…at the thought that commenter “Brian” from this post teaches undergraduates.
Scroll down a ways, and be amazed.
[Update a few minutes later]
I should add, that there’s another howler there:
Regarding Newton’s second law of motion, F=ma is just fine for all physics short of things traveling greater than 0.95 the speed of light, or quantum effects.
He’s apparently confused, thinking that I’m referring to Einstein’s Special Relativity version of Newton’s Second Law, in which rest mass is converted to true mass via the factor gamma, which is a function of velocity, or F = dp/dt where p = m*v, or in the Einsteinian version, p = gamma*m*v.
Gamma is a function of velocity. It’s 1/(1-v^2/c^2)^1/2 (or in words, it’s the inverse of the square root of one minus the ratio of velocity squared over the speed of light squared). For low velocities, it’s one divided by the square root of one minus a tiny number, or simply one, so at low velocities, mass equals mass. But for high velocities, you’re starting to divide one by a very tiny number (as the difference between 1 and velocity squared over c squared becomes infinitesimal), so gamma blows up to be a huge number. That’s why mass approaches infinity as its speed approaches that of light.
As I pointed out in the other thread, in the Newtonian case it is simple to take the derivative:
F = dp/dt = d(mv)/dt = m*dv/dt + v*dm/dt. But dv/dt is acceleration, so we get:
F = ma + v*dm/dt.
The Einsteinian case is a much more complicated derivative, because it’s a much more complicated function of velocity. But it’s not relevant, since we’re not talking about near-light speeds. The fact remains that Newton’s Second Law is F = ma + v*dm/dt. The only reason that we always see it as the more simple (and incorrect) F = ma, is that this is a special case in which the mass is constant (the derivative of a constant is zero, and the second term goes away). This is the case for most physics problems, but it certainly isn’t for rocketry, in which the vehicle is ejecting mass (that’s what makes it go).
Anyway, as I said, it’s very disturbing that this person is teaching anyone, let alone undergrads.
[Update at noon Eastern]
Professor Hall, who does teach undergrads as well as grads (and I’m glad of it), expands on his comment via email:
I think Brian’s a bit of a putz in his comments. However, he’s right about F=ma and F=dp/dt. Derivation of the rocket equation is a little tricky to work out, as you say. However, the chain rule does not lead to the correct equation.
In the 2nd law,
F = dp/dt
F is the sum of all applied forces, and p=mv is the linear momentum of the particle of mass m.
If you apply the chain rule to this equation, you get
F = m dv/dt + v dm/dt
as you noted.
However, in order for the chain rule to make any sense here, the two v’s must be the same v. What v is it?
If it’s the velocity of the particle, then this equation can’t apply to a rocket, since it couldn’t lift off the ground. On the ground, v is zero, and initially dv/dt is zero, so F is zero. If F is zero, the linear momentum cannot change, so v remains zero.
If it’s the velocity of the mass leaving the rocket, then initially F = v*dm/dt, which is essentially correct. However, the v in the dv/dt term is clearly not the velocity of the mass leaving the rocket. It’s supposed to be the velocity of the rocket.
The correct derivation of the rocket thrust equation uses a control volume approach, which is essentially a summation of Newton’s 2nd law over a continuum of particles of different velocities (the rocket and the propellant clearly have different velocities).
This leads to the following vector equation for rocket motion
F + ve dm/dt – vehat A (Pe-Pa) = ma
The terms on the left comprise the sum of all external forces acting on the
F includes all the forces such as gravity, drag, ….
The term ve dm/dt is the thrust due to the rocket, where ve is the exhaust velocity and dm/dt is the mass flow rate (negative number, since m is the mass of the rocket, which is decreasing). The vehat A (Pe-Pa) term is the pressure force. Vehat is a unit vector in the direction of ve, Pe is the exhaust pressure, and Pa is the atmospheric pressure.
I’ll just add that a) I’m glad that at least some of my readers and commenters are smarter than me and b) while I didn’t say that the chain rule led to the rocket equation, I did imply it, and that was a mistake, and c) I had known that at one time, but it’s been a long time.
And we are in agreement that Brian is a putz.