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In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested traveling at close to the speed of light. Suppose you wish to visit the red giant star Betelgeuse, which is 430 lyly away, and that you want your 20,000 kgkg rocket to move so fast that you age only 36 years during the round trip.

A. How fast (v) must the rocket travel relative to earth?
B. How much energy is needed to accelerate the rocket to this speed?
C. How many times larger is this energy than the total energy used by the United States in the year 2000, which was roughly 1.0 x 10^20 J?

User Avirup
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1 Answer

3 votes

Answer:

a)
v=0.999124c

b)
E=7.566*10^(22)

c)
E_a=760 times\ larger

Step-by-step explanation:

From the question we are told that

Distance to Betelgeuse
d_b=430ly

Mass of Rocket
M_r=20000

Total Time in years traveled
T_d=36years

Total energy used by the United States in the year 2000
E_(2000)=1.0*10^20

Generally the equation of speed of rocket v mathematically given by


v=(2d)/(\triangle t)


v=860ly/ \triangle t

where


\triangle t=(\triangle t')/((√(1-860/ \triangle t)^2))


\triangle t=(36)/((√(1-860/ \triangle t)^2))


\triangle t=√((860)^2+(36)^2)


\triangle t=860.7532

Therefore


v=(860ly)/( 860.7532)


v=0.999124c

b)

Generally the equation of the energy E required to attain prior speed mathematically given by


E=(1)/(√(1-(v/c)^2) )-1(20000kg)(3*10^8m/s)^2


E=7.566*10^(22)

c)Generally the equation of the energy
E_a required to accelerate the rocket mathematically given by


E_a=(E)/(E_(2000))


E_a=(7.566*10^(22))/(1.0*10^(20))


E_a=760 times\ larger

User Oliort UA
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