What speed is needed for a spacecraft to travel 10 light-years in one year?

[msimmons]

Homework Statement

Plans are made to send a spacecraft from Earth to a nearby start 10 light years away. The system support will last one year and one day. The trip is one way trip.

a) What speed must the craft travel to arrive at the star with battery power for one day to make the measurement? (aka, you have one year.)

b) For the reference frame with the spacecraft , what is the distance between the Earth and the star? (so how far does it go that it measures?)

Homework Equations

As far as I know only
$$t'=(t-vx/c^2)\gamma$$
$$x'=(x-vc)\gamma$$
$$\gamma=(1-v^2/c^2)^{-1/2}$$

The Attempt at a Solution

Well, from what I can gather the Earth is in the x' frame and the spacecraft is in the x frame.
So t=1, and x' = 10ly

solving the t' equation for t , then pushing it into the x' equation I get (assuming I didn't make any silly mistakes)
$$x'=(x(1+v^2/c^2))\gamma-vt'$$
which won't work because I have two unknowns, (x and v). solving the other way around (first x' for x, then putting it into the t' eqn) I get
$$t'=(t(1+xv^2/c^2))\gamma-x'v^2/c^2$$
which predictably arrives at the same conclusion.

So my approach is wrong, but I can't figure out how else to approach it.

 

[lightgrav]

no, I think the supplies are consumed in 1 year of *proper time* in the spaceship.
The duration t2' -t1' ... (=Dt') has to be gamma factor *shorter*
than the time in any other reference frame.

 

[msimmons]

I'm not sure I understand what you mean.
I know the 1 year is in proper time of the spaceship, but I thought that would make it the unprimed frame and the Earth the primed frame. is that backwards?

 

[lightgrav]

the time interval between any two events is *dilated* in all *non-proper* frames.
The first equation you have for your PRIMED frame has :
(t2' - t1') = (t2g -vxg ) - (t1g - vxg ) ... ~ (t2 - t1)g , so IT is dilated by gamma.

 

[astromaestro]

Ok, forget primed and unprimed and just think of SR principles. The distance to the star in the Earth frame is 10 LY. However, in the spaceship's frame, the ruler used to measure that distance will be length contracted, so the astronauts will say that they are traveling a shorter distance, namely $$\frac{10 LY}{\gamma}=10 LY\sqrt{1-\frac{v^2}{c^2}}$$. The fuel supply gives one year in the rocket's frame. So both the mission control on Earth and the astronauts on the rocket will agree on their relative speed, so let's just invoke the simple kinematic equation $$v=\frac{d}{\Delta t}$$ which applies in both frames but we must be sure to use $$d$$ and $$\Delta t$$ for the same frame. From earth, the rocket travels a longer distance for a longer time, and from the rocket, it travels a shorter distance for a shorter time, but still at the same speed. So we have $$v=\frac{10 LY\sqrt{1-\frac{v^2}{c^2}}}{1 yr}\Longrightarrow v=\sqrt{\frac{100}{101}}c$$. Does that make sense?