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A rocket ship flies past the earth at 87.0 % of the speed of light. Inside, an astronaut who is undergoing a physical examination is having his height measured while he is lying down parallel to the direction the rocket ship is moving.

If his height is measured to be 1.96 m by his doctor inside the ship, what height would a person watching this from earth measure for his height?

Respuesta :

Olajidey

Answer:

L=0.9664m

Explanation:

According to length the astronaut with respect to a person watching from earth is:

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

Given that,

87.0 % of the speed of light, v

height is measured to be 1.96 m,Lp

substitute the values in the equation

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

[tex]L = (1.96)\sqrt{1-0.87^2}[/tex]

[tex]L = (1.96)\sqrt{1 -0.7569}[/tex]

[tex]L =(1.96)\sqrt{0.2431}[/tex]

[tex]L = (1.96)\times0.4931[/tex]

[tex]L=0.9664m[/tex]

tochjosh

Answer: 0.967m

Explanation:

Rocket's speed = 0.87c

Where c is speed of light in space

Since he's lying vertical his height becomes a length

Lenght l' = Lenght in rocket = 1.96m

Percieved Lenght on earth l =?

Due to relativistic effect his length will be reduced to;

l = l'(1-(v/c)^2)^0.5

l = 1.96(1-0.87^2)^0.5

l = 1.96(1-0.7569)^0.5

l = 1.96(0.2431)^0.5

l = 1.96(0.4931) = 0.967m

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