Modeling Satellite Orbit with Sinusoidal Functions

AI Thread Summary
The discussion focuses on modeling a satellite's orbit using a sinusoidal function, with the satellite reaching 4500 km north and south of the equator every 15 minutes, completing an orbit in 2 hours. Participants suggest using the general form of a sinusoidal function, l(t) = A*sin(xt + a), where A represents the amplitude of 4500 km. The period of the function is calculated to be 8 radians, derived from the satellite's orbit duration. To find the satellite's distance from the equator one hour after launch, substituting 60 minutes into the function is recommended. The conversation emphasizes understanding the sinusoidal model to solve the homework problem effectively.
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Homework Statement


A satellite is deployed from a space shuttle into an orbit which goes alternately north and south
of the equator. Its distance from the equator over time can be approximated by a sine wave. It
reaches 4500 km, its farthest point north of the equator, 15 minutes after the launch. Half an
orbit later it is 4500 km south of the equator, its farthest point south. One complete orbit takes
2 hours.

a. Find an equation of a sinusoidal function that models the distance of the satellite from
the equator.
b. How far away from the equator is the satellite 1 hour after launch?
 
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If anyone has any clue to this question, please help, because I really need it soon.
 
You should start off with what you have:
sinusoidal function means a function off this sort: A*sin(xt+a), where a and A are constants.
The info gives you a couple of equations

<br /> (i)\ l(t)=Asin(xt+a)\leq4500km<br />
<br /> (ii)\ l(t)=l(t+2h)<br />
<br /> (iii)\ l(15min)=Asin(x*15+a)=4500km<br />
 
Assume the equator is 0, and the max/min is 4500. This is your amplitude. You can find your period by using period=2pi/k, which would equate to:

period= 360/45
period=8 (I believe your working in radians?)

I might be wrong about the period, but it's pretty easy to go on from there. To do question #2, just substitute 60 into time.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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