How Fast is a Spaceship Crossing Earth's Orbit?

[dave2001]

ive been practising relativity questions and there's one i can't do, I've tried loads of ways of approaching it and i can't go any further hence asking for help

it says " the diameter of the Earth's orbit is 3.0 x 10^8 m. a spaceship crosses the orbit in 750s, as measured in the spaceship. what is the speed of the spaceship relative to the earth?"

any guidance would be much appreciated,

 

[tiny-tim]

welcome to pf!

hi dave2001! welcome to pf!

(try using the X² button just above the Reply box )

" the diameter of the Earth's orbit is 3.0 x 10^8 m. a spaceship crosses the orbit in 750s, as measured in the spaceship. what is the speed of the spaceship relative to the earth?"

ok, so in the sun's frame x = vt = 3 10⁸,

and in the spaceship frame t' = 750,

sooo … ? ​

 

[dave2001]

oh sorry , just seen it now you have said that (x²)

 

[dave2001]

hi dave2001! welcome to pf!

(try using the X² button just above the Reply box )


ok, so in the sun's frame x = vt = 3 10⁸,

and in the spaceship frame t' = 750,

sooo … ? ​

sorry, that's how mashed my head is, its not 3 10⁸ that's the value for c,

the diameter of the orbit is given as 3.0 x 10¹¹ m

 

[dave2001]

the diameter of the Earth's orbit is 3.0 x 10¹¹ m. a spaceship crosses the orbit in 750s, as measured in the spaceship. what is the speed of the spaceship relative to the earth? (c = 3.0 x 10⁸ ms^-1

 

[dave2001]

sorry, that's how mashed my head is, its not 3 10⁸ that's the value for c,

the diameter of the orbit is given as 3.0 x 10¹¹ m

i got that far but v comes out at 4x10⁸ ms^-1 and it doesn't work

 

[dave2001]

the answer is 2.4x10⁸ ms^-1

cant figure out how to get to that

 

[tiny-tim]

hi dave2001!

you have been busy while i was out!

i got that far but v comes out at 4x10⁸ ms^-1 and it doesn't work

show us how you got that …

presumably you started by finding t = 1000 ?

 

[dave2001]

yeah that was the first assumption i made

distance x = 3x10¹¹ m
t'=750
t=1000

so in frame for t, v=3x10⁸
for frame t' v=(3x10¹¹)/750 gives 4x10⁸ms^-1

i got to feeling I am going about this the totally wrong way

 

[dave2001]

im getting nowere either with the f-l contraction

 

[tiny-tim]

forget contractions

go back to the Lorentz transformation …

ct' = … ? ​

 

[dave2001]

3x10⁸ x 750 = 2.25x10¹¹m

 

[tiny-tim]

3x10⁸ x 750 = 2.25x10¹¹m

(it'll be a lot easier if you stick to writing c instead of 3x10⁸ )

yes, but i mean what is the formula with ct' on the LHS and x and t on the RHS?

hint: in the sun's frame, t = x/v = 1000c/v, isn't it?

 

[dave2001]

you mean in the Earth's frame, yeah i get the time is 1000c/v,
as for the equation i don't know, i can't find anything in the textbook, I am sorry been working on this question since yesterday and its really getting to me, i keep getting stuck at this point :-(

 

[dave2001]

keep coming out with 1.98x10⁸ms^-1

 

[tiny-tim]

you mean in the Earth's frame

no, in the sun's frame

i can't find anything in the textbook

i keep telling you …

use the Lorentz transformation ! ​

 

[dave2001]

why the suns frame?

 

[tiny-tim]

why the suns frame?

it's "the diameter of the Earth's orbit",

which is stationary in the sun's frame,

but obviously not in the Earth's frame! ​

 

[dave2001]

so ct'=γ(ct-βx) with β=v/c, and γ=1/√1-β²

 

[tiny-tim]

so ct'=γ(ct-βx) with β=v/c, and γ=1/√1-β²

ok, now write that out in terms of β and x (with no γ or t), and solve

 

[dave2001]

x'=1/1-β² (x-βx)?

 

[tiny-tim]

how did you get that?

 

[dave2001]

dont know

 

[tiny-tim]

ok, this was correct …

so ct'=γ(ct-βx) with β=v/c, and γ=1/√1-β²

now rewrite that as t'=γ(t-βx/c), and use …

t = x/v = 1000c/v

… what do you get?

(show all the stages, so you don't make a mistake)

 

[dave2001]

ct'=γ(ct-βx) with β=v/c, and γ=1/√1-β2

t'=γ(t-βx/c)

using t = x/v = 1000c/v

t'=y(1000c/v - βx/c)

t'=y(1000c/v - vx/c²)

t'=y(1000c-x/c²)

 

[tiny-tim]

(i'm not getting email notifications today )

t'=y(1000c/v - vx/c²)

t'=y(1000c-x/c²)

nooo

 

[dave2001]

hahaha, this is the most aukward question i have ever come across, i just can't seem to get anything correct with it, haha

 

[tiny-tim]

hahaha, this is the most aukward question i have ever come across, i just can't seem to get anything correct with it, haha

dave2001, it's not awkward, you're the one making awkward mistakes

you don't seem able to apply equations, and to adjust them from one line to the next

if you're not normally like this, then get some sleep :zzz:

if you are normally like this, then seek some advice, in the academic guidance forum, or at your university​