# Relativity Physics (1)

1, A spaceship moves with speed

$vs = 0.80c$ directly towards a space station. It dispatches a food supply package (containing all the main courses) to the station with speed $vr = 0.40c$ with respect to itself. What is the supply package’s speed ′$vr$ with respect to the station? Your answer for ′$vr$ should be greater than either $vr$ or $vs$ ; explain why this is the case.

2, A second supply package (containing all the desserts) is mistakenly dispatched before the spaceship’s cargo ejection system has fully recharged, and consequently recedes from the spaceship at a speed of only

$vr =0.01c$. What is this second supply package’s speed with respect to the station?

3, If an inexperienced crew member on the space station erroneously used the Galilean velocity transformation, what results would he get for the speeds of the two supply packages? Comment on these answers in the context of causality, and in comparison to the values you calculated in parts (1) and (2)

4,If the first package travels with constant speed and takes time ΔT = 5.0 days to arrive as measured by the crew member on the space station, what distances it travel in the frame of reference of the space station? Give your answer in units of light-days, the distance traveled by light in one day.

5, How much time elapses in transit as measured by a clock traveling with the first supply package?

6, Assuming that the difference in times of dispatch, and the difference in distance traveled by the two consignments is negligible, calculate the difference in arrival times at the space station between the main course and the dessert consignments.

7, How much less fresh than the main course will the dessert be? (Hint: how much has each of the packages ‘aged’?)

1. Let S be a non moving inertial system and S’ an inertial system that moves with speed v relatively to system S. If a body moves with speed u in S then its speed in S’ will be

$u’ = (u-v)/(1-uv/c^2)$, where $c$ is the speed of light

in our case one can consider S the ship, and S’ the space station which moves with the speed

$V=-Vs$ towards the ship.

$Vr’ =(Vr+Vs)/(1+Vr*Vs/c^2) = c*(0.4+0.8)/(1+0.4*0.8) =0.90909*c$

The speed of the package is greater in the space station system because the space station moves with speed -vs relatively to the ship.

2.

$Vr2′ =(Vr2+Vs)/(1+Vr2*Vs/c^2) = c*(0.01+0.8)/(1+0.01*0.8) =0.8035*c$

3.

Applying the Galilean transformations in point 1 and 2 would get

$vr’ = vs+vr = (0.8+0.4)*c =1.2*c$ toward the space station

and $vr2′ = vs+vr2 =(0.8+0.01)*c =0.81*c$ toward the space station.

One can observe that in the case of small speeds (in the second case) the error is not extremely big. However with respect to the answers calculated relativistic in points 1 and 2 these answers are bigger, (even bigger than C for big enough speed which is false). This comes from the fact that the time frame in the view of an observer in system S (space ship) is different than the time frame in the view of an observer is system S’ (station) and hence the Galilean transformations are not valid.

4.

Because delta(T) is measured in the space station frame the distance traveled in the space station frame is simply

$\Delta(X) =\Delta(T)*Vr’ =0.909*c*5/(1*c) =5*0.909 =4.545$ light-days

$\Delta(X)= 4.545*3*10^8*24*3600 =1.178*10^14$ m

5. Let S be a non moving inertial system and S’ a moving inertial system with speed v relatively to S. Let delta(t) be the interval of time in the S frame and delta(t0) the interval of time in the S’ frame.

If S is the space station and S’ the package frame then $v =vr’$

$\Delta(t0) =[\Delta(t)-vr’*\Delta(X)/c^2]/\sqrt{1-vr’^2/c^2} =[5-0.909*4.545]/\sqrt{1-0.909^2} =$ $=0.868595/0.4168=2.084$ days

6. The time needed for the second package to arrive is

$T2 = distance/vr2′ = 4.545/0.8035 =5.6565$ days

The difference in time arrivals is just

$Time = T2-T1 =5.6565-5 =0.6565 days = 0.6565*24*3600 =488722$ sec

7. The same considerations as at point 5 for the second package

$\Delta(t2) =[\Delta(t)-v_{r2}’*\Delta(X)/c^2]/\sqrt{(1-v_{r2}’^2/c^2)} =$

$=[5.6565-0.8035*4.545]/\sqrt{(1-0.8035^2)} = 2.0045925/0.595=3.367$ days

The desert will be older than the main course with

$T= 3.367-2.084 =1.283 days$