# Tennis & Trajectories

There's many, many sports out in the world that involve sharply hitting a ball with something. Baseball, tennis, golf, cricket, polo, you name it. After being hit, a ball describes a trajectory determined by the gravity of the earth and the interaction of the ball with the atmosphere. This can be exploited in many sports, since different trajectories can be useful in different situations. Golf and baseball especially are built on tweaking air flow around the ball in order to make it do precisely what the athlete wants. Tennis too, which has been drawing a lot of interest with the dramatic underdog victories in the French Open. Fig. 1: A Google image search for "Rafael Nadal" will suggest that you might mean "Rafael Nadal shirtless". I am not making this up.

In the absence of an atmosphere (or local gravitational variations and things of that nature) it's generally said that a projectile will follow a parabola. This is true in a constant gravitational field, but the earth is a sphere of finite size and thus does not in fact have a uniform field. It's very close to uniform near the surface and so a parabola is an excellent approximation. But formally it's not quite right. Atmosphere aside, a thrown projectile is in a very extended orbit about the center of the earth, and thus the trajectory is a section of an ellipse.

So here's a challenge problem that's not so hard, but requires a little bit of mathematical fluency. It's a Fermi problem, so estimates rather than exact numbers are what we're looking for.

Assume a baseball is hit into the outfield with a typical angle and initial speed. What is the order-of-magnitude difference in range between the trajectory approximated as a parabola and the trajectory approximated as an ellipse?

As a hint that you don't need, the difference is very, very small.

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If the ball is hit 100 meters, the discrepancy is a whopping 70 micrometers.

By CRM-114 (not verified) on 03 Jun 2009 #permalink

without calculation, I'm guessing less than a micron (1000 nanometers)

Well, since I've just earned the distinction of being the only amateur to get an award at the annual gravitation contest in at least the past 30 years, I guess I have to step up to the plate for this one.

Of course, being a theoretician, I'll use units where the radius of the earth = 1. Turns out that the initial vertical velocity v is more important than the horizontal velocity h, assuming they're of the same magnitude. I get -4 v^4/g^2 where g is the acceleration of gravity at the surface (and has units of cm^3/sec^2, say).

Converting into regular units this gives
-4 v^4 / (r g^2),
where r is the earth's radius, v is the vertical velocity, and g is the acceleration at the surface.

Am I close? Usually I drop a few terms on the way to getting the right answer. And I think that the answer depends a little on how you interpret the question. If one were looking for the parabola which most closely matched the ellipse, one might get a closer value. Or would it be the same???

The problem is a little more complicated if you assume that the earth is rotating, LOL.

The ball will typically cover a horizontal distance of 100 m, which corresponds to a plane angle of (100 m / R_E) or about 16 microradians. Assuming that the error will be comparable to the quadratic term in the cosine approximation, 0.5*θ^2 = 1.3 * 10^-10, gives a net offset of about 13 nanometers.

In other words, we don't have to worry about this. The way the ball is spinning after the bat makes contact will have a much larger effect. In any case, the outfielder can adjust the position of his glove and still make the play (or finds that he started too far from where it's going and fails to make the play anyway).

By Eric Lund (not verified) on 04 Jun 2009 #permalink

Might I ask a counter question ? Or rather a supplementary question ?

In a given undergraduate mechanics course you take a more thorough treatment of projectile motion - factor in resistance that varies quadratically with the speed, trace non-parabolic trajectories, et cetera. But something occured to me as I was thinking about how animals move through different fluids (a ray as opposed to a bird): why do we never factor in the bouyant force of air (a fluid) on projectiles. The simple answer is it is small. Unless you're a balloon. What about a paper airplane? It seems a superfluous contribution perhaps, but is it? What is the bouyant force on your tennis ball? Could it be the difference between 40-love and game ? A few centimeters ?

Just the musings of an undergrad Physics major,

I enjoy your blog very much by the way.

By BandTheory (not verified) on 05 Jun 2009 #permalink

A prediction of that ball's path
using a parabola as an approximation
will produce an answer that is off

Here are the numbers:
It only takes a few minutes to program
a computer to do the integral. I used a
step size of 0.1 microseconds with 45,000,000 steps.
For input I used values from the Astronomical
Almanac and the values suggested above:

x0=0
y0=6378136.6 meters
Gm=398600441800000
vx0=22 meters per second
vy0=22 meters per second
t(start)=0
t(end)=4.5 seconds
thus g=9.7982867081714239872899723665106

The integration gave:
x(4.5 sec)= 98.99948671263155 meters
y(4.5 sec)= 6378136.392859302 meters(from earth's center)
=> y(4.5 sec)-y(0)= -0.207140698192688 meters
(those are accurate to a picometer as shown below)

Using the parabolic approximation in the
usual way gives:
x(4.5 sec)= x0 + vx0 * time
= 0 + 22 meters per second * 4.5 seconds
= 99 meters
y(4.5 sec)=
= x0 + vy0 * time + acceleration * time *time /2
= 0 + 99 meters - g * 4.5 seconds * 4.5 seconds /2
= -0.207652920235667871310970210919789 meters

Thus the parabola method is off by
0.725 millimeters

The answer can also be solved exactly using
kepler's equation on a calculator:
(measuring x and y from the apex)
The shape of the trajectory is exactly:

y=y (input values for y, trace curve with calculated x)
then any of the following gives the x coordinate:
x=sqrt(2*vx0*vx0/g*y - b*b*y*y)
=sqrt(2*RE*RE*22*22/Gm * y - b*b*y*y)
=sqrt(2*L*L/Gm *y - b*b*y*y)
=b*sqrt(y*(2*SMA-y))

where L is the angular momentum, RE is y0,
SMA is the semimajor axis of the orbit.
and b a measure of the amount the ellipse is squashed
from a circle:
b*b=RE*RE*22*22/Gm*(2*Gm/RE-22*(1+(tan(theta))^2))
"theta" is the angle the ball is hit at: 45 degrees.

the parabola method gives:
x=sqrt(2*vx0*vx0/g*y)
which is the same as the first one above except
the y*y term is left out.

However including a time coordinate in the above
equations makes them more complicated.

To check my integration above to see
how much trouble the summation and other errors
affected the computer's result,
I used the full kepler equation to get the
following exact results:

x(4.5 sec)=98.9994867126144376775986336787553 meters
y(4.5 sec)=-0.207140698198552983973423839038731

thus the integration was off by:
-5.8649839734238390387309547153534e-12 meters

and the parabola method is off by:
7.25145050265668701659636788550951e-4 meters

Sincerely,
Erik Bergren

By Erik Bergren (not verified) on 30 Jun 2009 #permalink