**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:

- The Moon is a flat plane with a uniform gravitational acceleration of 1.624 m/s^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.”

As someone that threw the shot put I think you don’t get the same magnitude velocity out of the launch at different angles. I am sure they have studied this to either confirm or deny my assumption.

As you look at the same issues with the golf club problem you have your highest velocity with a perpendicular angle. The more the club leans toward horizontal the more energy put into spin on the ball rather than velocity. Without air resistance spin is simply lost energy that doesn’t affect the flight path. For this reason the angle would again be less than 45 degrees.

You’re right, the dependence of velocity on launch angle is basically the key idea in both cases, and for the shot put is addressed in the referenced paper. For the golf ball, this is an issue even without considering spin; that is, we can simply assume the golf ball is a point mass, and velocity will still depend on launch angle. The problem is to determine what that optimal angle is.

Pingback: Hitting a Golf Ball on the Moon: Solution | Possibly Wrong