The Wikipedia article 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:

  1. The train's speed relative to the platform is 50 km/h. The ball, traveling in the opposite direction, 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):

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.

  1. 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?


This is kind of comical:

You're watching a tennis match. Player A serves the ball at 200 km/h. Player B returns the serve, the speed of his racket added to DOUBLE the speed of the ball. Player A then puts in a ground stroke, and the speed of the ball doubles again. If they keep this up, then, according to the above logic, the speed of the ball relative to the court will be approaching the speed of light before they know it.

Again: what am I missing?

  • $\begingroup$ There is no absolute state of rest. The point of view of the driver is a perfectly valid inertial frame of reference. In this frame the ball approaches at 80kph, and departs at 80kph. You assume that "it'll have the train's velocity (50 km/h) added to its own (30 km/h)" That is not correct. In fact, when measured in the reference frame of the platform it gets twice the train's velocity. This is a simple case of your assumptions being wrong. $\endgroup$ – James K Jan 21 '16 at 20:56
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    $\begingroup$ However I'm voting to close this question as off-topic as it is not about astronomy. $\endgroup$ – James K Jan 21 '16 at 20:57
  • $\begingroup$ I'd argue that it is since it's supposed to provide an analogy to the gravity boost, which is an astronomical concept. $\endgroup$ – user21 Jan 21 '16 at 21:05
  • $\begingroup$ @JamesKilfiger: You're very kind. You haven't explained anything, though. $\endgroup$ – Ricky Jan 21 '16 at 21:37
  • $\begingroup$ The question appears to be closer related to Space Exploration than to Astronomy. $\endgroup$ – Gerald Jan 22 '16 at 1:08

Since the collision is perfectly elastic, the ball's velocity goes from -80 km/h from the train's reference point (negative being towards the train) to +80 km/h from the train's reference point, a speed increase of 160 km/h.

For a stationery observer, therefore, the velocity goes from -30 km/h (towards the train) to 130 km/h, an increase of 160 km/h.

The situation you describe only applies if the train isn't moving.

I made an error in units below, assuming the train is moving at 50 km per second (not hour) and that the ball is moving at 30 km per second (not hour), am I'm too lazy to correct it. The general principle, however, still applies.

The confusion may occur because we're ignoring the train's loss of momentum, which means the train is going slower after the collision, and moving backwards in its own frame of reference. A slightly more detailed calculation:

  • Suppose the the train has mass M kg and the ball has mass m kg.

  • From the "fixed" observer's point of view, the initial momentum (in Newtons) is:

$\rho =50 M-30 m$

and the initial kinetic energy (in Joules) is:

$e=450 m+1250 M$

  • Let $v_t$ and $v_b$ be the train's and ball's velocities in meters/second after the collision. Since perfectly elastic collisions preserve both momentum and kinetic energy, we have:

$m v_b+M v_t=50 M-30 m$

$\frac{m v_b^2}{2}+\frac{M v_t^2}{2}=450 m+1250 M$

There are only two solutions to the equation above, one of which is the initial conditions. The other is:

$ \left\{{v_b}\to -\frac{10 (3 m-13 M)}{m+M},{v_t}\to -\frac{10 (11 m-5 M)}{m+M}\right\} $

Plugging in 0.0585 kg for the mass of a tennis ball and 640000 kg for the train, this becomes:

$\{{v_b}\to 129.9999854,{v_t}\to 49.99998538\}$

effectively confirming the calculation.

I'm not convinced this is a good analogy, however. Gravitational boost occurs when a planet's gravity almost captures a spacecraft, thus nearly making it a satellite, and giving it the same revolution velocity around the Sun as the planet itself. The wikipedia analogy bears only a passing resemblance to this.

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  • $\begingroup$ Suppose you're watching a tennis match. Player A serves the ball at 200 km/h. Player B then returns the ball at (speed of his racket + 200 + 200). Player A now puts in a ground stroke, and the ball is now traveling at (speed of his racket + 400 + 400). If they keep going like that, pretty soon the speed of the ball will be approaching the speed of light. Again: what am I missing? $\endgroup$ – Ricky Jan 21 '16 at 21:57
  • $\begingroup$ Remember, it's double the train's velocity, not double the ball's velocity. In this case, the ball's speed would increase by double the racket's velocity. Since this is relatively small and the collision isn't perfectly elastic and there's air resistance, this extra speed will soon diminish. $\endgroup$ – user21 Jan 21 '16 at 22:10
  • $\begingroup$ Isn't the racket's velocity normally greater than the speed of the ball at the moment of contact with the ball, though? 200-plus km/h is hardly small. $\endgroup$ – Ricky Jan 21 '16 at 22:14
  • $\begingroup$ If you're watching tennis matches where players are swinging tennis rackets at 200+ km/h both serving and returning in the exact opposite direction the ball is traveling, I'm impressed. $\endgroup$ – user21 Jan 21 '16 at 22:17
  • $\begingroup$ I know, right? 400 km/h added with each stroke, with the interval between strokes shortening rapidly, the speed of light is only a handful of hours away. $\endgroup$ – Ricky Jan 21 '16 at 22:21

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