Lee and Leigh are twins. At their first birthday party, Lee is placed on a spaceship that travels away from the earth and back at a steady 0.714 c . The spaceship eventually returns, landing in the swimming pool at Leigh's eleventh birthday party. When Lee emerges from the ship, how old is he?

a. He is still only 1 year old
b. He is 8 years old
c. He is also 11 years old
d. He is 18 years old

Respuesta :

Answer:

b. He is 8 years old

Explanation:

We will use Einstein's formula for time dilation, to calculate the age of Lee. Because Lee was traveling comparable to the speed of light, his age must be lesser than Leigh.

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

where,

T₀ = Time on Earth = ?

T = Relative Time = 10 years

v = relativistic speed of Lee = 0.714 c

c = speed of light = 3 x 10⁸ m/s

Therefore,

[tex]10\ years = \frac{T_o}{\sqrt{1-\frac{(0.714\ c)^2}{c^2} } } \\\\[/tex]

T₀ = 7 years

Hence, the age of Lee will be:

[tex]Lee's\ Age = 1\ year + 7\ years = 8\ years[/tex]

b. He is 8 years old