This is just a follow-up in response to a couple of requests for a solution to last week’s problem about hitting a golf ball on the Moon.  Given the speed of the club face m/s, and the acceleration due to gravity m/s^2, at what loft angle should a golfer hit a golf ball on the Moon to maximize the distance the ball travels?

It is helpful to consider two different frames of reference, one to handle the “bounce” of the ball off the club, and another to handle the resulting flight of the ball:

First, in the “club frame,” the club face is fixed, and the ball approaches the club at speed .  Assuming a perfectly elastic collision with a negligible mass ratio, the ball bounces off the club face with equal angles of incidence and reflection, leaving at the same speed.  As the following figure shows, the ball is launched at angle … in this frame.

In the moving frame of the club face, the ball is launched at twice the loft angle of the club.

However, in the “Moon frame,” the club face– and thus everything in the figure above– is also moving to the right at speed .  So the resulting velocity of the ball is

At this point, we have a straightforward projectile motion calculus problem.  Assuming this initial velocity, we compute the distance traveled, then maximize as a function of .

(An interesting but non-obvious side note: the “actual” launch angle of the ball– in the Moon frame– ends up equaling the loft angle!  (Why?)  That is, for example, if we hit the golf ball with a loft angle of 45 degrees, then the ball’s initial trajectory will also be 45 degrees.  However, it will not be traveling as fast as it would if we hit it at a lower angle.  The problem is to determine the “sweet spot” between lowering the angle to increase speed, and raising the angle to increase time of flight.)

I’ll skip most of the details.  The distance the ball travels is given by

The maximum distance is realized when the loft angle is 30 degrees, in which case the ball travels about 2560 meters, or about 1.6 miles.

Problem: How far can you hit a golf ball on the Moon?

Ok, that was intentionally vague.  I’ll make it more precise shortly.  But first, I took a slightly roundabout path in getting to this problem.  Last week, I read an interesting paper on “The Optimal Angle of Release in Shot Put.”  The trajectory of a shot (the heavy metal ball) can be modeled reasonably well assuming a uniform gravitational field and ignoring air resistance… exactly the “nice” situation used in introductory physics textbooks.  In that simple case, it is a common exercise to show that, for a given initial speed, the maximum range of a projectile launched from the ground is achieved with a launch angle of 45 degrees.

However, the best shot putters in the world release the shot at an angle of only 37 to 38 degrees.  Why is this angle so low?  The paper describes the necessary additional assumptions that yield a mathematical solution that agrees with observed practice.  I think this is a nice example of the need to be careful about which simplifying assumptions are “safe,” such as neglecting air resistance, and which are not, such as the need to consider the height of release– although this is not the only important additional factor, as the paper shows.

From there, let’s move on to a blog post by Ethan Siegel from a couple of years ago, titled “Could you really hit a golf ball ‘miles and miles’ on the Moon?”  (The “miles and miles” is a reference to Alan Shepard’s comment after hitting a golf ball on the Moon during the Apollo 14 mission.)  Here on Earth, even the best golfers hit the ball only a few hundred yards.  On the Moon, the force of gravity is weaker, and there is no air resistance, so we should be able to hit the ball much farther.  The question is, how much farther?

So now let’s make the problem more precise.  As in the blog post, our golfer can swing the club so that the club face strikes the ball at 40 meters per second (about 90 miles per hour).  Also, we are allowed the same simplifying assumptions:

1. The Moon is a flat plane with a uniform gravitational acceleration of 1.624 m/s^2.
2. The club-ball collision is perfectly elastic, and the mass of the ball is negligible compared to the mass of the club.

The problem: at what angle should the golfer hit the ball (i.e., what is the loft angle of the club) to maximize the distance the ball travels?  Hint: the answer is not 45 degrees as suggested in the blog post, and accordingly, the maximum distance is significantly less than “nearly two and a half miles.”