The Wikipedia article on Gravity assist has this terrestrial analogy:

Imagine standing on a train platform, and throwing a tennis ball at 30 km/h toward a train approaching at 50 km/h. The driver of the train sees the ball approaching at 80 km/h and then departing at 80 km/h after the ball bounces elastically off the front of the train. Because of the train's motion, however, that departure is at 130 km/h relative to the train platform; the ball has added twice the train's velocity to its own.

I have two problems with this:

The train's speed relative to the platform is 50 km/h. The ball, traveling in the opposite direction relative to the same platform, is going at 30 km/h. Thus, the train operator will indeed see the ball at

50 + 30 = 80 km/h.

So far, so good. However, once the ball has hit the front of the train (let's assume that no speed at all is lost in the process), it'll have the train's velocity (50 km/h) added to its own (30 km/h) relative to the platform:

50 + 30 = 80 km/h.

Because the train's velocity relative to the platform is 50 km/h, and the ball's now 80 km/h, the operator will see it departing from him at

80 - 50 = 30 km/h.

As already mentioned above, the ball's velocity relative to the platform after the bounce will be:

The train's velocity (50 km/h) + the ball's initial velocity (30 km/h) = 80 km/h, and not twice the train's velocity.

What am I missing?

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    $\begingroup$ As I understand it, the ball bouncing gains the trains momentum during the elastic bounce, so it leaves the train at the relative velocity it impacted with (50km/h + 30km/h) plus the relative velocity of the object it bounced off of (50km/h) $\endgroup$ – Sarah Bailey Jan 21 '16 at 22:17
  • $\begingroup$ @SarahBailey: Okay, so if for argument's sake the ball were stationary relative of the platform at the moment of contact, the train, traveling at 50 km/h would kick it to 100 km/h (twice it's own speed)? $\endgroup$ – Ricky Jan 21 '16 at 22:26
  • $\begingroup$ See also astronomy.stackexchange.com/questions/13302/… -- can a mod merge these? $\endgroup$ – user7073 Jan 22 '16 at 16:39
  • $\begingroup$ @barrycarter Only if they're on the same site $\endgroup$ – TildalWave Jan 22 '16 at 17:21
  • $\begingroup$ What if that question gets migrated here. Then? $\endgroup$ – user7073 Jan 22 '16 at 19:13

However, once the ball has hit the front of the train (let's assume that no speed at all is lost in the process), it'll have the train's velocity (50 km/h) added to its own (30 km/h) relative of the platform

No, the platform is not participating in this collision. You could have just as well asked what the speeds were relative to a faster train on another track and gotten yet another answer. That frame of reference would be just as relevant as the frame of reference of the platform, which is not relevant at all, and the resulting answer would be equally wrong.

The only things participating are the front of the train and the ball. So all that matters is the speed of the front of the train relative to the ball, or the speed of the ball relative to the front of the train. That's 80 km/hr. Since the collision has been stated to be elastic (which we will take to be 100% elastic), and since the train weighs much, much more than the ball, then the train won't change speed at all (ok, it will change by the ratio of the mass of the ball to the mass of the train, out in the ninth decimal place), whereas the ball will depart the front of the train at 80 km/hr in the opposite direction, relative to the front of the train.

The rest follows.

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In your step two, you're considering the speed relative to the station for an interaction that doesn't involve the station any more. The ball travels towards the train at 80 km/h, and it will bounce off it at 80 km/h with respect to the train just like a ball that drops towards the floor at 80 km/h rebounds at 80 km/h (in a fully elastic collision at least, but that's the assumption of the analogy in your question). That is then 130 km/h with respect to the station, assuming a head-on collision.

See A Gravity Assist Primer for additional explanation and where that analogy likely originates from (just in mph instead of km/h):

                     enter image description here

                 Image credit: Charles Kohlhase (Concept) & Gary Hovland (Artwork). Source: NASA/JPL

TL;DR is that in the reference frame of the gravity assisting body, the assisted body neither gains nor loses any speed. That is the same for the ball with respect to the train in your analogy.

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    $\begingroup$ I know Charley well (he was my first boss at JPL), and I recognize in the cartoon his super-careful attention to being accurate. You will note that in the cartoon, the ball hits the front of the train and departs at an angle. That is in fact exactly like a planetary gravity assist, where the spacecraft arrives and departs at the same speed, but its direction has been changed, but never by 180°. Furthermore, he was careful to say in the cartoon "up to" 130 mph. Charley would have never let that cartoon go out saying just 130 mph, since it couldn't be with that angle. $\endgroup$ – Mark Adler Jan 21 '16 at 22:39
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    $\begingroup$ @MarkAdler Yes, I did add "in a fully elastic collision" and "assuming a head-on collision" for simplicity's sake, since it's irrelevant to answering the question as it was asked. $\endgroup$ – TildalWave Jan 21 '16 at 22:45

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