Calculating the Speed of a Moving Ship in Special Relativity

In summary, the captain of a moving ship measures the speed of her ship relative to Earth by observing an identical ship at rest relative to Earth. Using Lorentz transformations and time dilation/length contraction formulas, the captain calculates the speed to be approximately 0.866c. This is done by finding the proper length of the other ship from the captain's viewpoint and using it to determine the relative velocity between the two ships.
  • #1
matt14690
11
0
2. When traveling in space, a ship (length 1000m in its rest frame) passes an identical ship, which is at rest
relative to Earth. The captain of the moving ship decides to measure the speed of her ship. She looks out of
a window and uses her clock to measure the time between the moments when the head and the tail of the
other ship pass. The result of the measurement is 1 μs. Calculate the speed of the “moving” ship relative to
Earth.
Attempted using both Lorentz transformations and time dilation/length contraction formulas. Tried to solve simultaneously but equation turned out to be ridiculously complex for an introductory physics course question.
 
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  • #2
Hint: According to the captain, how far did the other ship move during that 1 μs?
 
  • #3
1000m??
 
  • #4
wouldn't that require finding the length of the captains ship from within the Earth's reference frame. I am just confused as to how the length contraction affects this situation
 
  • #5
matt14690 said:
1000m??
Nope. The distance traveled must equal the length of the ship (from head to tail) according to the captain.
matt14690 said:
wouldn't that require finding the length of the captains ship from within the Earth's reference frame.
Since the captain is doing the observing, her ship is at rest. But you'll need to find the length of the other ship according to the captain.
I am just confused as to how the length contraction affects this situation
See above.
 
  • #6
i got a value of around 1006 metres for the length of other ship from the captains reference. to do so i had to solve a quartic equation. i used v=(length of other ship from captains reference)/(the time interval from captains reference of 1 microsecond)
i then substituted this into L=L'/gamma where gamma is 1/sqrt(1-v^2/c^2)
and solved for v to find the speed.

it was something like
v^2-(v^4/c^2) = 10^15

does that seem right at all?
 
  • #7
matt14690 said:
i got a value of around 1006 metres for the length of other ship from the captains reference.
Both ships have a proper (rest) length of 1000m. From the reference frame of one ship, will the other ship be longer or shorter?

Also, until you've figured out the speed and thus gamma, how can you get a numerical answer for this length? Instead, just express it using gamma.
i used v=(length of other ship from captains reference)/(the time interval from captains reference of 1 microsecond)
Excellent.
i then substituted this into L=L'/gamma where gamma is 1/sqrt(1-v^2/c^2)
and solved for v to find the speed.
I think you're on the right track, but just got messed up somewhere. L' is the rest length of the ship.
Do it over carefully and you'll get an easy equation to solve. (No quartics!)
 
  • #8
ah can you spell it out a bit more clearly. i can only seem to get the quartic equation
 
  • #9
i just can't find how i can get the velocity of the first ship, when i need the relative velocity to get gamma, and i need gamma to get the proper time and length of other ship from captains viewpoint. then i need to use these to find the velocity. it seems like its a catch 22
 
  • #10
matt14690 said:
i just can't find how i can get the velocity of the first ship, when i need the relative velocity to get gamma, and i need gamma to get the proper time and length of other ship from captains viewpoint. then i need to use these to find the velocity. it seems like its a catch 22

The whole point of special relativity is that you can calculate by picking any inertial frame and taking it to be 'at rest'. Try solving the problem assuming the ship taking the measurement is at rest. Then there is only one gamma.
 
  • #11
matt14690 said:
i just can't find how i can get the velocity of the first ship, when i need the relative velocity to get gamma, and i need gamma to get the proper time and length of other ship from captains viewpoint. then i need to use these to find the velocity. it seems like its a catch 22
There's only one speed in this problem--the relative speed of the ships, which is what you're asked to find. And that's the same speed you need to get gamma.

Start as you did by using that equation for speed:
matt14690 said:
i used v=(length of other ship from captains reference)/(the time interval from captains reference of 1 microsecond)
All you have to do is rewrite "length of other ship from captain's reference" in terms of the ship's known rest length. (Hint: That's where gamma comes in.)

Once you write that equation, just rearrange and solve for v. Do it step by step, showing each step if you need to.
 
  • #12
thanks for your help, i got it out
 

Related to Calculating the Speed of a Moving Ship in Special Relativity

What is special relativity?

Special relativity is a theory developed by Albert Einstein that describes the relationship between space and time. It explains how the laws of physics are the same for all observers in uniform motion and how the speed of light is constant in all reference frames.

What is the difference between special relativity and general relativity?

Special relativity deals with the relationship between space and time, while general relativity also includes the effects of gravity on space and time. General relativity is a more comprehensive theory, but special relativity is still important for understanding the fundamentals of space and time.

What is the concept of time dilation in special relativity?

Time dilation is the phenomenon where time seems to pass at different rates for different observers depending on their relative speeds. According to special relativity, time moves slower for an observer who is moving at high speeds compared to an observer at rest.

How does special relativity explain the twin paradox?

The twin paradox is a thought experiment where one twin travels at high speeds while the other stays on Earth. When the traveling twin returns, they have aged less than the twin on Earth. Special relativity explains this by showing that the traveling twin experienced time dilation due to their high speed.

What is the significance of the speed of light in special relativity?

The speed of light, denoted by the symbol "c", is a fundamental constant in special relativity. It is the maximum speed at which all objects and information can travel in the universe. This constant plays a crucial role in the equations and principles of special relativity.

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