Relativistic Time for spaceship movement

In summary, the time it would take for a spacecraft traveling at 0.946c to reach a star 76.6 light-years away from Earth, as measured by observers on Earth, is 81 years. The time it would take for the spacecraft to reach the star from Earth, as measured by observers on the spacecraft, is 250 years. This can be calculated using the equation t = t0 / sqrt[1 - (v/c)^2], where t0 represents the proper time and t represents the relativistic time experienced by the observer.
  • #1
skibum143
112
0

Homework Statement


A certain star is 76.6 light-years away. How many years would it take a spacecraft traveling 0.946c to reach that star from Earth, as measured by observers on Earth? How many years would it take to reach that star from Earth, as measured by observers on the spacecraft ?


Homework Equations


Relativistic Time = Original Change in Time / sqrt(1 - v/c)^2


The Attempt at a Solution


For the first question, 1/.946 = 1.057
1.057 * 76.6 light years = 81 years

For the second question, I can't get the right answer. What I did was:
81 / sqrt[1- (.946)^2] and got 250, but that is wrong. not sure what I'm doing wrong...
 
Physics news on Phys.org
  • #2
The factor is the inverse, that is with your names it should be "Relativistic Time = Original Change in Time * sqrt(1-(v/c)^2)". Usually one writes it as

[tex]\tau = \frac{t}{\gamma}[/tex]

where [itex]\tau[/itex] is the proper time for the spacecraft and

[tex]\gamma = \frac{1}{\sqrt{1-(v/c)^2}}[/tex]
 
  • #3
Good to know. Our professor gave us the wrong equation in notes. Thanks for your help!
 
  • #4
It could also be that your professor gave the correct equation, but somewhere along the way in your notes the meaning of the two times got swapped, especially since the first equation I gave above very often is written with gamma on the other side, i.e.

[tex]\tau \gamma = t[/tex]

or, using unprimed time for the spacecraft and primed for the Earth time,

[tex]\Delta t' = \Delta t \gamma[/tex]

As you can probably gather it is easy to get these times mixed up, and one way to keep them straight is to remember, that proper time, i.e. the time interval as measured in a frame at rest relative to the object in question which is the spacecraft in this case, will always be the smallest time interval and since gamma always is equal or bigger than one you can easily deduce with time interval is which in the last equation above.
 
Last edited:
  • #5
I did know that relativistic time (where observer is moving) should be shorter than proper time (where observer is stationary), but our professor gave us this:
proper time = t0
relativistic time = t
relativistic time = t0 / sqrt[1 - (v/c)^2], which doesn't give the right relationship...
so I can just change my equation to t = t0*sqrt[1 - (v/c)^2]
or
t = t0 / 1 / sqrt[1 - (v/c)^2]
but i guess the biggest clue is that unprimed time or relativistic time should be less than primed time or proper time?
 
  • #6
skibum143 said:
I did know that relativistic time (where observer is moving) should be shorter than proper time (where observer is stationary), but our professor gave us this:
proper time = t0
relativistic time = t
relativistic time = t0 / sqrt[1 - (v/c)^2], which doesn't give the right relationship...

This is the correct relationship. Notice, that t = t0 / sqrt(...) = t0 * 1/sqrt(..) = t0*gamma, and that t > t0 for v > 0.

In the context of your original question, the time you find in the first part is what is called relativistic time in your original post, i.e. time t, and you then need to find the proper time t0 (called Original Change in Time in your original post). I guess you just got the times swapped at some point along the way.
 
  • #7
Ah, I see. I was confused about the wording, because I thought "proper time" (t0) was the time measured by the person on earth, but "proper time" is actually the time that is measured by the person that is moving on the spaceship. Thank you so much for your help!
 

1. What is relativistic time for spaceship movement?

Relativistic time is the concept that time can be experienced differently by different observers depending on their relative velocities. This theory is based on Albert Einstein's theory of relativity.

2. How does relativistic time affect a spaceship's movement?

Relativistic time dilation occurs when an object moves at a significant fraction of the speed of light. This means that time will appear to pass slower for an observer on the spaceship compared to an observer on Earth.

3. Can a spaceship travel at the speed of light?

According to the theory of relativity, the speed of light is the maximum speed at which any object can travel. Therefore, a spaceship cannot travel at the speed of light, but it can approach it.

4. What is time dilation in relation to spaceship movement?

Time dilation is the phenomenon where time appears to pass slower for an object traveling at high speeds compared to an object at rest. This is due to the effects of relativity on the perception of time.

5. How is the concept of relativistic time used in space exploration?

The concept of relativistic time is crucial in space exploration as it allows scientists to make accurate calculations and predictions about the movement of objects traveling at high speeds. It also plays a role in the design and operation of spacecrafts.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
754
  • Introductory Physics Homework Help
Replies
6
Views
575
  • Introductory Physics Homework Help
Replies
1
Views
603
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
740
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
Back
Top