# Who gets there first? Speed of light.

1. Sep 11, 2008

### Tomtom

Say you launch a rocket from earth, towards planet Zulu. The rocket travels at approximately c, relative to an observer on Zulu. Now, the rocket is pretty large, and fires a rocket from this rocket - a tiny rocket (so there's only a miniscule change in momentum). This rocket travels at a velocity almost c, towards Zulu - relative to the first rocket!
Now, the observer at Zulu sees two rockets of almost equal magnitude of velocity heading towards it. But who arrives first? They both leave from the same spot at the same velocity relative to their target, but different velocity relative to each other. This is my problem: Who arrives first, and why?

Edit: After thinking a bit about this, I realize that I'm probably going to get a couple of answers stating "massed objects can't reach the speed of light", it would take infinite energy. Well, that's fine with me, those who have a problem with that aspect can instead consider a situation where the first rocket travels at 0.4c, and the second one travels at 0.4c relative to the first rocket. According to my calculations, Zulu will see the second rocket travelling at 0.698c.
The problem -for me- with this, is that for a given set of time, the distances travelled are different. If it takes 1 year for the first rocket to arrive at Zulu, the second rocket should use half that time, according to the first rocket, but Zulu will be expecting it somewhere between 1/2 year and 1 year. (can't be bothered to multiply).

Where's the problem? Is there something about time dilation I should look into?

Last edited: Sep 11, 2008
2. Sep 11, 2008

### JesseM

An object with mass like a rocket can't travel at precisely c, so presumably when you say "approximately c" you mean something like 0.999999c? In this case, the answer is that the observers on Earth will always see the tiny rocket traveling at a velocity slightly closer to c, say 0.9999999999c, so it does arrive first. If you want exact numbers, you can use the formula for relativistic velocity addition here--if the tiny rocket travels at speed v in the rest frame of the larger rocket, and the larger rocket travels at speed u (in the same direction) in the frame of the Earth, then in the Earth's frame the tiny rocket travels at (u + v)/(1 + uv/c^2). So for example if v and u are both 0.999c, then in the Earth's frame the tiny rocket travels at (0.999c + 0.999c)/(1 + 0.999^2) = 1.998c/1.998001 which is equal to about 0.9999995c.

3. Sep 11, 2008

### HallsofIvy

Staff Emeritus
What do you mean "the same velocity relative to their target"? The first rocket travels at "approximately c" but NOT at c because that is impossible. It travels at some speed slightly less than c. Now it launches a smaller rocket. This new rocket adds it speed (almost c) relative to the first rocket to the first rockets speed relative to the planet. No, not "add" in the simple sense of v1+ v2, but adding in the sense of the formula that Tomtom gave. The smaller rocket will, relative to the planet be moving at a speed slightly higher (slightly closer to c) that the large rocket and will, as you would assume, before the larger rocket.

4. Sep 11, 2008

### Tomtom

This response is the same as my edit above:

After thinking a bit about this, I realize that I'm probably going to get a couple of answers stating "massed objects can't reach the speed of light", it would take infinite energy. Well, that's fine with me, those who have a problem with that aspect can instead consider a situation where the first rocket travels at 0.4c, and the second one travels at 0.4c relative to the first rocket. According to my calculations, Zulu will see the second rocket travelling at 0.698c.
The problem -for me- with this, is that for a given set of time, the distances travelled are different. If it takes 1 year for the first rocket to arrive at Zulu, the second rocket should use half that time, according to the first rocket, but Zulu will be expecting it somewhere between 1/2 year and 1 year. (can't be bothered to multiply).

Where's the problem? Is there something about time dilation I should look into?

5. Sep 11, 2008

### JesseM

You have to take into account both time dilation and length contraction if you want to make predictions in both the rest frame of Zulu and the rest frame of either of the rockets. Suppose there's a buoy in space which is at rest with respect to planet Zulu and exactly 1 light year away from it, in the rest frame of Zulu and the buoy. At the moment the large rocket passes next to the buoy, it fires the smaller rocket. Now, in the Zulu/buoy rest frame, the large rocket moves at 0.4c and the smaller one moves at 0.689655c, so the larger rocket takes 1/0.4 = 2.5 years to reach Zulu in this frame, and the smaller one takes 1/0.689655 = 1.45 years to reach Zulu in this frame. But the time dilation formula says that each rocket's clock is slowed down by a factor of $$\sqrt{1 - v^2/c^2}$$, so in the Zulu/buoy frame the larger rocket's clock is only ticking at $$\sqrt{1 - 0.16}$$ = 0.916515 the normal rate, and the smaller rocket's clock is only ticking at $$\sqrt{1 - 0.475624}$$ = 0.724138 the normal rate. So, in the 2.5 years it takes the larger rocket to reach Zulu after passing the buoy in this frame, the larger rocket's clock only measures a time of 2.5*0.916515 = 2.2913 years; and in the 1.45 years it takes the smaller rocket to reach Zulu after passing the buoy in this frame, the smaller rocket's clock only measures a time of 1.45*0.724138 = 1.05 years.

You can also consider things from the rest frame of either the larger rocket or the smaller one; I'll just consider the larger one's frame here. In this frame, the distance between the buoy and Zulu is not 1 light year; since they are moving at 0.4c in this frame, the distance is shrunk by a factor of $$\sqrt{1 - v^2/c^2}$$, in this case $$\sqrt{1 - 0.16}$$ = 0.916515, so the distance is 1*0.916515 = 0.916515 light years. After the large rocket passes the buoy, Zulu is 0.916515 light-years away and approaching at a speed of 0.4c, so it will take a time of 0.916515/0.4 = 2.2913 light years for Zulu to reach the larger rocket in the rocket's rest frame (which was the same as the time on the large rocket's clock when it reached Zulu when we calculated it in the Zulu rest frame). Meanwhile the smaller rocket is moving towards Zulu at 0.4c in the large rocket's frame, and Zulu is moving towards the smaller rocket at 0.4c, so the distance between them is shrinking at a rate of 0.8c (this is sometimes called the 'closing velocity', i.e. the rate at which the one moving object is closing in on another moving object in the frame of a third observer, which is different from the velocity of one of the two objects in the other object's rest frame, for which you'd need to use the relativistic velocity addition formula...note that the closing velocity is allowed to be faster than c!) So if Zulu was 0.916515 light years away at the moment the smaller rocket was fired, the smaller rocket will reach Zulu in 0.916515/0.8 = 1.14564 years in the larger rocket's frame, which is indeed precisely half the time it takes for Zulu to reach the larger rocket in this frame. But in this frame the smaller rocket is moving at 0.4c so its clock is slowed down by a factor of $$\sqrt{1 - 0.16}$$ = 0.916515, so in the larger rocket's rest frame we calculate that the smaller rocket's clock will only measure a time of 1.14564*0.916515 = 1.05 years between leaving the larger rocket and reaching Zulu. And of course, this is also identical to the time we calculated would elapse on the smaller rocket's clock in the Zulu rest frame.

Last edited: Sep 11, 2008
6. Sep 11, 2008

### paw

This spacetime diagram might help. It's in the Zulu frame of reference. Zulu sees the small rocket fired at the first event, the small rocket arrives at Zulu at the second event and the big rocket arrives at the third event.

The big rocket is travelling 0.4c. The small rocket is fired at 0.4c relative to the big rocket.

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7. Sep 11, 2008

### Tomtom

Ah, this is grand! JesseM, that was brilliantly explained, thank you very much!
I'm really grasping this now! :)