Thread: Another Relativety Question

A rocketship passes extremely close to Earth with a speed of .800c. Just as the ship passes Eath, observers on the ship and on Earth synchronize their clocks to 3pm. At 3:40 pm, as indicated on the rocketship clock, the ship passes a small space station stationary relative to Earth. The station clocks read Earth time.
(a) What time is it on the station clocks when the ship passes?
(b) How far from Eath is the station (as measured in the reference frame of Earth)?
(c) At 3:40 pm the rocketship time, the ship sends a radio signal back to Earth. When (by Earth time) is the signal recieved at Earth?

Originally Posted by Mtl

A rocketship passes extremely close to Earth with a speed of .800c. Just as the ship passes Earth, observers on the ship and on Earth synchronize their clocks to 3pm. At 3:40 pm, as indicated on the rocketship clock, the ship passes a small space station stationary relative to Earth. The station clocks read Earth time.
(a) What time is it on the station clocks when the ship passes?
(b) How far from Eath is the station (as measured in the reference frame of Earth)?
(c) At 3:40 pm the rocketship time, the ship sends a radio signal back to Earth. When (by Earth time) is the signal recieved at Earth?

Set up a stationary (Earth bound) coordinate system where the origin is at the Earth's position and a +x direction in the direction of travel of the rocketship.

For future reference
$\displaystyle \gamma = \frac{1}{\sqrt{1 - \left ( \frac{0.800c}{c} \right )^2}} = \frac{5}{3}$

a) The time that has passed on the rocketship before reaching the station is 40 minutes. We know that
$\displaystyle t^{\prime} = \gamma t$

So
$\displaystyle t = \frac{t^{\prime}}{\gamma} = 24~\text{minutes}$

b) We want to know the value of x, the distance from the station to the Earth. If we stay in the Earth's frame, we don't need relativity any more:
$\displaystyle x = ut = 0.800c \cdot 1440 = 3.456 \times 10^{11}~m$.

c) At 3:24 (Earth time) the rocketship is sending a radio signal to the Earth from a position of $\displaystyle 3.456 \times 10^{11}~m$. Radio signals are a form of light emission, and thus travel at a speed of c in any reference frame, so the amount of time it takes for the light signal to get to the Earth from this position is
$\displaystyle t = \frac{x}{c} = \frac{3.456 \times 10^{11}~m}{c} = 1152~s = 19~min~12~s$

So the clock reading on Earth when the signal reaches Earth is 3:43:12.

-Dan