What is the orbital period of a satellite in a low-Earth orbit?

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The discussion focuses on deriving the orbital period of a satellite in a low-Earth orbit, emphasizing that the elevation above Earth's surface is much smaller than Earth's radius. The key equation used is P² = (4π²)/(GM) * a³, where the semi-major axis a is expressed as a = h + R⊕. By applying a Taylor series expansion, the orbital period P is approximated as P = C(1 + 3h/[2R⊕]). The simplification to the linear term is justified because the higher-order terms become negligible when h is much smaller than R⊕. This approach clarifies the relationship between orbital period and altitude for satellites in low-Earth orbit.
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Homework Statement


Consider a satellite in a circular, low-Earth orbit; that is, its
elevation above the Earth’s surface is h ≪ R⊕. Show that the orbital period P for such a satellite is approximately P=C(1+ 3h/[2R⊕]).


Homework Equations


P2 = (4pi2)/(GM) * a^3. (G - gravitational constant, M - mass of the Earth (in this case) and a = semi-major axis)


The Attempt at a Solution


Well, the semi-major axis will be: a = h + R⊕.
I've also picked up that a useful representation of a will be: a = R⊕(1 + h/R⊕)
This means our equation because P2 = (4pi2)/(GM) * (R⊕(1 + h/R⊕))^3.
Now we just want P, so P = (2pi/√GM) * (R⊕(1 + h/R⊕))^(3/2).

This obviously doesn't leave me with much. I've picked up from a few lectures that it may have something to do with Taylor series, but I'm severely stumped.
 
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Good guess. It is taylor series. (1+x)^(3/2)=1+3x/2+3x^2/8+... for x small. Truncate to the linear term. Does that help?
 
Oh yes, I do see. That is almost exactly what is required.
I don't suppose you'd know any reason for keeping it to the linear term?
Perhaps because h << R the other terms become negligible. I think that is reasonable!
Thank you very much.
That was much simpler than I thought.
 
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