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Finger [1]
3 years ago
12

A spaceship moves radially away from Earth with acceleration 29.4 m/s 2 (about 3g). How much time does it take for sodium street

lamps (λ = 589 nm) on Earth to be invisible to the astronauts who look with a powerful telescope upon the city streets of Earth?
Physics
1 answer:
Gemiola [76]3 years ago
7 0

Answer:

doppler shift's formula for source and receiver moving away from each other:

<em>λ'=λ°√(1+β/1-β)</em>

Explanation:

acceleration of spaceship=α=29.4m/s²

wavelength of sodium lamp=λ°=589nm

as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm

using doppler shift's formula:

<em>λ'=λ°√(1+β/1-β)</em>

putting the values:

700nm=589nm√(1+β/1-β)

after simplifying:

<em>β=0.17</em>

by this we can say that speed at that time is: v=0.17c

to calculate velocity at an acceleration of a=29.4m/s²

we suppose that spaceship started from rest so,

<em>v=v₀+at</em>

where v₀=0

so<em> v=at</em>

as we want to calculate t so:-

<em>t=v/a</em>                                                v=0.17c      ,c=3x10⁸           ,a=29.4m/s²

putting values:

=0.17(3x10⁸m/s)/29.4m/s²

<em>t=1.73x10⁶</em>

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