Maths for relativity question. (1 Viewer)

[chingyloke]

Qu 18, Chapter 5 Review, Jacaranda

If our galaxy, the milky way, is 20 kiloparsecs or 65000 light years in radius, calculate how fast a spacecraft would need to travel so that its occupants could travel right across it in 45 years.

Okay...i've gotten this far...

(see attached)

i just can't solve for v...

can someone show me how?


the question has been driving me mad. lol

 

[Trebla]

Assuming you did everything correctly, simply square both sides and collect like terms for v.

 

[chingyloke]

Assuming you did everything correctly, simply square both sides and collect like terms for v.

but do i turn c^2 into 3x10^8^2?

and when i do i have trouble getting it back in terms of c... cause i get an answer but can't turn it into 0.9999994c or whatever the book says

 

[Trebla]

but do i turn c^2 into 3x10^8^2?

and when i do i have trouble getting it back in terms of c... cause i get an answer but can't turn it into 0.9999994c or whatever the book says

If your answer has to be in terms of c, just leave the c as it is.

 

[shady145]

1st i changed the light years to metres.
1 light year = 3x10^8x365x24x60x60
= 9.4608x10^15

diameter = 65000x(9.4608x10^15)x2
= 1.229904x10^21

Lv=(1.229904x10^21)xroot(1-v^2/c^2)
45 years (in seconds) = 45x365x24x60x60
= 1.41912x10^9
v=(1.229904x10^21)xroot(1-v^2/c^2)
1.41912x10^9
1.41912x10^9v=(1.229904x10^21)xroot(1-v^2/c^2)
square both sides
(2.013901574x10^18)(v^2)=1.512663849x10^42(1-v^2/c^2) (2.013901574x10^18)(v^2)=1.512663849x10^42-(1.512663849x10^42)v^2
(c^2)= 9x10^16
(2.013901574x10^18)(v^2)= 1.512663849x10^42-(1.6807361x10^25)(v^2)
(2.013901574x10^18)(v^2)+1.68073761x10^25(v^2)=1.512663849x10^42
1.680737811x10^25(v^2)=1.512663849x10^42
v^2=1.512663849x10^42
1.680737811x10^25
v^2=8.999998922x10^18
root both sides
v=299999982
in terms of c
v(in terms of c)=299999982
3x10^8
=0.99999994c

sorry if my working is confusing

 

[shady145]

waaaa some of the things didnt stay under the divided by sign, but instead moved to the left

the
(c^2)= 9x10^16 (line 15) is supposed to be to the right a little, under the underlined part.
and
3x10^8 (second last line of working) is supposed to be to the right under the underlined stuff

 

[chingyloke]

with this line

v^2=1.512663849x10^42
1.680737811x10^25

I get an answer > 3x10^8....is that just my calculator shitting itself at the sight of dividing such big numbers? where did u get your answer from?

thankyou so much for the working. lifesaver.

 

[shady145]

with this line

v^2=1.512663849x10^42
1.680737811x10^25

I get an answer > 3x10^8....is that just my calculator shitting itself at the sight of dividing such big numbers? where did u get your answer from?

thankyou so much for the working. lifesaver.

yea thats the line.

i knew the answer was right coz i have that text book too. the key is to not do any rounding up at all.
one of the last steps i got the velocity to be 299999982m/s and i rounded it up to 300000000m/s (this was the first time i rounded up). i then saw it was c(3x10^8) so i went back and kept it as exact value and got the right answer

 

[shady145]

o srry i read ur post wrong.
from that line u got c as ur answer.

what calculator do u use?
i got a casio, just a normal standard type one.

a check if u calc is on the right mode.

 

[chingyloke]

o srry i read ur post wrong.
from that line u got c as ur answer.

what calculator do u use?
i got a casio, just a normal standard type one.

a check if u calc is on the right mode.

lol. i did a reset all and got the right answer. musta been in some weird mode. thanks heaps bro

 

[spyro14]

Times like this im glad 3 unit Physics isn't running anymore

 

[shady145]

lol. i did a reset all and got the right answer. musta been in some weird mode. thanks heaps bro

no worries.

spyro when did they take 3u phy out