A 1.00*10^6 Kg has 1.00*10^3 Kg of fuel on board. The rocket is parked in space

Question

A 1.00*10^6 Kg has 1.00*10^3 Kg of fuel on board. The rocket is parked in space when it suddenly becomes necessary to accelerate. The rocket engines ignite, and the 1.00*10^3 Kg of fuel are consumed. The exhaust (spent fuel) is ejected during a very short time interval at speed of 0.500c relative to S-the inertial reference frame in which the rocket is initially at rest. (a) calculate the change in the amss of the rocket-fuel system. (b) calculate the final speed of the rocket Ur relative to S. (c) Again, calculate the final speed of the rocket relative to S, this time using classical (Newtonian) mechanics.

Explanation / Answer

a)

vf = ve ln(Mi/Mf) = (0.5 c) * ln((1e6+1e3)/1e6) = 4.9975e-4 c

m-m0 = -1 = 1/(1-(v/c)2) -1 = 1/(1-(4.9975e-4)2) -1 = 1.249 * 10^-7 Kg

b)

vf = 4.9975e-4 c = 4.9975e-4 *3e8 = 149925 m/s

c)

vf = 4.9975e-4 c = 4.9975e-4 *3e8 = 149925 m/s

(unfortunately, I don't know if these answers are correct or not)

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