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 Special and General Relativity Special and General Relativity Physics Help Forum Aug 29th 2008, 09:46 PM #1 Junior Member   Join Date: Apr 2008 Posts: 5 Special Relativity Q's Our galaxy is about 10^5 light years in diameter. (a) How fast would a spaceship have to travel in order to cross the galaxy in 300 years as measured from within the spaceship? (b) How much time would elapse on Earth during the transversal? For this question i seem to have got that the velocity of the spaceship to be the speed of light. Seems kinda strange, and by using that velocity, the answer for part (b) is infinite, so theres gotta be something wrong there. Two inertial observers O and O' are moving away from each other at a relative speed of 0.69 c. A pair of events is measured by O to have a spatial separation of 200 m and a temporal separation of 7.2^10-6 s, with the event occurring at the later time being more distant from O. What is the spatial separation of the events as measured by observer O'? For this one here, im having problems with the spatial separation measured by 0, a little confusing by the fact that i dont know what O and O' are moving at. If they are both going at 0.69/2 c then the spatial separation of O would equal the spatial separation of O'. I was thinking of assuming that O is stationary relative to O' and let v(O')=0.69c   Aug 31st 2008, 09:54 AM   #2
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 Originally Posted by ah-bee Our galaxy is about 10^5 light years in diameter. (a) How fast would a spaceship have to travel in order to cross the galaxy in 300 years as measured from within the spaceship?
$\displaystyle t' = \gamma t = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}}$
where t' = 300 years. Notice that the ship is going at a speed v in the stationary frame so we also know that
$\displaystyle t = \frac{x}{v}$
where x is the length of the Galaxy.

Thus
$\displaystyle t' = \frac{\frac{x}{v}}{\sqrt{1 - \frac{v^2}{c^2}}}$

$\displaystyle t' = \frac{x}{v \sqrt{1 - \frac{v^2}{c^2}}}$

 (b) How much time would elapse on Earth during the transversal?
As I mentioned above
$\displaystyle t = \frac{x}{v}$

-Dan
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See the forum rules here.   Aug 31st 2008, 10:03 AM   #3
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If you are going to post more than one question per post (and please don't do this) then at least make it clear where one question ends and another begins. Originally Posted by ah-bee Two inertial observers O and O' are moving away from each other at a relative speed of 0.69 c. A pair of events is measured by O to have a spatial separation of 200 m and a temporal separation of 7.2^10-6 s, with the event occurring at the later time being more distant from O. What is the spatial separation of the events as measured by observer O'? For this one here, im having problems with the spatial separation measured by 0, a little confusing by the fact that i dont know what O and O' are moving at. If they are both going at 0.69/2 c then the spatial separation of O would equal the spatial separation of O'. I was thinking of assuming that O is stationary relative to O' and let v(O')=0.69c
You know that, according to O', O' itself is not moving. So the speed of O in the O' frame is v = 0.69c.

Now you know
$\displaystyle x' = \gamma (x - vt)$

-Dan
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See the forum rules here.   Aug 31st 2008, 04:49 PM #4 Junior Member   Join Date: Apr 2008 Posts: 5 sorry i didnt set my post up properly and thanks a lot guys ill take a look into that. im sure that it will help.  Tags relativity, special Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post maple_tree Special and General Relativity 2 Oct 12th 2010 08:37 AM mars shaw Special and General Relativity 1 Sep 5th 2009 08:10 AM brentwoodbc Special and General Relativity 0 Jan 19th 2009 05:02 PM ah-bee Special and General Relativity 0 Nov 10th 2008 02:43 AM evabern Special and General Relativity 2 Oct 6th 2008 04:26 AM