Einstein's equivalence principle says that you cannot distinguish between an accelerating frame or a gravitational field. However, in an gravitational field, if I drop a tennis ball, it will bounce, but I don't think that it will in the accelerated rocket. Will it bounce? If so, how?
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1$\begingroup$ But It will not bounce because It will "go" with the floor of the rocket $\endgroup$– user189508Mar 19, 2018 at 2:39
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15$\begingroup$ Why do you assume that? Imagine you hitting a ball with a tennis racket. The ball is reflected by an accelerating object (in this case, the racket). In the rest frame of the tennis racket, the ball "bounces". $\endgroup$– PeaBraneMar 19, 2018 at 2:40
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24$\begingroup$ @JoséGarcía Why would it not bounce? $\endgroup$– MichaelKMar 19, 2018 at 10:53
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1$\begingroup$ Let us continue this discussion in chat. $\endgroup$– can-ned_foodMar 19, 2018 at 16:17
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$\begingroup$ Gentle reminder that comments are for improving the question. There's a chat room for extended discussion on this question. Further back-and-forth here will probably be deleted. $\endgroup$– rob ♦Mar 19, 2018 at 23:03
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3 Answers
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This is one of those things that should become clear once you see it, so I made an animation:
As you can see, the ball simply bounces off the back of the rocket once the rocket catches up with it, just like a tennis ball bouncing off the racket during a serve. In the comoving frame (i.e. if we are accelerating along with the rocket), this amounts to the ball bouncing off the floor.
Since the rocket is still accelerating but the ball is not, the rocket will eventually catch up with the ball again and it will bounce a second time. Here is a bonus animation showing multiple bounces:
In this version the ball bounces elastically, and it starts at a lower height, so that several bounces can be observed before the rocket reaches the side of the image. It's a little hard for the eye to see, but in between collisions the ball moves at a constant speed, while the rocket accelerates to catch up with it.
Finally, here's another bonus animation to show that if the ball doesn't bounce elastically then it will stop bouncing and start just moving along with the rocket:
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3$\begingroup$ Based on OP's comment (
But It will not bounce because It will "go" with the floor of the rocket) I think that if the rocket is accelerating fast enough the ball wouldn't bounce off the ground. It would still compress a little but the ball would be touching the ground the entire time. Maybe that is what they are confused about? $\endgroup$ Mar 19, 2018 at 13:02 -
50$\begingroup$ @CaptainMan the faster the rocket accelerates, the faster it will be moving when it hits the ball, so the faster the ball will bounce off it. The ball will stick to the ground if it's really squishy and absorbs all the extra energy from being hit by the rocket, but that's just exactly what happens when you drop a really squishy ball on the ground and it doesn't bounce. $\endgroup$– N. VirgoMar 19, 2018 at 13:11
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$\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$– rob ♦Mar 20, 2018 at 18:34
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The ball will bounce exactly as it would on the surface of a planet with local gravitational acceleration equal to the rocket's acceleration.
The physics really does play out exactly as in Einstein's accelerating rocket thought experiment, and not even bouncing balls will tell the accelerating frame apart from a planet's surface for you in this regard.
I suggest that you work the problem out from a inertial frame removed from the rocket. You let the ball go, and then the floor of the rocket accelerates towards the ball whilst the ball coasts with the velocity it has at the instant of release. Calculate the effects of collision from there, and then transform the ball's position versus time path back to the accelerated co-ordinates; you will find my first two paragraphs to be true.
answered Mar 19, 2018 at 6:05
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$\begingroup$ I suspect your first sentence is the key issue. Most people finding it counterintuitive are probably using their everyday experience at 9.8 m/s² for one frame but not the other. $\endgroup$– GossarMar 19, 2018 at 23:20
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1$\begingroup$ @Gossar Yes I think you're right. Nathaniel's answer a killer for that one, so wonderful! $\endgroup$ Mar 20, 2018 at 0:06
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$\begingroup$ Thanks for the response, but Why the floor of the rocket will hit the ball? And if that happens Why the ball will go at more speed than the rocket? $\endgroup$– user189508Mar 20, 2018 at 0:51
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2$\begingroup$ @JoséGarcía: Try reasoning as in IanF1's answer. The rocket accelerates whilst the ball, after release, coasts, as in Nathaniel's simulation. So the floor has some velocity $\Delta$ relative to the ball when it hits. When the two impact, model them as a collision between two masses and make the rocket mass big. This means in the elastic case, the ball bounces back off the rocket at a velocity $\Delta$ relative to the rocket (if the rocket's mass is much greater than that of the ball), but the latter keeps accelerating and the process repeats. $\endgroup$ Mar 20, 2018 at 1:01
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7$\begingroup$ @JoséGarcía, "Why the ball will go at more speed than the rocket?" Why does a tennis ball travel faster than the racket after it has been hit? $\endgroup$– JustsaltMar 20, 2018 at 19:53
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Let's say both the rocket and the ball start at zero velocity and the rocket accelerates at a constant rate.
The ball starts at some distance $s$ from the floor.
In the time it takes for the rocket to travel $s$ (the first bounce), it will have accelerated to a certain speed, $v$ let's say.
Assuming a perfectly elastic collision, the ball (which was stationary with respect to the starting frame, and is now travelling at $-v$ relative to the rocket) will now be travelling at $+v$ relative to the rocket or $+2v$ relative to the starting frame.
The ball will continue travelling forward at $2v$, and the rocket will continue accelerating towards it. By the time it catches the ball again, it works out the rocket is travelling $3v$ (this can be seen either by symmetry arguments or explicitly working out the equations of motion). Relative to the rocket the picture is identical to the first bounce: the ball is at $-v$ and will bounce to $+v$ again.
Following it through in this way, we find that the behaviour of the ball relative to the (accelerating) frame of the rocket really is the same as if the ball was in a gravitational field.
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$\begingroup$ But why will travel +v or +2v ? The ball will go with the same speed of the rocket at the moment of the impact, no? For example if the velocity of the rocket in the moment of impact is 40 m/s, the ball will go at that speed. I think It won't bounce, except if the ball reduces the speed of the rocket wich its impossible. $\endgroup$– user189508Mar 19, 2018 at 21:17
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4$\begingroup$ @JoséGarcía: Why would it be "impossible" for the ball to reduce the speed of the rocket? The rocket need to exert a force on the ball to keep it from moving through the floor; the reaction to that force is going to slow the rocket down (compared to a situation where the it didn't hit the ball). $\endgroup$ Mar 20, 2018 at 1:52
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6$\begingroup$ @JoséGarcía In the reference frame of the Earth, the ground isn't moving. When you drop the ball and i bounces off the ground, does it end up moving at the same speed as the ground at the moment of impact (which is 0)? No; we're supposing it rebounds elastically, which means all of its relative velocity towards the surface is redirected into velocity away from the surface. The same thing happens in an outside reference frame when the rocket strikes the ball; the ball doesn't gain exactly the velocity of the rocket, the ball's relative velocity towards the rocket is redirected away from it. $\endgroup$– BenMar 20, 2018 at 2:22
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1$\begingroup$ This answer may not have fancy animations, but I feel it explains what actually happens far better than the fancy animations do. $\endgroup$– arothMar 20, 2018 at 12:54
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$\begingroup$ @JoséGarcía for the same reason that a rubber ball doesn't just stop when it lands on the ground - because it's an elastic collision so the relative velocity before and after are the same magnitude but opposite directionn. The physics are identical in both cases. $\endgroup$– IanF1Mar 20, 2018 at 12:56