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On the Moon’s surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distance to the Moon is calculated from the round-trip time. What percent correction is needed to account for the delay in time due to the slowing of light in Earth’s atmosphere? Assume the distance to the Moon is precisely 3.84×108 m , and Earth’s atmosphere (which varies in density with altitude) is equivalent to a layer 30.0 km thick with a constant index of refraction n = 1.000293.

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okpalawalter8

Answer:

[tex]T_d=2.8*10^-^6\%[/tex]

Explanation:

From the question we are told that

Distance b/w Earth and the moon [tex]d_e=3.84×10^8 m[/tex]

Thickness of Earth's atmosphere[tex]d_t=30km[/tex]

Constant index of refraction [tex]n = 1.000293.[/tex]

Generally the equation for percentage of time delay [tex]T_d[/tex] is mathematically given as

[tex]T_d=\frac{d(n-1)}{rm}*100\%[/tex]

[tex]T_d=\frac{(35*10^3)(1.000293-1)()100)}{3.674*10^8} *\%[/tex]

[tex]T_d=2.8*10^-^6 \%[/tex]

Therefore the general percentage correction needed because of tym deley is given as

[tex]T_d=2.8*10^-^6\%[/tex]

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