# Romulans, Klingons and Earth

Being at quarrel since over two centuries, Romulans and Klingons have decided that it is now time to settle their differences once and for all. They have sent their admiral space1ships to fight the ultimate battle. From the Southern quadrant the Romulan ship is approaching at $0.82c$, while from the Northern quadrant of the galaxy, the Klingon ship is approaching at $0.88c$. At the center of this battle stands the Earth…

a) At what speed does the Klingon ship sees the Romulan ship approaching?

b) The Klingons fire a massive photon torpedo at Earth, which people on Earth see approaching at $0.98c$. What is the speed of the photon torpedo in the Klingon ship reference frame?

c) Humans defend the Earth by firing a powerful Laser towards the Klingons. The Laser ray leaves Earth, of course at the speed of light $c$. What is the speed of the Laser ray, as seen by the Klingonians?

The relativistic velocity addition relation is

$S = (U + V)/[1+(UV/C^2)]$

where $U$ and $V$ are the speeds of objects 1 and 2 in the rest frame

and S is the speed of 1 in the moving frame of 2.

The convention used is that if both 1 and 2 are approaching each other both $U$ and $V$ are positive.

a) $U =+0.82C$ , $V =+0.88C$

$S =(0.82+0.88)*C/(1+0.82*0.88) =0.987C$

b) $U =0.98C V= -0.88C$

$S =(0.98-0.88)*C/(1+0.88*0.98) =0.0537*C$

c) The speed of light is the same in all reference systems. Thus the speed of laser in the reference frame of Klingons will still be $C$.

Around the Earth at 400 km altitude there is placed a sttellite. Light of 500 nm wavelength is shining upon the licence plate of a car on the Earth. What diameter is necessary for the satelite mirror to resolve the image of the licence plate?

The image of the mirror needs to be separated. The equation for the resolution of an optical intrument ($D$ is the aperture-diameter of the instrument),

$\theta=1.220(\lambda/D)$

the angle on which is seen from $H=400 km$ distance a letter having length $d= 10 cm$ is

$\theta=\tan(\theta)=(d/H)$

It follows that the diameter of the optical instrument is

$D=1.220*(\lambda H/D)=1.220 \frac{500*10^{-9}*400*10^3}{0.1}=2.44 m$