- #1
- 726
- 166
Homework Statement
This is from chapter 1 (2nd edition) of Taylor and Wheeler's 'Spacetime Physics.'
It's a four part problem, and I didn't have any troubles with parts a, b, or c. I'm stuck on part d. I'm supposed to determine if we could make it to Andromeda within a lifetime.
Assume that the Andromeda galaxy is 2 million light years distant from Earth. Treat both Earth and Andromeda as points, and neglect any relative motion between them.
d. Trip 3: Now set the rocket time for the one-way trip to 20 years, which is all the time you want to spend getting to Andromeda. In this case, what is your speed as a decimal fraction of the speed of light?
Discussion: Solutions to many exercises in this text are simplified by using the following approximation, which is the first two terms in the binomial expansion
$$(1+z)^n~\approx~1+nz$$
if
$$z\ll1$$
Here n can be positive or negative, a fraction or an integer; z can be positive or negative, as long as its magnitude is very much smaller than unity. This approximation can be used twice in the solution to part d.
Homework Equations
$$(spacetime~interval)^2=(difference~in~time~readings)^2-(difference~in~space~readings)^2$$
$$I^2=(\Delta t)^2-(\Delta x)^2$$
3. My attempt at a solution
Following the above equation, I assumed that ##\Delta t=20~yrs## and that ##\Delta x=2.00\times10^6~yrs##
This gives me
$$I^2=(20~yrs)^2-(2.00\times10^6~yrs)^2$$
Which led me to
$$I^2=-4\times 10^{12}~ yrs$$
Which is a negative number. This solution didn't make sense to me.
Then I tried calculating it directly by converting the distance of ##2.00\times 10^6~yrs## into meters, and 20 years into seconds. This gave me
$$2.00\times 10^6~yrs ~ \times ~ \frac{31,536,000~s}{1~yr}~\times~3.00\times 10^8~\frac{m}{s}=1.89216\times10^{22}~m$$
And
$$20~yrs~\times~\frac{31,536,000~s}{1~yr}=630,720,000~s$$
Then dividing distance by time I get ##3.00\times10^{13}~ m/s##
Which is faster than c, and clearly wrong.
Since they mention the binomial expansion, I'm assuming that I have to incorporate it in this problem somehow. I'm not seeing how to go about doing that. Do I need to somehow incorporate the difference between the Earth frame and the rocket frame in calculating my time interval? Is the space interval going to be different than the 2 million years? Any help would be much appreciated. :)