Hitting a Golf Ball on the Moon: Solution

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 v = 40 m/s, and the acceleration due to gravity g = 1.624 m/s^2, at what loft angle \theta 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 v.  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 2\thetain 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 v.  So the resulting velocity of the ball is

(v, 0) + v(\cos 2\theta, \sin 2\theta) = v(1 + \cos 2\theta, \sin 2\theta)

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 \theta.

(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

d = \frac{8 v^2 \cos^3 \theta \sin \theta}{g}

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

 

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5 Responses to Hitting a Golf Ball on the Moon: Solution

  1. Jewels Vern says:

    This trade off between force and speed is called impedance, and it is the reason a baseball bat is marked with the exact spot so you don’t break it. Hit the ball at the tip of the bat and you have high speed but low force. The ball is not accelerated to full speed and the unused energy is reflected down the bat, breaking it. If you hit the ball near the grip, force is high but speed is low. Again, the bat is broken. Hit the ball with the sweet spot and force is just right to accelerate the ball to full speed. Maximum energy is transferred to the ball and the bat simply stops with no reflected energy.

  2. Pingback: Optimal angle on a swing | Possibly Wrong

  3. bolt says:

    Concerning your collision model, I think it would be fairer to assume that the club has a head-on collision with the ball. The club can be thought of as a rod rotating around an axis, in which case its velocity is always perpendicular to its length or parallel to the normal on the club’s face. However, in your solution, the club’s normal and the velocity are inclined at an angle. In a non-ideal world, the kind of motion you described could be very much possible, but since you’ve taken everything else to ideal, it’s only fair the motion of the club be ideal as well.

    • This “loft angle” business is something that I realize I never (strictly speaking) made explicit, either in my original post or in this solution. By this I meant essentially, “Which club should you select from your bag?” e.g., a 45-degree wedge, a 35-degree 7-iron, etc.

      Note that the resulting “inclination” of the ball’s velocity is not an idealization, but a very real effect, that can be due solely to the loft angle of the club, even if the *motion* of the club is exactly horizontal when it strikes the ball, as I have assumed here. (Essentially, the golfer strikes the ball at the lowest point of his swing.)

      Actually, I don’t think there is a lot that has been “idealized” here, is there? The reason we do this on the Moon is to neglect air resistance and the complex boundary layer effects of a dimpled golf ball. Other than the perfectly elastic collision (which is a safer assumption for, say, a driver than a smaller club), and the uniform “flat-Earth” gravity field (instead of a large spherical body), I’m not sure what effects would yield significantly different results than those posted here?

      • bolt says:

        In retrospect, I realize your model is perfectly accurate. I was under the assumption that the face of a golf club is perfectly aligned with the shaft and by the loft angle you meant the angle of swing at which the golfer hits the ball. So I looked up loft angle and what I found agrees with your model.
        Great post, by the way. Cheers!

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