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# Tutor profile: Hunter H.

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Hunter H.
Researcher looking to help with online classes
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## Questions

### Subject:Astrophysics

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Question:

You are using a radio telescope to observe Mars from Earth. Your telescope is running on a band at 430 KHz ($$f_0$$). You measure one edge of Mars to give a signal of 429.654 KHz ($$f_1$$)and the opposite edge of Mars to give a signal of 430.346 KHz ($$f_2$$). You know what Mars' radius is ($$r_M = 3389.5 km$$) and you are tasked with trying to find the length of its day ($$t_{Sol}$$). Assume that you are observing Mars in a way that makes your orbit and its orbit around the sun negligible, as well as Earth's own rotation.

Inactive
Hunter H.

First, we recognize that there is a doppler shift due to the planet's rotation and we get the difference in frequency via the following: $$f_2-f_0 = f_0-f_1 = \Delta f = 0.346$$. We also know that the basic formula for doppler shift is the following: $$\frac{v_0}{c} = \frac{\Delta f}{f}$$. We then solve for $$v_0$$ since we know that the speed of light (c) is 3E8 m/s. We get a value of 241.39 m/s for $$v_0$$. We also know Mars' radius ($$r_M = 3389.5 km$$), which when multiplied by $$2 \pi$$ gives us the circumference of Mars. We know that $$velocity = \frac{distance}{time}$$, so when we plug in $$v_0$$ for $$velocity$$ and $$circumference$$ for $$distance$$ we can solve for time. We end up getting a value of ~ 24.53 hours for time, which is within 20 minutes of Mars' real day, $$t_{Sol}$$.

### Subject:Astronomy

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Question:

You are an astronaut on the near side of the moon. From mission control in Houston, TX, USA on Earth, your ground team tells you that the moon is currently in its waxing crescent phase from their point of view. They want you to photograph the Earth when the Earth is in its waning crescent phase from your point of view. Approximately how long do you have to wait before photographing the Earth?

Inactive
Hunter H.

The first thing I would do is draw a lunar phase diagram that has the sun on the right side of the picture, the Earth in the center and the 8 lunar phases around the Earth. If you draw this correctly, the new moon should be directly at 3 o'clock (to the right of the Earth), the full moon being directly at 9 o'clock (to the left of the Earth), the first quarter moon at 12 o'clock (above the Earth), the third quarter moon at 6 o'clock (beneath the earth), the waxing crescent moon at 1:30 o'clock, the waxing gibbous at 10:30 o'clock, the waning gibbous at 7:30 o'clock and the waning crescent at 4:30 o'clock. You know that the moon is in its waxing crescent phase from Earth's point of view (the 1:30 o'clock position in the diagram), and the ground team wants a photograph of Earth when the Earth is in its waning crescent phase from your point of view, so you find where the waning crescent moon is in the diagram and go to the opposite side of the diagram from there since that phase is what the Earth looks like from your point of view on the moon (this would be the 10:30 o'clock phase). You see that you have moved 2 out of 8 phases in the counterclockwise direction from where you started, which is a quarter of the whole cycle and you know that the cycle is approximately one month long, so one quarter of approximately one month is approximately one week. The answer is that you have to wait approximately one week before photographing the Earth.

### Subject:Physics

TutorMe
Question:

Say you are at the coastline of a large ocean and you are at sea level with a radio that you intend to use to communicate with a friend who is on a very far away island in the ocean (beyond the horizon). You remember that the ionosphere's index of refraction ($$n_{ionosphere} = 1$$) is slightly less than air's ($$n_{air} = 1.0003$$. You know the ionosphere is 100 km in altitude, but you don't know exactly how far your friend's island is from you. What is the minimum ground distance ($$d$$) between you and your friend on the island who will be receiving your radio transmissions?

Inactive
Hunter H.

This is a two-step question. The first step is to recognize that because the ionosphere has a lower index of refraction than air ($$n_{ionosphere} < n_{air}$$), it can be used as a medium to bounce your radio signal off of. You use Snell's Law to find what the critical angle ($$\theta_{critical}$$) is between the two mediums: $$\theta_{critical} = arcsin(\frac{n_{ionosphere}}{n_{air}}) = 88.5967°$$. The second step is setting up a diagram that shows the triangle of the ground, the ionosphere, and the path the radio signal takes and then using the height of 100 km and the angle of 88.5967° to find the distance between you and the island. You can get one half of the length of the distance by the following: $$tan(88.5967°) = \frac{0.5 * d}{100}$$. You then solve for d and get d = 8164.35 km. You know this is the minimum distance between you and your friend because 88.5967° is the critical angle and any angle larger will result in a further distance between you two.

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