Light travels to a certain star in 221 y. An astronaut makesthe trip in 14 y in

Question

Light travels to a certain star in 221 y. An astronaut makesthe trip in 14 y in his frame of reference. a) How fast do we see him going (as a multiple of c)? Answerin units of c. b) If he just signs the guest book and comes right back,in how many years will he return as seen in the earth frame ofreference? Answer in units of y. c) How much energy would it take to accelerate a 0.8 Mgpayload to this speed? The speed of light is 3E8 m/s. Answer inunits of J. d) How many times greater than the classical solution for theenergy is the relativistic solution? e) How much mass must be annihilated in our matter-antimatterengines to achieve this speed if we assume 100% efficiency? Answerin units of kg. Light travels to a certain star in 221 y. An astronaut makesthe trip in 14 y in his frame of reference. a) How fast do we see him going (as a multiple of c)? Answerin units of c. b) If he just signs the guest book and comes right back,in how many years will he return as seen in the earth frame ofreference? Answer in units of y. c) How much energy would it take to accelerate a 0.8 Mgpayload to this speed? The speed of light is 3E8 m/s. Answer inunits of J. d) How many times greater than the classical solution for theenergy is the relativistic solution? e) How much mass must be annihilated in our matter-antimatterengines to achieve this speed if we assume 100% efficiency? Answerin units of kg.

Explanation / Answer

a)the time dilation is

t = (to/(1 -v2/c2)1/2)

solving the above equation for v we get

v = c * (1 -to2/t2)1/2

to = 14 y and t = 221 y

b)the speed of the astronaut when he goes and comes back are vand -v respectively

therefore,the speed of the astronaut during the entire trip is V= (-v - v) = -2v

the distance between the earth and star is d = c * t

c = 3 * 108 m/s

t = 221 y = 221 * 365 * 24 * 60 * 60 s = 6969456000s

the time taken by the astronaut to return to the earth is

t1 = (d/V) = (d/2v) s = (d/2v) * (1/365 * 24 * 60 *60) y = (d/2v) * (1/31536000) y

c)the energy needed to accelerate a 0.8 Mg payload to this speedis

K = (1/2)m * v2 ---------(1)

m = 0.8 Mg = 0.8 * 106 * 10-3 kg = 0.8 *103 kg = 800 kg

d)the classical solution for the energy is

Ec = m * c2 ----------(2)

(1)/(2)

(K/Ec) = ((1/2)m * v2/m * c2) =(1/2) * ((c * (1 -to2/t2)1/2)2/c2)= (1/2) * (c2 * (1 -to2/t2)/c2) = (1/2) *(1 - to2/t2)

or K = (1/2) * (1 - to2/t2) *Ec

e)the mass that must be annihilated in our matter-antimatterengines to achieve this speed if we assume 100% efficiency is

m = (mo/(1 -v2/c2)1/2)

mo = 800 kg

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