Section II, Chapter Seven: Acceleration, Force, and Momentum
But if we can apply Newtonian mechanics to each object independently, why bother considering the set of objects to be a system at all?
Note—this is a draft of Chapter Ten of a book I’m writing. The book is going to cover humanity’s deepest ideas from philosophy, physics, epistemology, and economics. Each chapter is meant to be short and digestible. Most of the chapters will explain just one or two ideas.
‘Chapter []’ indicates a future chapter that I’ve not yet written.
Chapter Ten: Acceleration, Force, and Momentum
Consider dropping a stone from a rooftop 100 meters above the ground such that the Earth’s gravity is the only force acting on it. What is its acceleration?
We may combine Newton’s Second Law, F = m*a(t), with his Law of Universal Gravitation, F = -G*m*M / r2, so that m*a = F = -G*m*M / r2 (see Chapter Eight). The m’s cancel out, leaving us with a(t) = -G*M / r2, where the negative sign indicates that the stone accelerates towards the Earth’s surface.
G is a constant of Nature, M is the planet’s mass, and r is the Earth’s radius. Since these are all constants, so is the stone’s acceleration due to gravity. Therefore, the stone’s acceleration as it falls from the rooftop to the ground is simply a(t) = -g, where ‘g’ is the conventional label for acceleration of objects near the Earth’s surface.
Although the stone’s acceleration is simply a constant, I’ve written it as ‘a(t)’, or ‘acceleration as a function of time’, for completeness’s sake.
(Numerically, g = (6.67 x 10^-11 N·m²/kg²) * (5.972 x 10^24 kg) / (6,371,000 m)² = 9.8 m / s2.)
Can we also calculate the stone’s velocity and position at any arbitrary time t along its trajectory? We can—provided we also know the stone’s initial position and velocity (in fact, the stone’s position and velocity at any other arbitrary point in time would also suffice—see Chapter []).
To obtain the stone’s velocity as a function of time, we take the integral of the acceleration function with respect to time (see Chapter Nine): v(t) = a(t)*dt = g*dt = g*t+C, where C is a constant. To solve for C, we take advantage of the fact that the stone’s velocity at t = 0, or its initial velocity, is zero. Plugging in 0 for v(t) and 0 for t, we obtain C = 0. The stone’s velocity function is therefore v(t) = g*t.
To obtain the stone’s position as a function of time, we take the integral of the velocity function with respect to time: x(t) = v(t)*dt = g*t*dt = (½)g*t² + C, where C is a constant. We solve for C by taking advantage of the fact that the stone’s position at t = 0, or its initial position, is the rooftop 100 meters above the ground (we’ve arbitrarily set the Earth’s surface to be x = 0). Plugging in 100 for x(t) and 0 for t, we obtain C = 100. The stone’s position function is therefore x(t) = (½)g*t² + 100.
In our example, there was only one force acting on the system in question, namely the force due to gravity (see Chapter Eleven). Moreover, it yielded an extremely simple expression for the resultant acceleration: a(t) = g. In general, the net force acting on a system may be the sum of a number of different forces, and some of them may depend on time (or, indeed, other parameters). Regardless, the procedure for determining the system’s velocity and position as functions of time is the same as in our example: first, determine the acceleration function by summing the forces acting on the system and then dividing by the system’s mass. Then, to deduce the velocity function, integrate the acceleration function with respect to time and impose supplementary data to determine the value of the emergent constant. Finally, to deduce the position function, integrate the velocity function with respect to time and impose supplementary data to determine the value of the emergent constant.
Momentum
Although our example consisted of a single object, the laws of Newtonian mechanics generalize to systems consisting of an arbitrary number of subsystems: pairs of rotating stars (binary star systems) and sets of billiard balls alike. We may apply the procedure outlined above to each subsystem independently to determine its acceleration, velocity, and position as functions of time.
But if we can apply Newtonian mechanics to each object independently, why bother considering the set of objects to be a system at all?
An isolated system is one in which there is no net force acting on it. In the example of dropping a stone from a rooftop, our ‘system’ consisted of only one object—the stone. And there was only one force acting on it—that of Earth’s gravity.
But consider a system consisting of two billiard balls on course to collide with one another (we’ll ignore friction). Before they collide, each has some initial velocity. During their collision, each exerts a force on the other, and, owing to Newton’s Third Law, those forces are equal in magnitude but opposite in direction. Therefore, there is no net force acting on the system.
Meanwhile, Newton’s Second Law tells us how each billiard ball changes velocity following their collision: for each ball, we simply divide the force exerted upon it by the other ball and divide by the receiving ball’s mass.
Although the billiard balls might each undergo acceleration, the fact that the system is isolated—that there is no net force acting on it—implies that the total momentum of the system, defined as the sum of each subsystem’s mass times its velocity, is conserved.
How can the total momentum be conserved if each subsystem accelerates, as in our billiard ball example? Their collision is such that each ball changes velocity precisely so that the sum of their momenta before the collision is exactly the same as the sum of their momenta after the collision.
This is emphatically not the case for systems in which there is an overall net force. That is, the total momentum of a system is not conserved for a system on which a net force is exerted. Consider again our example of a rock that falls from a rooftop. Its initial momentum is m*v(t=0) = 0, since its initial velocity, vi, is zero. It accelerates at a constant rate g until it reaches the ground, at which point its final velocity, vf, is greater than zero. (I leave it as an exercise for the reader to use the equations in the first subsection of this chapter to work out the rock’s final velocity.) Clearly, then, the rock’s momentum was not conserved during its trajectory—in fact, it increased in magnitude continuously from beginning to end.
We could also show that the momentum of an isolated system is conserved by analyzing Newton’s Second Law directly: F = m*a. Recalling that a = dv / dt, we may write F = m*(dv / dt). But we’d said that momentum is just mass times velocity, and so we have F = dp / dt, where p represents momentum. So, the net force acting on a system is equal to the system’s change in momentum over the change in time. If there is no net force, then F = 0 = dp / dt. In other words, the momentum of an isolated system does not change with time—it is conserved.
To be sure, isolated systems are always idealizations, simplifications that remove unnecessary complications. In reality, the only isolated system is the universe as a whole. The falling rock, for example, is not isolated from the Earth itself—in fact, the Earth gains momentum in the upward direction as the rock gains momentum in the downward direction during its accelerating descent, leaving the total momentum of Earth-rock nearly conserved—‘nearly’ because the Earth-rock system is itself not isolated from the rest of the subsystems in the universe, which themselves change momentum in accordance with the changes of those of the rock and the Earth (and every other subsystem). We may therefore conclude that the universe itself is the only system whose momentum is truly conserved.
Thanks to Dennis Hackethal for early feedback.