Answer:
T = 92.8 min
Explanation:
Given:
The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:
![A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})](https://tex.z-dn.net/?f=A%28t%29%20%3D%20415%20-%20sin%28%5Cfrac%7B2%2A%5Cpi%20%28t%2B23.2%29%7D%7B92.8%7D%29)
Find:
- How long does the International Space Station take to orbit the earth? Give an exact answer.
Solution:
- Using the the expression given we can extract the angular speed of the International Space Station orbit:
![A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )](https://tex.z-dn.net/?f=A%28t%29%20%3D%20415%20-%20sin%28%7B%5Cfrac%7B2%2A%5Cpi%2At%20%7D%7B92.8%7D%20%2B%20%5Cfrac%7B23.2%2A2%2A%5Cpi%20%7D%7B92.8%7D%20%29)
- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8
- We know that the relation between angular speed w and time period T of an orbit is related by:
T = 2*p / w
T = 2*p / (2*p / 92.8)
Hence, T = 92.8 min
Cell, tissue, organ, organ sytem, organism, population, community, ecosystem, biome.
Matter, substance. Material howya call it.
250 m. for a longer explanation or solution look at this article, i’m sorry.
https://www.quora.com/A-projectile-is-thrown-so-it-travels-a-maximum-range-of-1000m-How-high-will-it-rise