Topic 7.1 key terms
- differential equation
- An equation that relates a function of an independent variable and the function's derivatives, such as .
- proportional to
- Equal to a constant times. "The rate is proportional to the difference between and 20" means .
- solution of a differential equation
- A function that, together with its derivative(s), makes both sides of the equation equal for every input in an interval.
- verify a solution
- Differentiate the proposed function (left side), substitute it into the other side (right side), and show the two are equal for all inputs; check any initial condition separately.
- terminal velocity
- The speed where : . In still air (constant and ) a body falling at that speed stays at it; when the air changes, it marks the instant where .
- family of solutions
- One differential equation can have infinitely many solutions, such as for every constant . A starting value picks one.
Transcript
956 words · 3 parts
Thirty-nine kilometers up
On the fourteenth of October, 2012, Felix Baumgartner stood on a step outside a tiny capsule, hanging under a balloon about thirty-nine kilometers above New Mexico. Up there the sky is black at midday, and the air is less than a hundredth as thick as the air you’re breathing. He leaned forward and let go.
Thirty-four seconds later he was falling faster than the speed of sound. At fifty seconds he reached his fastest, thirteen hundred fifty-eight kilometers an hour, with nothing around him but a pressure suit. A skydiver falling belly-down from an ordinary plane levels off around two hundred kilometers an hour, however long she falls. Felix went almost seven times faster, with the same gravity and the same kind of body, so what was different?
By the end of this lesson, you’ll be able to write down the one equation that ran his entire fall, and use it to work out his top speed and even design his parachute, without ever solving it. To get there, we need to look closely at what the air was doing to him the whole way down.
Gravity minus drag
At the instant he stepped off he wasn’t moving yet, so the air couldn’t push on him. The only thing acting on him was gravity, adding about ten meters per second to his speed every second.
But the faster he fell, the harder the air pushed back, and that push grows fast, because if k stays fixed, doubling your speed makes the air push four times as hard. At every instant, the rate his speed changes is gravity minus drag, and the drag part grows like the square of his speed. That one sentence runs his whole fall.
Let’s turn that sentence into math. The rate of change of his speed is d v d t, gravity is g, and drag is some number k times v squared. Put together, that’s d v d t equals g minus k v squared. We call an equation like this a differential equation, an equation that connects a function, here his speed v of t, with its own derivative.
It never says what his speed is at a given time, but once you know his speed right now, it tells you how fast that speed is changing. Here v is his speed in meters per second, counting down as positive. The number k is bigger for thicker air and for a bigger, lighter body, and since k v squared has to come out in meters per second squared, k comes out in units of one per meter.
Let’s use friendly numbers. Near the ground a skydiver’s k is about point zero zero four per meter, almost exactly what Felix had near the end of his fall. With g rounded to ten, the rule is d v d t equals ten minus point zero zero four v squared. At zero speed there’s no drag, so her speed grows at ten meters per second every second. At twenty-five, drag takes away point zero zero four times twenty-five squared, which is two and a half, so her speed still grows, at seven and a half.
Now imagine a skydiver who somehow got going seventy-five meters per second in this air, say she dove head-first and just flattened out. At that moment, what is her speed doing?
Top speed without solving
Look at the three rates. At twenty-five she’s speeding up and at seventy-five she’s slowing down, and since the rate changes continuously as her speed changes, somewhere in between there has to be a speed where she does neither.
That’s where the rate is zero, so let’s set ten minus point zero zero four v squared equal to zero. Moving the drag over, point zero zero four v squared equals ten, so v squared is twenty-five hundred, and v is fifty meters per second.
That’s about a hundred and eighty kilometers an hour, right around the top speed real skydivers reach, and we found it without ever finding a formula for her speed. Skydivers call this terminal velocity. At exactly fifty, drag balances gravity, the rate is zero, and she just stays at fifty. In general, setting g minus k v squared to zero gives a top speed of the square root of g over k.
Now you can do what I promised. In our model of his jump, at the moment he was fastest, high in the stratosphere, k is only about sixty-eight millionths per meter, and gravity is nine point seven. What top speed does that air allow, in meters per second?
That’s his record, and it’s no coincidence, because at his fastest instant his speed had stopped changing, so his record is exactly the top speed for that thin air. Now you know what was different. Up there the air is so thin that k is about sixty times smaller than near the ground. Since the top speed is the square root of g over k, it comes out more than seven times bigger.
Let’s go back to the real jump and add one more curve to his speed graph. It’s the top speed the air around him allows at each moment. At the start the air is so thin that the air’s top speed is enormous, way off the chart, but it keeps dropping as he falls into thicker air. At fifty seconds his speed catches it, exactly where the rate is zero.
After that the air keeps thickening, so the air’s top speed keeps dropping, and now he’s going faster than that top speed. The right side of his equation turns negative and he slows down. You may have watched him slow down in our motion lesson, and this is the rule that did it.