Solve Relativity Questions: KE, Time, and Distance

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Homework Help Overview

The discussion revolves around a relativity problem involving kinetic energy, time dilation, and distance traveled by a spaceship moving at a significant fraction of the speed of light. The original poster presents a series of questions related to the spaceship's velocity, time experienced by the spaceship versus observers on Earth, and the distances traveled in both reference frames.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the velocity at which the kinetic energy of the spaceship equals two-thirds of its rest energy, and subsequently explores time dilation and distance calculations.
  • Some participants question the validity of using the distance formula d=vt at relativistic speeds and discuss the relationship between distance and time in different frames.
  • Others suggest reconsidering the use of length contraction and time dilation equations, noting that these formulas relate quantities in different frames.

Discussion Status

Participants are actively exploring various approaches to the problem, with some confirming the correctness of calculations while others express confusion about the application of certain formulas. There is a productive exchange regarding the relationship between time and distance in relativistic contexts, though no consensus on a single method has been reached.

Contextual Notes

There is an acknowledgment of the original poster's uncertainty regarding the use of certain equations and the distinction between time and length in the context of relativity. The discussion reflects a learning environment where participants are clarifying concepts without providing definitive solutions.

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Homework Statement


A) at what velocity will be the KE of a spaceship 2/3 of its rest energy?
B) If the Spaceship travels at this velocity relative to the Earth and its trip takes 10 years as measured in the earthbound system, how much time has elapsed on the spaceship clock?
C) What is the distance (in lightyears) it travels as seen from the Earth and as seen from the spaceship?


Homework Equations


m=mo/(\sqrt{}1- (v^2/c^2)
\Deltat= \Deltato/(\sqrt{}1- (v^2/c^2)
L=Lo\sqrt{}1-(v^2]/c^2)

Sorry for some reason the subscript isn't working or I'm doing it completely wrong...

m2


The Attempt at a Solution


A)
KE = (m-mo)c^2 = 2/3moc^2 ... m=1.67o(1+2/3 = 1.67)

1.67 = 1/(\sqrt{}1- (v^2/c^2)
2.7889 = 1/(1-(v^2/c^2))
1/2.7889 = (1-(v^2/c^2))
0.6414 = v^2/c^2
v= 0.80c

B)
\Deltat= \Deltato/(\sqrt{}1- (v^2/c^2)
10 yrs = \Deltato/\sqrt{}1-.80^2
\Deltato = 6 years

C)I didn't really know how to do this one so I just tried and equation:
L=Lo\sqrt{}1-(v^2/c^2)
L=10yrs\sqrt{}1-.80^2
L=61 lightyears for the people on earth

L=Lo\sqrt{}1-(v^2/c^2)
10years = Lo\sqrt{}1-.80^2
Lo16.67 Lightyears for the people in the spaceship

Unfortunately I don't have the answers to these so I can't check my own work, any help would be greatly appreciated..

Thanks




 
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To find the distance, just multiply the speed of the ship by the time elapsed.
 
vela said:
To find the distance, just multiply the speed of the ship by the time elapsed.

Distance on Earth inertial frame = 0.80c(10years) = 8yc (does yc = light years?)

Distance on spaceships inertial frame 0.80c(6years) = 4.8yc

Thanks... does all the other stuff look correct?

Thanks for your help
 
Yes, yc is a light-year. Your other work is fine. You might also note that the two distances you found are related via length contraction.
 
vela said:
Yes, yc is a light-year. Your other work is fine. You might also note that the two distances you found are related via length contraction.

The only formula I thought fit after doing multiple relativity questions was Length contraction for a question like this. Is there another way to solve it?

I tried time dilation first thinking time/years seems reasonable, but my answers were wacky. But it may have been a mathematical error, can it be done using time dilation?

Thank you for all your help, i greatly appreciate it.
 
The length contraction and time dilation formulas merely relate quantities in one frame to those in another. You need an independent way to determine the distance in one of the frames.
 
If I use d=vt I get:
d=0.80c(10years) = 8 Lightyears for the people on earth

and

d=0.80(6years) = 4.8 Lightyears for the astronauts

I was under the impression that d=vt was not valid for speeds close to the speed of light?

(I'm a little confused was finding the distance via the length contraction equation incorrect?)
 
Your impression is incorrect; d=vt is valid at all speeds.

Unless you have the length in one of the frames, how can you use the length contraction formula?

(Just to be clear, those answers you got using d=vt are correct.)
 
now this makes total sense. Since I have no length in either frame I cannot use the Length contraction equation. The 10 years I plugged into the length contraction equation is not a length but a time...

Thank you for helping me with this Vela. I greatly appreciate it...
 

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