Relativity Question

  • Thread starter iurod
  • Start date
  • #1
51
0

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/([tex]\sqrt{}1- (v^2/c^2)[/tex]
[tex]\Delta[/tex]t= [tex]\Delta[/tex]to/([tex]\sqrt{}1- (v^2/c^2)[/tex]
L=Lo[tex]\sqrt{}1-(v^2]/c^2)[/tex]

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/([tex]\sqrt{}1- (v^2/c^2)[/tex]
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)
[tex]\Delta[/tex]t= [tex]\Delta[/tex]to/([tex]\sqrt{}1- (v^2/c^2)[/tex]
10 yrs = [tex]\Delta[/tex]to/[tex]\sqrt{}1-.80^2[/tex]
[tex]\Delta[/tex]to = 6 years

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

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

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

Thanks

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,164
1,763
To find the distance, just multiply the speed of the ship by the time elapsed.
 
  • #3
51
0
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
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,164
1,763
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.
 
  • #5
51
0
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.
 
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,164
1,763
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.
 
  • #7
51
0
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?)
 
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,164
1,763
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.)
 
  • #9
51
0
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...
 

Related Threads on Relativity Question

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
904
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
954
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
0
Views
955
Top