The Physics of a Falling Apple, and a DE We Can’t Solve

According to the story, Newton discovered the Law of Universal Gravity after watching an apple fall from a tree on his country estate. Gravity is the force that pulls the apple toward the earth and causes it to fall. It also acts on the moon to hold it in its orbit. Newton knew the force acting on the apple, but what he wanted to understand was the apple’s trajectory as it falls, that is, he wanted to know the position function for a falling apple. To this end, he also discovered his Second Law of Motion, which is the equation force equals mass times acceleration, or F = M * A, which can be rewritten A = F / M.

For fun let’s use some real numbers: the Law of Universal Gravity states that the force F on a falling apple is

F = G * M_earth * M_apple / (R * R), where

G is the gravitational constant ( = 6.67 * 10-11 m³/(kg sec2) )

M_earth is the mass of the earth ( = 5.9742 * 1024 kg )

M_apple is the mass of the apple ( = 0.25 kg )

R is the distance between the centers of earth and apple ( = 6,378 * 10⁶ m )

So A = F / M_apple = (G * M_earth * M_apple / (R * R)) / M_apple

= G * M_earth / (R * R)
Note that the mass of the apple divides out of the expression for A. We can (only this one time) use our calculators to calculate A = 10 m/sec*sec (9.8…..)

Acceleration is the rate of change of velocity, so, if v(t) is the apple's velocity function, we have the differential equation
v’(t) = -10 This is the type of differential equation we solved in Vol. 1, and the solution is
v(t) = -10 * t + V₀ where V₀ is the apple’s initial velocity.
The solution is shown graphed to the left.

So, we can solve the differential equation for the apples velocity v(t), and since p'(t) = v(t), we can write a differential equation for the apple's position,
p'(t) = -10*t

Can we solve this DE? Ans: no, solving this DE will occupy us for the remainder of Volume 2.

Graphing Solutions to DEs – Linear Approximation

We can’t write the exact solution to p’(t) = -10 * t, but we can graph an approximate solution. Suppose we want to graph the first 10 seconds of the apple’s fall. We will draw a graph made up of straight line segments (and called piecewise linear) to approximate the apple’s position function. How to do it? We'll start with a linear graph. The apple’s starting altitude is 0, so, now, all we need is the slope of the line. We know the apple’s velocity at time t = 0 is 0 (m/sec), and its velocity at time t=10 is –100. So, we could use an ‘average’ velocity, but, let’s do the least amount of work possible, and just use the velocity at the start of the interval, v(0) = 0!. This gives us the linear graph shown to the left.

A lousy approximation to the true trajectory of a falling apple, but, hey, we're just getting started.

But, notice what happens when we divide the 10 seconds into two 5 second intervals and use a linear approximation for each subinterval. We get the same graph for the first 5 seconds. At t=5 our approximate solution is still at altitude 0, so the solution for the second interval starts at 0, and the slope of the solution is given by the velocity at the start of the second 5 seconds, which is v(5) = -50, giving the graph to the left.

Not a good approximation, but at least better than our first one.

If we divide the 10 seconds into 10 one second intervals, and follow the method used above to produce a linear approximation for each subinterval, we get the graph to the left. This is a pretty good approximation to the true solution to the DE. Compare it to the exact solution below.

This is the technique engineers most often use to solve differential equations! It’s just this simple, and it can be used to obtain a solution to any DE.

Defining 'Instantaneous Velocity' and Differentiating t²

In Volume 1, we learned that the velocity function for p(t) = V * t + P₀ is p’(t) = V. We were a bit cavalier there, not really bothering with definitions, and when velocity is not constant we need to be more careful. The velocity function p’(t) calculates the runner’s velocity at the instant when time equals t – what does that mean? Here is the problem: the formula that calculates velocity for the interval t₁ to t₂ is velocity = distance / (t₂ – t₁). This formula is more than a way of calculating velocity, it is the definition of velocity. And, it works fine for intervals, but not for instants. Why not? For an instant at time t, t₁ = t and t₂ = t, so, the formula becomes distance / (t₂ – t₁) = 0/0. This is a bit of a sticky wicket.

With this in mind, let’s set out to define and calculate the velocity for p(t) = t², plotted to the left. What is the velocity at time t=4, for example? Intuitively, we can approximate the velocity at the instant t=4 by calculating the velocity for a short interval starting at 4. The distance traveled in the interval 4 to 4 + h is given by
p(4+h) – p(4) = (4+h)² - 4² = 4² + 2*4*h + h² - 4² = 2*4*h + h²
and the for velocity for the interval (velocity = distance / time) is
(2*4*h + h²) / h = 2 * 4 + h
As h becomes smaller, the approximation improves, and the velocity for the interval approaches 2*4. We define the instantaneous velocity of any function at time t to be the limit of the velocity for a interval of length h starting at time t, as h approaches 0. So, we see that p’(4) = 2*4 = 8.

For the interval t to t+h, the distance traveled is given by
p(t+h) – p(t) = (t+h)² - t² = 4t² + 2*t*h + h² - t² = 2*t*h + h²
and hence the velocity is
(2*t*h + h²) / h = 2 * t + h
As h becomes smaller, the velocity for the interval approaches 2*t, and thus we have
p’(t) = 2*t

That's calculus, folks.

Integrating p’(t) = A * t

We will evaluate the integral
when p(t) = A*t² / 2 and p’(t) = A*t, geometrically. The formula for the area of a right triangle with base b and height h is one-half the area of a rectangle with base b and height t, or b*h/2. So, the area under the curve A*t for the interval 0 to t₂ is t₂ * (A* t₂) / 2 = A* t₂² / 2.

We can calculate the area under the curve for the interval t₁ to t₂ by subtracting the area of the smaller triangle A* t₁² from A*t₂² giving A* t₂² / 2 – A* t₁² / 2.

Recall that the Fundamental Theorem of Calculus states that
= p(t₂) – p(t₁)
The calculation above shows that it is true when p(t) = A*t² / 2 and p'(t) = A*t.

The Trajectory of a Falling Object – Solving the Differential Equation F = M * A when F is the Force of Gravity

We can differentiate p(t) = A*t² + V₀*t + P₀ by adding the derivatives of each term, giving p’(t) = 2*A*t + V₀ + 0. Then, differentiating p’(t) gives p’’(t) = 2*A.

So, if A = -5, then p(t) is a solution to the DE p’’(t) = -10 for any P₀ and V₀. Thus we have solved the DE p’’(t) = -10. The method used was 'by inspection', that is, we looked at the DE and recognized is as one for which we know the solution. Then p(0) = P₀, so P₀ is the object's initial position, and p’(0) = V₀, so V₀ is the object's initial velocity.

If an apple falls from an initial altitude of 0 meters, then its initial position P₀ is 0, its initial velocity V₀ is 0, and its trajectory is given by p(t) = -5*t²

If an apple is thrown upwards from an initial altitude of 0 meters, with an initial upward velocity of 50 meters/sec, then P₀ = 0, V₀ = 50, and its trajectory is given by p(t) = -5*t² + 50*t + 0.

Air resistance opposes the flight of any object, and is proportional to the square of the velocity. If we include drag in the differential equation for the trajectory of a falling apple we have G – D* p’(t) * p’(t) = p’’(t) where D is the ‘drag coefficient’ of the apple. This DE is difficult to solve analytically (that is, with a mathematical epxression for p), however, we can easily determine a graphical solution using the method of piecewise linear approximation discussed above.