<p><a href="https://en.wikipedia.org/wiki/Gravity_assist" rel="nofollow noreferrer">The Wikipedia article</a> has this terrestrial analogy:</p>
<p>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.</p>
<p>I have two problems with this:</p>
<ol>
<li>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</li>
</ol>
<p>50 + 30 = 80 km/h.</p>
<p>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):</p>
<p>50 + 30 = 80 km/h.</p>
<p>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</p>
<p>80 - 50 = 30 km/h.</p>
<ol start="2">
<li>As already mentioned above, the ball's velocity relative to the platform after the bounce will be</li>
</ol>
<p>The train's velocity (50 km/h) + the ball's initial velocity (30 km/h) = 80 km/h, and <em>not</em> twice the train's velocity.</p>
<p>What am I missing?</p>
<p><strong>Addendum:</strong></p>
<p>This is kind of comical:</p>
<p>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.</p>
<p>Again: what am I missing?</p>

CC BY-SA 3.0
2016-01-21T20:56:41.117Z
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.
James K
CC BY-SA 3.0
2016-01-21T20:57:32.613Z
However I'm voting to close this question as off-topic as it is not about astronomy.
James K
CC BY-SA 3.0
2016-01-21T21:37:31.120Z
@JamesKilfiger: You're very kind. You haven't explained anything, though.
Ricky
CC BY-SA 3.0
2016-01-22T01:08:50.740Z
The question appears to be closer related to Space Exploration than to Astronomy.
Gerald
<gravity><general-relativity><special-relativity>
Gravity assist: why is the velocity doubled?
Ricky
# Gravity assist: why is the velocity doubled?

(Asked by: username_0 on 2016-01-21)

[The Wikipedia article](https://en.wikipedia.org/wiki/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:

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.

2. 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?

**Addendum:**

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?

**Tags:** gravity, general-relativity, special-relativity

> **Comments:**
> **username_1**: 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.  
> **username_1**: However I'm voting to close this question as off-topic as it is not about astronomy.  
> **username_0**: @JamesKilfiger: You're very kind. You haven't explained anything, though.  
> **username_2**: The question appears to be closer related to Space Exploration than to Astronomy.  

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