# Help me use mathematics to resolve this 'paradox'

1. May 5, 2013

### Jarvis323

In order to attempt to understand relativity, I thought of the following 'paradox', which I hope to resolve mathematically.

An astronomer makes a terrifying observation. An asteroid is on a crash course towards Earth at almost the speed of light. An ultra advanced nuclear rocket is launched to intercept and break up the asteroid before it is able to reach Earth. The rocket accelerates to almost the speed of light relative to Earth in the direction of the asteroid.

Observers on Earth see the rocket destroy the asteroid at somewhere close to where the midpoint
between where the rocket and the asteroid were when the rocket reached it's max velocity. The Earth is saved.

To an observer on the Asteroid, the rocket, like the Earth, is already heading towards it at nearly the speed of light. The rocket is only able to gain an indiscernible amount of velocity relative to the Asteroid after it has been launched. The observer is unable to notice any difference in speed between the rocket and the Earth. Both the Earth and the rocket crash into it at nearly the same moment. The Earth is destroyed.

Of course this cannot be what actually happens.

How can you solve this problem mathematically? Say you use the following set of initial conditions:

The Earth and the Rocket were converging at a speed S = C - 10^-(10^10) m/s. At the moment the Rocket was fired, they were 1 light year apart. For the sake of this problem say the Rocket were able to reach the same speed S (relative to Earth) in 1 second.

Feel free to change the values to make the problem easier to solve.

2. May 5, 2013

### Mentz114

The easiest way to analyse this scenario is with space-time diagrams. These show the position of the players ( along the horizontal axis ) at all times ( along the vertical axis ). The first diagram shows the setup from the earth frame, the second one from the missiles frame.

The fact that the missile hits the asteroid before the asteroid reaches earth is unalterable.

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3. May 5, 2013

### Staff: Mentor

There is some difference between the speed of earth and the speed of the rocket, that is sufficient. In the frame of the asteroid, the collision takes place closer to earth (due to length contraction).

4. May 5, 2013

### Jarvis323

What if the rocket were fired from deep underground, and the conditions were such that in the asteroids point of view, the rocket and the asteroid collide within the interior of the Earth?

5. May 5, 2013

### phinds

To back off a bit from your speed assumptions, look at it this way (I'll get back to your speed) and I think you'll see where you are going wrong:

Let's say the asteroid is a light-year from Earth and moving at 1/10 the speed of light and at that moment, a spaceship flies past Earth, going towards the asteroid at 1/10 the speed of light. Clearly the asteroid and the spaceship are going to meet up when the asteroid is 1/2 light year from Earth.

NOW ... increasing the speed of each (by the same amount to keep things simple) does NOTHING to change the fact that they are going to meet up when the asteroid is 1/2 LY from Earth, so there cannot be any argument about WHERE the event happens only about when it is SEEN to happen.

Under the scenario I presents (speed = 1/10 c) the collision will occur, according to the folks on Earth, about 5 years after the rocket passes Earth.

NOW ... up the speed to .9c and the collision takes place at the same location but folks on Earth see it as happening about a year after the rocket passes Earth.

Up the speed to .999c and everything is the same except that folks on Earth see the collision happening a tad over 1 year after the rocket goes by.

You can use the same analysis to show that the asteroid "sees" the event happening a different times depending on the speeds, but it still happens at the same location.

Your problem basically is that you have glossed over the fact that the asteroid sees a difference in speed between the spaceship and Earth that IS enough to cause the even to occur 1/2 LY out from Earth.

EDIT: and by the way, Mentz's analysis is really a better approach, I just like to see if I can explain things as simply as possible and if you don't know space-time diagrams, I don't want to force you to learn them --- but you SHOULD if you are going to pursue this area of study.

Last edited: May 5, 2013
6. May 5, 2013

### Staff: Mentor

The earth gets flattened as well due to length contraction. Oh, and the rocket starts earlier (in the frame of the asteroid), due to relativity of simultaneity.

7. May 5, 2013

### Jarvis323

I'm going to try and wrap my head around your answers. I'll respond later if I have more questions.

Thank you.

8. May 5, 2013

### LURCH

You could also consider that, form the asteroid's reference frame the collision with the rocket takes place just an instant before colliison with earth, but everything else is equally rapid. The collision with the rocket, the detonation of the warhead, and the subsequent scattering of asteroidal debris all happen very quickly. In fact, the ratio of "how long it takes the pieces to scatter"/"How long till we hit Earth" remains the same. Some outside observer sees the asteroid getting hit 1 year before hitting Earth, and it takes about a week for the debris field to disperse enough that it's no longer a threat. But the asteroid sees itself getting hit just 1 second before reaching Earth. However, from this perspective the dispersal takes just over 1/50th of a second.