Light travels to a certain star in 146 y. An astronaut makes thetrip in 25 y in

Question

Light travels to a certain star in 146 y. An astronaut makes thetrip in 25 y in his frame of reference. (a) How fast do we him going (as a multiple of c)?Answer in units of c. (b) If he just signs the guest book and comes right back, inhow 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 0.7 Mgpayload to this speed? The speed of light is 3 x108 m/s. Answer in units of J. (d) How many times greater than the classical solution for theenergy is the relavistic solution? (e) How much mass must by annihilated in our matter-antimatterengines to achieve this speed if we assume 100% efficiency? Answerin units of kg. (a) How fast do we him going (as a multiple of c)?Answer in units of c. (b) If he just signs the guest book and comes right back, inhow 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 0.7 Mgpayload to this speed? The speed of light is 3 x108 m/s. Answer in units of J. (d) How many times greater than the classical solution for theenergy is the relavistic solution? (e) How much mass must by annihilated in our matter-antimatterengines to achieve this speed if we assume 100% efficiency? Answerin units of kg.

Explanation / Answer

(a) How fast do we him going (as a multiple of c)?Answer in units of c.
answer: let his speed in our frame be
v = c
gamma factor = = 1/(1-^2)
distance to star = 146 ly
in his frame, it is contracted to
146/ ly
v = c = (146/)c/25
= 146/(25) = 5.84/
^2 = 34.1056(1-^2)
= (34.1056/35.1056) = .9857
v = .9857c
(b) If he just signs the guest book and comes right back, inhow many years will he return as seen in the earth frame ofreference? Answer in units of y.
= 5.925
trip time in our frame = 2*25 years = 296.25 y
(c) How much energy would it take to accelerate a 0.7 Mgpayload to this speed? The speed of light is 3 x108 m/s. Answer in units of J.
mass m = m0
energy needed = m0 c^2 - m0 c^2 =m0 (-1)c^2 = 700(5.925-1)(3 * 10^8)^2 = .31 x10^21 J
(d) How many times greater than the classical solution for theenergy is the relavistic solution?
classically, the energy needed would be given by the KE formula
KE = (1/2)m0 v^2 = (.5)(700)(.9857 * 3* 10^8)^2 = .03 x10^21 J
relativistic E/ KE = 10.3
(e) How much mass must by annihilated in our matter-antimatterengines to achieve this speed if we assume 100% efficiency? Answerin units of kg.
Mc^2 = m0 (-1)c^2
M = m0 (-1) = 3447.5 kg

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