Geostationary Earth Orbit Satellite

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Part 1: Physics derivation of GEO orbit radius

Step 1: Equate gravitational force and centripetal force:

$$m\frac{v^{2}}{r}=G\frac{mM_{e}}{r^{2}}$$

where:

• $m$ = satellite mass (will cancel out),

• $v$ = orbital speed,

• $r$ = distance from Earth's center,

• $G=6.67430\times 10^{-11}(m^{3}.kg^{-1}.s^{-2})$ = gravitational constant,

• $M_{e}\simeq 5.972\times 10^{24}$(kg)= mass of Earth

Step 2: Solve for $v$:

$$v^{2}=\frac{GM_{e}}{r}$$

Step 3: Use orbital period formula.

For circular orbit, period $T$ is:

$$v=r\omega=r\frac{2\pi}{T}\Rightarrow T=\frac{2r\pi}{v}$$

We want geostationary orbit, so:

$$T=24×60×60=86400\;\;seconds$$

Step 4: Substitute $v$:

$$v=\frac{2r\pi}{T}$$

Now, from Step 2:

$$\left( \frac{2r\pi}{T} \right)^{2}=\frac{GM_{e}}{r}$$

Step 5: Solve for $r$

$$r^{3}=\frac{G.M_{e}.T^{2}}{4\pi^{2}}\Rightarrow r≈42,164 km$$

Result: The geostationary orbit radius ≈ 42164 km from Earth’s center.

Step 6: Altitude above Earth’s surface

Earth’s mean radius ≈ 6378 km.

$$Altitude=42164−6378=35786 km$$

Therefore, GEO altitude ≈ 35786 km.

Part 2: Communication delay calculation

Now that we proved the altitude, let’s do the signal delay:

Step 1: Distance from surface ≈ 35786 km.

$$d=35786,000 m$$

Step 2: Speed of light

$$c≈299792458 m/s$$

Step 3: Time = distance / speed

$$t=\frac{d}{c}=\frac{35786000}{299792458}≈0.1194 seconds$$

• One-way delay ≈ 119 ms.

• Round-trip delay ≈ 238 ms.