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Q14PE

Expert-verifiedFound in: Page 930

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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 ${\mathbf{3}}{\mathbf{.}}{\mathbf{84}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{8}}}{\mathit{m}}$****, 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.**

The percentage difference is then obtained as: $2.28906\phantom{\rule{0ex}{0ex}}$$\times {10}^{-6}\%$

**Geometrical optics, often known as ray optics, is an optics model that describes light propagation using rays. In geometric optics, a ray is an abstraction that can be used to approximate the routes along which light propagates under particular conditions.**

If the distance between the Earth and the Moon is known, and the distance between them is assumed to be a vacuum, we can calculate the time it would take for a light signal to travel this distance using the following formula: If the distance between them is known, and the velocity of the light is known, we can calculate the time it would take for the light to travel this distance using the following formula:

$t=\frac{d}{v}$

And if a portion of the distance between the Earth and the Moon is not a pure vacuum, we can expect an additional delay for the light signal to return to the Earth because light travels at the fastest in a vacuum, so we can expect the light to travel at a slower speed in this medium between the Earth and the Moon. Because there is a medium between the Earth and the moon, the computed number for time would be a more accurate depiction of the predicted value of the round trip time duration in this scenario (which is the atmosphere).

We can find this time duration in the presence *t *and absence *t ^{' }* of the atmosphere if we know the equivalent thickness of the atmosphere and its corresponding refractive index. After that, the percentage difference is determined using the formula:

$\u2206=\frac{True-False}{True}$

Where true is the most exact predicted value, while false denotes a value with a lower degree of precision.

Calculating the time of a round trip which would take to reach back to Earth in the absence of the atmosphere we obtain is:

$t=\frac{2\times 3.84\times {10}^{8}m}{3\times {10}^{8}m\xb7{s}^{-1}}\phantom{\rule{0ex}{0ex}}=2.56s$

In the absence of the atmosphere, we then obtain:

${t}^{\text{'}}=\frac{2\times \left(3.84\times {10}^{8}-3\times {10}^{8}\right)m}{3\times {10}^{8}}-\frac{2\times 3\times {10}^{-4}}{3\times {10}^{8}}\times 1.000293$

Here in the second term, we multiply it by the refractive index of the atmosphere to account for the delay in the speed of the light inside the atmosphere. So, we obtain:

=2.5600000586 s

The percentage difference is then evaluated as:

$\u2206=\frac{2.5600000586-2.56}{2.5600000586}\times 100\phantom{\rule{0ex}{0ex}}=2.28906\times {10}^{-6}$

This indicates that the acquired number for the time length of a round journey in the absence of the vacuum must be multiplied 1.0000000022806 to get the right result.

Therefore, the percentage difference is $2.28906\times {10}^{-6}\%$.

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