroup$ – user36790 Dec 7 '14 at 8:32
  • $\begingroup$ I would need to see the derivation of the original book, both assumtions will give rise to the same result (because $(M - dM)=M$). Could you scan the page of the book, or tell me the name author and page, I might find it online and tell you how the author reasoned about it. $\endgroup$ – Wolphram jonny Dec 7 '14 at 18:34
  • $\begingroup$ Principles Of Physics by Walker,Resnick,Halliday; page no. 226 $\endgroup$ – user36790 Dec 7 '14 at 18:55

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