Why is $M\frac{dv}{dt} = v_{rel} .\frac{dm}{dt}$ correct and $(M - dm)\frac{dv}{dt} = v_{rel} .\frac{dm}{dt}$ wrong?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $Mv$.

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 $(M-\Delta m)(v+\Delta v)$.

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:

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

Answer 4

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 $dc$