What Is the Relative Speed Between Two Spaceships Approaching Earth?

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Homework Statement



Two spaceships approach the Earth from opposite directions. According to an observer on the Earth, ship A is moving at a speed of .753c and ship B at a speed of .851c. What is the speed of ship A as observed from ship B? Of ship B as observed from ship A?


Homework Equations



v'=(v-u)/(1-u*v/c^2)

The Attempt at a Solution



For the speed of ship A with respect to B I tried
(.851c-.753c)/(1-.851c*.753c/c^2) = .2728c
For the speed of ship B with respect to A i tried
(.851c+.753c)/(1+.851c*.753c/c^2) = .9776c
I am not sure of the correct answers, However I'm pretty sure these are wrong. Thanks.
 
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Let [tex]S[/tex] be a coordinate system fixed on the observer on earth, oriented such that the positive x-axis is in the direction of the motion of ship A. Then in that frame ship A has velocity [tex]0.753c[/tex] and ship B has velocity [tex]-0.851c[/tex].

Now, we set up a fram [tex]S'[/tex] in which ship A is stationary and which has x-axis oriented in the same way as fram [tex]S[/tex]. Then in your formula [tex]v'[/tex] is the velocity of ship B as seen from ship A, [tex]u[/tex] is the velocity of ship A w.r.t. [tex]S[/tex] (earth) and [tex]v[/tex] is the velocity of ship B w.r.t. earth.

Then the velocity of ship B as seen from ship A is

[tex]v' = \frac{-0.851-0.753}{1+0.753\times 0.851} c[/tex]

If you instead consider an observer in ship B then you will get ([tex]v'[/tex] is the velocity of ship A as seen from ship B, [tex]v[/tex] is the velocity of ship A as seen from Earth and [tex]u[/tex] is the velocity of ship B as seen from Earth

[tex]v' = \frac{0.753+0.851}{1+0.753\times 0.851} c[/tex]

and both these are equal to (except for signs) [tex]0.9776 c[/tex] .
 
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