Homework Help: Basic SR question?

1. Feb 8, 2008

magicuniverse

1. The problem statement, all variables and given/known data

a ufo travels from one planet to another (distance measured 384000km) at v=0.8c.

1)How long does the trip take according to observer on planet left?

2)How long does the trip take according to observer in ufo?

3) What is the two planets distance measusured my aliens in the ufo?

2. Relevant equations

I think I have to use the lorrentx transformations.

t'=alpha(u/c)(t-ux/c^2)

t=alpha(u/c)(t'+ux'/c^2)

x=alpha(u/c)(x'+u't)

3. The attempt at a solution

This is my first ever SR question I have done. So I wasnt sure.

I thought that from the original planet the apparent velocity is going to be 0?

so i said the time takes was = 1.83 s?

So then i said using the above formula that alien man thought it would be = 2.69s

As this is wrong I havnt proceeded to the next part. Can you please tell me where i am going wrong and possibly give me some advice as to how to set up these types of questions.

thanks

2. Feb 9, 2008

magicuniverse

come on surely someone can help me with this???

3. Feb 9, 2008

magicuniverse

can someone tell me if im posting this in the wrong place at least lol?

4. Feb 9, 2008

chocokat

I can try to help. First, you should convert the distance in km to s, i.e. 384,000 km = ?s.
I used $$c = 3 * 10^8$$ and got d = 1.28s.

Now, I know that the aliens travelling in the ufo will have the shortest time. I used the spacetime equation

$$\Delta s^2 = \Delta t^2 - \Delta d^2$$

in this particular case, d = 0, (this has to do with the distance the 'clock' is from the person (or in this case alien) using the 'clock') I calculated t as distance/velocity so t = 1.6s (this is what I got for part B or 2 of your problem). Do you happen to know if this is the correct answer? If this is right, I may be able to get you to the first part.

Also, let me know if you're using a different value for c (such as 2.998).

5. Feb 9, 2008

Staff: Mentor

You are told that the ufo travels at 0.8 c.

How did you arrive at this number?

6. Feb 9, 2008

magicuniverse

kk so I know that speed =distance/time

so time = distance/speed = 4.8 seconds. Is that right?

7. Feb 9, 2008

Staff: Mentor

Good.

No. What's the distance? What's the speed?

8. Feb 9, 2008

magicuniverse

No. What's the distance? What's the speed?

the distacne is 384000000m
and speed is 0.8*3ee9

when i put this into my calulator i get 1.6

Last edited: Feb 9, 2008
9. Feb 9, 2008

Staff: Mentor

Don't worry about the time according to the ufo until you've nailed down the time according to the planet.

10. Feb 9, 2008

magicuniverse

and for the other time i get 2.67s is that right?

11. Feb 9, 2008

Staff: Mentor

That's better.

Will the time for the trip as measured by the ufo be shorter or longer? By what factor? (Hint: The ufo can be treated as a single moving clock.)

Do moving clocks run slow or fast?

Last edited: Feb 9, 2008
12. Feb 9, 2008

magicuniverse

Simply from the twins paradox thing im gonna say that the time is shorter for the moving ufo.
So the time will be 1.25s right?

13. Feb 9, 2008

Staff: Mentor

What you need is the formula for the time dilation of a moving clock. But yes, the moving ufo clock runs slow (according to the planet clocks)--so the time measured by the ufo will be shorter than that measured by the planet.
How did you get this number?

14. Feb 9, 2008

magicuniverse

I plugged numbers into a fomula in my book.

15. Feb 9, 2008

Staff: Mentor

What formula?

16. Feb 9, 2008

magicuniverse

t' = (t-ux/c^2)/root(1+u^2/c^2)

is it wrong?

17. Feb 10, 2008

Staff: Mentor

That's one of the Lorentz transforms, but with a typo. Here's the corrected version:
$$\Delta t' = \gamma(\Delta t - v\Delta x/c^2)$$

Where:
$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$

But even simpler is to recognize that moving clocks run slow by a factor of $\gamma$:
$$\Delta T = \gamma \Delta T_0$$

(Either way will work.)

18. Feb 10, 2008

magicuniverse

but putting all the numbers into the formula you gave me gives 2.67s? what am i doing wrong?

19. Feb 10, 2008

Staff: Mentor

I gave two formulas. Which one are you using? (Not that it matters--both will give the same answer.)

Looks to me like you are somehow reversing the primed and unprimed coordinates. What are you plugging in for $\Delta t$ and $\Delta x$? What do you get for $\gamma$?

20. Feb 10, 2008

magicuniverse

i get 5/3 for lambda. is that wrong?

21. Feb 10, 2008

magicuniverse

no sorry im lying i get lambda to be 0.6 so t =0.96s is this right?

and i also get the distance to be 2.09m is that correct?

Last edited: Feb 10, 2008
22. Feb 10, 2008

Staff: Mentor

You mean gamma, not lambda. That's correct.
Gamma is always greater than 1; but, yes, that time is correct.

Redo that calculation.