Special relativity - compute relative velocity

[Cameron95]

Homework Statement

A spaceship travels from Earth to the vicinity of the star that is measured
by astronomers on Earth to be six light-years away. The spaceship
and its occupants have a total rest mass of 32 000 kg. Assume that the
spaceship travels at constant velocity. The time taken as measured by
clocks on the spaceship is 2.5 years.
(i) Compute the velocity of the spaceship. [3 marks]
(ii) How long does the trip take as measured by clocks in the Earth's
inertial rest frame? [3 marks]
(iii) What distance does the spaceship travel as measured in its own
reference frame? [3 marks]
(iv) Compute the kinetic energy of the spaceship in the Earth's rest
frame. [2 marks]

Homework Equations

(Lorentz transforms)

The Attempt at a Solution

x=6 light years, t'=2.5 years

I'm stuck on the first part of the problem. I know u=x/t, but the second part of the question asks for t, so their must be a way to find u without explicitly calculating t. Alternatively, u=x'/t', but the third part asks for x'. If anyone could give me a quick pointer that would be great. Thanks.

 

[Chestermiller]

Homework Statement

A spaceship travels from Earth to the vicinity of the star that is measured
by astronomers on Earth to be six light-years away. The spaceship
and its occupants have a total rest mass of 32 000 kg. Assume that the
spaceship travels at constant velocity. The time taken as measured by
clocks on the spaceship is 2.5 years.
(i) Compute the velocity of the spaceship. [3 marks]
(ii) How long does the trip take as measured by clocks in the Earth's
inertial rest frame? [3 marks]
(iii) What distance does the spaceship travel as measured in its own
reference frame? [3 marks]
(iv) Compute the kinetic energy of the spaceship in the Earth's rest
frame. [2 marks]

Homework Equations

(Lorentz transforms)

The Attempt at a Solution

x=6 light years, t'=2.5 years

I'm stuck on the first part of the problem. I know u=x/t, but the second part of the question asks for t, so their must be a way to find u without explicitly calculating t. Alternatively, u=x'/t', but the third part asks for x'. If anyone could give me a quick pointer that would be great. Thanks.

The easiest thing to do is to use the Lorentz Transform. There are two events to consider.

Event 1: Spaceship leaves earth
x = 0, x' = 0, t = 0, t' = 0

Event 2: Spaceship arrives at star
x = 6c, x' = 0, t = ?, t' = 2.5

Event 2 has enough information for you to determine the velocity:
x = γ(x'+vt')

Chet

 

[Cameron95]

The easiest thing to do is to use the Lorentz Transform. There are two events to consider.

Event 1: Spaceship leaves earth
x = 0, x' = 0, t = 0, t' = 0

Event 2: Spaceship arrives at star
x = 6c, x' = 0, t = ?, t' = 2.5

Event 2 has enough information for you to determine the velocity:
x = γ(x'+vt')

Chet

I'm not sure why x' is 0?

 

[dauto]

The problem gives you x and t' so neither v = x/t nor v = x'/t' will do. But keep in mind that time dilation and space contraction relate t to t' and x to x' respectively so using either of them allows you to solve the problem

 

[Chestermiller]

I'm not sure why x' is 0?

In the spaceship frame of reference, its coordinate doesn't change.

Chet

 

[Cameron95]

The problem gives you x and t' so neither v = x/t nor v = x'/t' will do. But keep in mind that time dilation and space contraction relate t to t' and x to x' respectively so using either of them allows you to solve the problem

I tried using time dilation, but I got a very small fraction of the speed of light, so I must have done something wrong. If it's not too much trouble, could you give a quick outline of the solution? I think I've confused myself too much thinking about it.

 

[Cameron95]

I tried using time dilation, but I got a very small fraction of the speed of light, so I must have done something wrong. If it's not too much trouble, could you give a quick outline of the solution? I think I've confused myself too much thinking about it.

In fact, don't worry - I think I understand now. Thank you both for your help.