# Special Relativity Problem [closed]

I am having trouble with the following problem:

Fry travels in a rocket ship towards Leela, at constant relative speed $v$: Fry is delivering a pizza, which in its rest frame stays hot for exactly another 2 minutes. If Leela measures that Fry is 27 million kilometers away, then calculate the minimal value of $v$ for which the pizza is hot when delivered.

My approach was to use speed=distance/time and account for time dilation but I cannot figure out how to eliminate the Lorentz Factor. The answer is meant to be $3c/5$. Any pointers would be appreciated! Thanks.

• you are not meant to get rid of the Gamma factor. It has v in it. It is the only thing that ensures the problem is relativistic instead of just Newtonian – SuperCiocia May 18 '14 at 14:05

Since an answer has already been posted using time dilation and $\gamma$, here's an alternative method employing the invariant interval.

$$(c\tau)^2 = (c \Delta t)^2 - \Delta x^2$$

As specified, the proper time for the pizza is $\tau = 120s$

The displacement of the pizza is $\Delta x = 27 \cdot 10^6 km = 90$ light-seconds.

Solving for the elapsed coordinate time $\Delta t$ yields

$$\Delta t = \sqrt{ \tau^2 + \frac{\Delta x^2}{c^2}} = \sqrt{120^2 + 90^2} = 150s$$

Thus,

$$v = \frac{\Delta x}{\Delta t} = \frac{90}{150}c= \frac{3}{5}c$$

Let the distance between Leela and Fry be $d$, as measured by Leela.

So Leela calculates that it takes $t_L = d/v$ for Fry to get to her.

In Fry's frame of reference, $t_F = \gamma t_L$ because, being the moving frame, he experiences time dilation.

You want $t_F < 2$ minutes.

So $t_F = \gamma t_L = \gamma d/v< 2$ minutes = 120 seconds

where $\gamma = \frac{1}{\sqrt{1+\frac{v^2}{c^2}}}$

Solve for v:

$d^2 < (120)^2 v^2(1+\frac{v^2}{c^2})$

• Maybe you meant 120 seconds? I can't seem to get the required answer out anywho. Hmm. – S Valera May 18 '14 at 15:30
• Yes sorry 120 seconds... (I'm going to change it now) – SuperCiocia May 18 '14 at 17:29