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Introduction
The applicability of Newton’s second law in the oft-quoted “general form” $$\begin{align}\frac{d\mathbf{P}}{dt}=\mathbf{F}_{\text{ext}}\end{align}$$ was an issue in a recent thread (see post #4) in cases of systems with variable mass. The following example illustrates the kind of confusion that could arise from the (mis)application of Equation (1):
A rocket is hovering in place above ground near the Earth’s surface. Assume that the combustion gases are expelled at constant rate ##\beta=dm/dt## with velocity ##w## relative to the rocket. What condition must hold for the rocket to hover in place?
A novice might start with Equation (1) and go down the garden path only to reach a quick impasse as shown below.
Attempted solution
We start with the general form of Newton’s second law, Equation (1) $$\frac{dP}{dt}=M\frac{dV}{dt}+V\frac{dM}{dt}=-Mg$$ If...
Newton's Second Law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). For variable mass systems, where the mass changes over time (such as rockets losing fuel), the law is modified to account for the changing mass. The generalized form is F = d(mv)/dt, where m is the mass, v is the velocity, and d(mv)/dt is the time derivative of the momentum.
In a variable mass system, the rate of change of mass must be included in the analysis. This is typically done by using the generalized form of Newton's Second Law, which includes the term dm/dt (the rate of change of mass). The equation becomes F = ma + v(dm/dt), where v is the velocity of the mass being ejected or added.
The ejection or addition of mass in variable mass systems directly affects the momentum of the system. For instance, in a rocket, fuel is ejected at high velocity, which changes the mass of the rocket and provides thrust. The momentum of the ejected mass must be considered in the overall momentum balance of the system.
A common example is a rocket. As the rocket burns fuel, its mass decreases. The thrust force (F) is equal to the rate of change of momentum. Using the equation F = d(mv)/dt, and considering that dm/dt is negative (since mass is decreasing), we can derive the rocket equation: F = ma + v_e(dm/dt), where v_e is the exhaust velocity of the ejected mass.
One limitation is that the system must be carefully defined to include all forces and mass flows. Additionally, the analysis assumes that the mass changes smoothly and continuously. Discrete changes in mass or external forces not accounted for can lead to inaccuracies. Finally, relativistic effects are not considered in classical mechanics, which can be significant at very high velocities.