A geostationary satellite completes one circular orbit in exactly one sidereal day ($T = 86{,}164\text{ seconds}$). Earth's gravitational parameter is $\mu = 3.986 \times 10^{14}\text{ m}^3/\text{s}^2$ ($398{,}600\text{ km}^3/\text{s}^2$). The orbital radius $r_{\text{GEO}}$ (in $\text{km}$) of the geostationary orbit is closest to:

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### 1. Strategy & Theoretical Principles From Kepler's Third Law for a circular orbit of period $T$: $$T = 2\pi \sqrt{\frac{r^3}{\mu}} \implies r^3 = \mu \left(\frac{T}{2\pi}\right)^2$$ ### 2. Analytical Formulation & Calculation Given: - $\mu = 398{,}600\text{ km}^3/\text{s}^2$ - $T = 86{,}164\text{ s}$ $$\frac{T}{2\pi} = \frac{86{,}164}{2 \times 3.14159265} \approx 13{,}713.43\text{ s}$$ $$\left(\frac{T}{2\pi}\right)^2 = (13{,}713.43)^2 \approx 1.88058 \times 10^8\text{ s}^2$$ $$r^3 = 398{,}600 \times 1.88058 \times 10^8 \approx 7.4960 \times 10^{13}\text{ km}^3$$ Taking the cube root: $$r = (7.4960 \times 10^{13})^{1/3} \approx 42{,}164.17\text{ km} \approx 42{,}164\text{ km}$$ ### 3. Distractor Forensics - GEO altitude above surface is $h = r - R_E = 42{,}164 - 6{,}378 = 35{,}786\text{ km}$. - Candidates frequently confuse orbit radius ($42{,}164\text{ km}$) with altitude above surface ($35{,}786\text{ km}$). ### 4. Exam Trap & Standard Reference - **Trap Alert:** Always verify whether the question asks for *orbital radius* from Earth's center or *altitude* above the Earth's surface. - **Standard Reference:** Curtis, H.D., *Orbital Mechanics for Engineering Students*, Chapter 3.

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