light years (1 Viewer)

[shkspeare]

"If our galaxy, the Milky Way, is 20kiloparsecs or 65 000 light years in radius. How fast would a spacecraft need to travel so that it's occupants can travel riht across it in 45 years"

ok i'm not really sure how ur meant to get 0.99999994c

anyone explain plz

 

[Xayma]

I think it is talking about relativity. By the look of the question. Ill do the working in a sec. Edit: I really cant be stuffed simultaneous equations now since the time to an outside observer is dependent upon the speed of the craft, so is the time to the inside observer.

 

[tooheyz]

what topic is this? i tried to solve it out and paper and didnt get anywhere...

hmm i think im forgetting how to do physics.. hahahah

 

[shkspeare]

space of course

 

[Xayma]

Originally posted by ToOhEyZ
what topic is this? i tried to solve it out and paper and didnt get anywhere...

hmm i think im forgetting how to do physics.. hahahah

I got lost when I started playing around with simultaneous equations and getting square roots and squares with 8 decimal places on my calc.

 

[Xayma]

When I do it I get above c, I know thats wrong though, hmmm. Nm I got it Ill post it soon.

 

[Ragerunner]

I have never seen a question like that in any past paper.

Your school must be crazy to ask that question because got no idea.

 

[shkspeare]

lol... ok

i think its right ur meant to use simultaneous equations tho

i just keep fuking it up... -.-

 

[Xayma]

Ok using the formula t=t_0/rt(1-v^2/c^2)
Since Time=distance/speed
t=1.230719443x10^21/v (I converted to SI units, I wasnt getting the answer any other way) (oh yeah the distance across is 130 000light years as 65 000 is the radius) anyway let 1.230719443x10^21=R

t_0=1420060896seconds
let t_0=E
And c^2=9*10^16ms^-1

therfore we get R/v=E/sqrt(1-v^2/(9x10^16))
Rsqrt(1-v^2/9x10^16)=Ev (cross multiplying)
R^2-(R^2xV^2)/(9x10^16)=E^2v^2 (squaring)
R^2=(R^2xV^2+(9x10^16xE^2xV^2))/9x10^16 (getting v^2 on one side)
9x10^16xR^2/(9x10^16xE^2+R^2)=v^2
Then calculator job to get 299 999 982ms^-1 or 0.99999994c

Bit messy but yeah. Yay I get to do this next year when we start relativity

 

[tooheyz]

man, thats depressing... i forgotten how to do everything already!

hahaha yeah that question seems a bit odd... never seen problems like that in past papers or specimen papers... its an ugly questions man

 

[:: ck ::]

here u go...

kinda stuffed up with calculator but shud b rite working out newaiz i hope ...

mm and last two or three lines except the answer u can ignore .. was tired at the time

 

[Xayma]

Originally posted by :: ryan.cck ::
mm and last two or three lines except the answer u can ignore .. was tired at the time

Yeah I was about to ask where the 45 dissapeared to. Looks 2 posts up. Fair bit of whiteout used there

Originally posted by ToOhEyZ
man, thats depressing... i forgotten how to do everything already!

I havent learnt to do it yet, but I read a brief history of time and the excel book does that count?

 

[:: ck ::]

ahha yeah was 12 oclcok n i just went for some exercise in the arvo

too tired to help ppl but meh ... gotta put wot i learned in ma holz into practice

 

[Xayma]

Originally posted by :: ryan.cck ::
ahha yeah was 12 oclcok n i just went for some exercise in the arvo

too tired to help ppl but meh ... gotta put wot i learned in ma holz into practice

So the motto of this story is either: Don't help people, or Exercise is bad.

 

[shkspeare]

thx guys

one more question

"Calculate the travel time to the following destinations travelling at 0.999c.
a)Pluto when at a distance of 39.1 AU from the Earth"

1 AU = 1.5813 x 10^-5 light years = 1.496 x 10⁸ km

i cant get 15 minutes

 

[Constip8edSkunk]

travel time from whose perspective?

v=0.999 c
l=39.1*1.5813*10^-5 = 6.1829*10^-4 l.y.

l=v*t_v
so t_v = 6.1829*10^-4/0.999
=6.1891*10^-4 years
=5 hr 25 min 18 sec

t_o=t_v*sqrt[1-(v/c)^2]
=6.1891*10^-4 * 0.4471
= approx. 14 min 33 sec