Geostationary Satellite

Derivation of Geostationary Orbital Radius

Conditions: Circular, equatorial, synchronous orbit around a spherically symmetric Earth.

• Equate the gravitational force providing the necessary centripetal force: \frac{GMm}{$r^2 = m\omega^2 r$
• Express angular velocity in terms of the time period T: \omega = \frac{2\pi}{T}
• Substitute \omega into the force equation: \frac{GM}{$r^2 = \left(\frac{2\pi}{T}\right)^2 r$
• Rearrange to solve for the orbital radius r: $r^3 = \frac{GMT^2}{4\pi^2}$

Result: $r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3}$

Derivation of Orbital Velocity for Geostationary Orbit

Conditions: Circular equatorial orbit at radius r.

• Start with the balance of gravitational and centripetal acceleration: \frac{v^2}{r} = \frac{GM}{r^2}
• Express orbital velocity v in terms of gravitational constant G, Earth mass M, and radius r: v = \sqrt{\frac{GM}{r}}
• Alternatively, relate v to the angular velocity \omega and radius r: v = \omega r
• Substitute numerical values for \omega and r to find the linear speed.

Result: v = \omega r = \sqrt{\frac{GM}{r}}

Effective Potential Energy vs. Orbital Radius

Draw: A plot of effective potential energy U_eff(r) showing a minimum corresponding to stable circular orbit at the geostationary radius.

Interpret: The minimum of the effective potential well denotes the radius where centripetal and gravitational forces balance, providing stable equilibrium for the geostationary orbit.

Exam use: Tested in PGT to explain orbital stability under central forces.

Ground Trace of Satellite (Inclined vs. Equatorial)

Draw: A world map projection showing a point for a true geostationary satellite ($i = 0\langle deg \rangle)$ and a figure-eight analemma for an inclined geosynchronous satellite (i > 0<deg>).

Interpret: Demonstrates why zero inclination is mandatory for a truly stationary position relative to the Earth's surface.

Exam use: Conceptual TGT/PGT multiple-choice question on ground track behavior.

Signal Coverage Footprint (Zenith Angle vs. Latitude)

Draw: A schematic diagram showing the Earth's curvature limiting line-of-sight communication beyond approximately 81<deg> latitude from a geostationary satellite.

Interpret: Explains polar coverage gaps for geostationary satellites, necessitating polar or inclined orbits for high-latitude communication.

Exam use: Applied physics application questions in communication satellite logistics.

Calculate the height of a geostationary satellite above the Earth's surface. Given: Radius of Earth $R_e = 6400 km$, Mass of Earth $M = 6 \times 10^{24} kg$, $G = 6.67 \times 10^{-11} N m^2 kg^{-2}$, and T = 24 hours.

• Time period T = 24 \times 3600 = 86400 s.
• Using formula $r^3 = \frac{GMT^2}{4\pi^2}$, substitute the values: $r^3 = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times (86400)^2}{4 \times (3.1416)^2}$.
• Calculate r &大约 4.22 \times 10^7 m = 42200 km.
• Height $h = r - R_e = 42200 km - 6400 km = 35800 km$.

Answer: 35800 km

Find the orbital velocity and the binding energy of a geostationary satellite of mass 500 kg. Take orbital radius $r = 4.2 \times 10^7 m$, $M = 6 \times 10^{24} kg$, and $G = 6.67 \times 10^{-11} N m^2 kg^{-2}$.

• Orbital velocity $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{4.2 \times 10^7}}$.
• $v = \sqrt{\frac{4.002 \times 10^{14}}{4.2 \times 10^7}} = \sqrt{9.528 \times 10^6} \approx 3086 m/s$.
• Binding energy $E_b = \frac{GMm}{2r} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 500}{2 \times 4.2 \times 10^7}$.
• $E_b = \frac{2.001 \times 10^{17}}{8.4 \times 10^7} \approx 2.38 \times 10^9 J$.

Answer: $v = 3086 m/s$, Binding Energ$y = 2.38 \times 10^9 J$

What is the minimum extra velocity (impulse delta-v) that must be imparted to a satellite in a circular low earth orbit (radius R_e) to transfer it to a geostationary transfer orbit whose apogee is at the geostationary radius $r_g = 7 R_e?$

• Low Earth Orbit (LEO) velocity $v_1 = \sqrt{\frac{GM}{R_e}}$.
• Semi-major axis of transfer ellipse $a = \frac{R_e + r_g}{2} = \frac{R_e + 7R_e}{2} = 4R_e$.
• Velocity at perigee of transfer ellipse $v_p = \sqrt{GM \left(\frac{2}{R_e} - \frac{1}{a}\right)} = \sqrt{GM \left(\frac{2}{R_e} - \frac{1}{4R_e}\right)} = \sqrt{\frac{7GM}{4R_e}} = \sqrt{\frac{7}{4}} v_1$.
• Delta-$v = v_p - v_1 = v_1 \left(\sqrt{\frac{7}{4}} - 1\right) = v_1 (1.3229 - 1) = 0.3229 v_1$.

Answer: 0.3229 \sqrt{\frac{GM}{R_e}}