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Wewaii [24]
3 years ago
7

HELP PLEASE A satellite rotates 50 miles above Earth's atmosphere. An astronaut works on the satellite and sees the sun rise ove

r Earth. To the nearest mile, what is the distance from the astronaut to the horizon? (HINT: Earth's radius is about 4,000 miles.) PLEASE HELP ME PLEASE AND THANK YOU, SHOW ALL WORK.
Mathematics
1 answer:
dusya [7]3 years ago
8 0
I must warn you I am not certain about this answer but this is the best I can come up with. Let's assume that the astronaut is 50 miles above the surface of the earth, due east of the centre of the Earth. He will therefore see the satellite when it is 50 miles above the surface of the earth, due north of the centre of the Earth. You can draw a triangle ABC where AB is the distance from the earth's centre to the astronaut, AC is the distance from the centre to the horizon, and BC is the distance from the astronaut to the horizon. The radius of the earth is 4000 miles, so the distance from the centre to the astronaut is 4050miles. This is the same as the distance from the centre to the horizon. This means AB and AC are 4050 miles. You can then use pythagoras' theorem to find out the length from the astronaut to the horizon:
<span>√(4050^2+4050^2)=5727.56
</span>So the distance is 5728 miles from the astronaut to the horizon.
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Diameter of semicircle is equal to 2 km.

<h3>Explanation:</h3>

Given: Perimeter of semicircle = P = 5.14 km

d=?

Perimeter of semicircle is given by

P = (\pi \times r ) + d\\5.14 = (\pi \times r ) + d\\\\Since \ d=2r; \ r=\frac{d}{2} \\5.14 = (\pi \times \frac{d}{2}) + d\\

Simplifying the above equation

5.14 = ( \frac{(\pi \times d) + 2d}{2}) \\\\5.14 \times 2 = d(\pi +2)\\\\d=\frac{5.14 \times 2}{\pi +2 }\\

As value of pi = 3.14, equation becomes

d=\frac{5.14 \times 2}{\ 3.14 +2} = \frac{10.28}{5.14} =2 \ km

Diameter of semicircle is equal to 2 km.

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