# Special Relativity -- two ships moving in one dimension

## Homework Statement

You approach an enemy ship at a speed of 0.5c measured by you, and the ship fires a missile toward your ship at a speed of 0.7c relative to the enemy ship. What speed of the missile do you measure, and how much time do you have measured by you and the enemy ship before the missile hits you if you are 10^ km away?

## Homework Equations

##\gamma = \frac{1}{\sqrt{1-\beta^2}}##

##\Delta X = \gamma \Delta X'##

##V_x' = \frac{V_x+u}{1+\frac{uv_x}{c^2}}## where ##u## is the speed of the frame of reference, and x denoted movement along one dimension/axis.

## The Attempt at a Solution

I have been working at this problem for way too long now (many hours). I assumed the enemy ship was at "rest" and the missile and your ship were moving. Just telling me whether im on the right track or not would be very helpful and appreciated. My work is shown with clear english explanations and the question is written on a high quality photo. This is the best way to explain, but if you cannot read my writing, just let me know and I will be happy to type it all out for you. Im not sure if i am right or wrong in this, but i think my answers seem reasonable

PeroK
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I agree with you first answer about the speed of the rocket in your frame, but not your other answers. The problem says you are approaching the enemy ship at ##0.5c## and that it is ##10^7 km## away. You should assume that this is in your frame of reference.

In any case, to get the time you have until impact, you should analyse the problem in your frame.

To get the measurements in the enemy frame, you need to analyse the problem in that frame.

First, I would try doing part b) using your frame.

llatosz
So in my frame, im moving at 0.5c and the missile is moving at 0.8888c correct? That would mean that I could travel ##\frac{.5}{.8888+.5} = 36\%## of the total initial distance before getting hit?

PeroK
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So in my frame, im moving at 0.5c and the missile is moving at 0.8888c correct? That would mean that I could travel ##\frac{.5}{.8888+.5} = 36\%## of the total initial distance before getting hit?

In your frame, you are (as always) stationary! That's what defines "your frame".

llatosz
OHHHHHH! This whole 'frame' stuff makes so much more sense now!!! Thank you so much!!!

PeroK
Ah but then in the enemy ship's frame, both my ship and the missile are moving. What ##\beta## would i use to find ##\gamma##, 0.5c or 0.7c? I plan on using the equation here for solving for change in time when changing from frame S' to S where my frame is S' and the "truly" stationary frame is the enemy ship's frame S
##c\Delta t = \gamma (c\Delta t' + \beta \Delta x')##
and solve for ##\Delta t## to find the time i have before getting hit measured by the enemy ship

PeroK
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2020 Award
Ah but then in the enemy ship's frame, both my ship and the missile are moving. What ##\beta## would i use to find ##\gamma##, 0.5c or 0.7c? I plan on using the equation here for solving for change in time when changing from frame S' to S where my frame is S' and the "truly" stationary frame is the enemy ship's frame S
##c\Delta t = \gamma (c\Delta t' + \beta \Delta x')##
and solve for ##\Delta t## to find the time i have measured by the enemy ship

You're jumping ahead. There is no "truly" stationary frame. There is your (rest) frame where you are not moving; and, there is the enemy ship's (rest) frame, where it is not moving.

First, you need to convert everything to your frame (which, in fact, you have already done as the only thing missing was the speed of the rocket, which was given in the enemy frame). Then, you need to analyse the motion in your frame. Hint: this should be easy.

You cannot use time dilation here, as events are separated by distance as well as time. To analyse the motion in the enemy frame, you could use the Lorentz Transformation. But, it should be simpler to convert everything to the enemy frame (hint: all your are missing is the distance to your ship when the rocket was fired). Then analyse the motion in the enemy frame.

Okay so yeah I got ##\frac{10^{10}m}{0.8888*3*10^8 m/s} = 37.5## seconds before the missile hits me in my frame, which im convinced is correct.

But then with the clock on the enemy ship, I cannot simply model the missile hitting me because I am not at rest. My ship and the missile must intersect so since my speed of 0.5c and the missile's speed of 0.7c are both in the frame of reference of the enemy ship. So then could I say that the distance of collision away from the enemy ship is
##\frac{.7}{.5+.7} = 58.33 \;\; \Rightarrow \;\; .5833 * 10^{10} m = 5.833*10^9 m##?

Next I would simply find the time it takes for the 0.7c missile to travel 5.833*10^9m

PeroK
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Okay so yeah I got 1010m0.8888∗3∗108m/s=37.5\frac{10^{10}m}{0.8888*3*10^8 m/s} = 37.5 seconds before the missile hits me in my frame, which im convinced is correct.

Yes, good.

But then with the clock on the enemy ship, I cannot simply model the missile hitting me because I am not at rest. My ship and the missile must intersect so since my speed of 0.5c and the missile's speed of 0.7c are both in the frame of reference of the enemy ship. So then could I say that the distance of collision away from the enemy ship is
.7.5+.7=58.33⇒.5833∗1010m=5.833∗109m\frac{.7}{.5+.7} = 58.33 \;\; \Rightarrow \;\; .5833 * 10^{10} m = 5.833*10^9 m?

Next I would simply find the time it takes for the 0.7c missile to travel 5.833*10^9m

This is not correct, I'm sorry to say.

One approach is to find the distance between the enemy ship and your ship when the missile is fired in the enemy ship's frame.

That will also help find the time in the enemy ship's frame between the launch and missile strike.

Let me help by explaining how I would think about it.

First, I would imagine that you have a marker at ##10^{10}m##. In your frame, the distance between you and the marker is fixed and represents a rest length (proper length) of ##10^{10}m##. But, in the enemy ship frame that length is effectively a moving object ...

Does that give you any ideas?

Ohhh the length between me and the enemy ship would no longer be 10^10 m, length will change. And to find the distance from me and the enemy ship from the enemy's frame I wouldn't use a lorentz transform? I would just do
##\Delta x' = \gamma \Delta x## where ##\gamma## is found by using 0.5 as ##\beta## ?

PeroK