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

Respuesta :

Answer: 0.99973 c

Explanation:

Betelgeuse is 430 ly away from earth = 4.07 ×10¹⁸ m

You want to age only 20 years during the round trip. It means you want to cover one way distance in 10 years relative to Earth.

10 years = 3.15 × 10⁸ s

This is possible only when rocket travels close to speed of the light.

We will use time dilation formula:

[tex]t_o=t' \sqrt {1-\frac{v^2}{c^2}}[/tex]

where, [tex]t_o[/tex] is the time elapsed in Earth's frame of reference, t' is the time elapsed in rocket's frame of reference, v is the velocity of the rocket and c is the speed of light.

[tex]t_o=\frac{d}{v} \sqrt {1-\frac{v^2}{c^2}}[/tex]

where d is the distance of Betelgeuse from Earth.

[tex]\Rightarrow t_o\times v=d \sqrt {1-\frac{v^2}{c^2}}[/tex]

[tex]\Rightarrow t_o^2v^2=d^2-\frac{d^2v^2}{c^2}[/tex]

[tex](t_o^2+\frac{d^2}{c^2})v^2=d^2\\ \Rightarrow v=\frac{d}{\sqrt {t_o^2+\frac{d^2}{c^2}}}[/tex]

[tex]\Rightarrow v = \frac{4.07\times10^{18} m}{\sqrt{(3.15\times10^8 s)^2+\frac{(4.07\times 10^{18}m)^2}{(3\times 10^8m/s)^2}}}=\frac{4.07\times 10^{18} m}{1.36\times 10^{10}s}=2.99\times10^8m/s = 0.99973 c[/tex]