Answer:
a) [tex]v=0.999124c[/tex]
b) [tex]E=7.566*10^{22}[/tex]
c) [tex]E_a=760 times\ larger[/tex]
Explanation:
From the question we are told that
Distance to Betelgeuse [tex]d_b=430ly[/tex]
Mass of Rocket [tex]M_r=20000[/tex]
Total Time in years traveled [tex]T_d=36years[/tex]
Total energy used by the United States in the year 2000 [tex]E_{2000}=1.0*10^20[/tex]
Generally the equation of speed of rocket v mathematically given by
[tex]v=\frac{2d}{\triangle t}[/tex]
[tex]v=860ly/ \triangle t[/tex]
where
[tex]\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}[/tex]
[tex]\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}[/tex]
[tex]\triangle t=\sqrt{(860)^2+(36)^2}[/tex]
[tex]\triangle t=860.7532[/tex]
Therefore
[tex]v=\frac{860ly}{ 860.7532}[/tex]
[tex]v=0.999124c[/tex]
b)
Generally the equation of the energy E required to attain prior speed mathematically given by
[tex]E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2[/tex]
[tex]E=7.566*10^{22}[/tex]
c)Generally the equation of the energy [tex]E_a[/tex] required to accelerate the rocket mathematically given by
[tex]E_a=\frac{E}{E_{2000}}[/tex]
[tex]E_a=\frac{7.566*10^{22}}{1.0*10^{20}}[/tex]
[tex]E_a=760 times\ larger[/tex]
