# 15.4 GEOSTATIONARY SATELLITES A geostationary satellite appears to remain stationary above a fixed point on Earth's surface. This occurs because its orbital period is exactly 24 hours, matching Earth's rotation period. The specific circular orbit at which a geostationary satellite revolves around Earth is called the geostationary orbit. This section explores the physics behind this special orbit, showing how to calculate the exact height and speed required for a satellite to remain "stationary" above a point on Earth. A **geostationary satellite** is a satellite that appears to remain stationary above a fixed point on Earth's surface. This occurs because its orbital period is exactly 24 hours, matching Earth's rotation period. The **geostationary orbit** is the specific circular orbit at which a geostationary satellite revolves around Earth. The **orbital velocity** ($v_0$) is the velocity required for a satellite to maintain a stable orbit. For a satellite of mass $m$ revolving around Earth of mass $M_E$ at a distance $r$ from the center, the orbital velocity is given by: $$v_{0}=\sqrt{\frac{G M_{E}}{r}}$$ It can also be expressed in terms of the time period $T$ and orbital radius $r$: $$v_{0}=\frac{2 \pi r}{T}$$ This derivation of orbital velocity is based on applying to circular satellite motion. The centripetal force required for circular motion is provided by the gravitational force. ## Derivation of Geostationary Orbital Radius By equating the two expressions for orbital velocity: $$\frac{2 \pi r}{T}=\sqrt{\frac{G M_{E}}{r}}$$ Squaring both sides and rearranging to solve for $r^3$: $$r^{3}=\frac{G M_{E} T^{2}}{4 \pi^{2}}$$ Taking the cube root gives the orbital radius: $$r=\left(\frac{G M_{E} T^{2}}{4 \pi^{2}}\right)^{1 / 3}$$ This equation is crucial for determining the radius of a geostationary orbit for any celestial body given its mass and the desired synchronous period. This same formula applies when analyzing in general. ## Calculations for Earth's Geostationary Orbit **Given constants:** - Gravitational constant ($G$) = $6.67 \times 10^{-11}\ \mathrm{N\ m^2\ kg^{-2}}$ - Mass of Earth ($M_E$) = $6 \times 10^{24}\ \mathrm{kg}$ - Period ($T$) = 24 hours = 86400 s - $\pi$ = 3.14 **Orbital radius ($r$):** $$r=\left(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times (86400)^{2}}{4 \times (3.14)^{2}}\right)^{1 / 3}$$ $$r = 4.23 \times 10^{7}\ \mathrm{m}\quad\text{or}\quad r = 4.23 \times 10^{4}\ \mathrm{km}$$ This is the distance from the center of the Earth. **Orbital speed ($v_0$):** $$v_{0}=\sqrt{6.67 \times 10^{-11} \times \frac{6 \times 10^{24}}{4.23 \times 10^{7}}} = 3.07 \times 10^{3}\ \mathrm{m\ s^{-1}}$$ ## Limitations of Synchronous Orbits A synchronous satellite requires its orbital radius to be within the celestial body's **Hill Sphere**. The Hill Sphere is a region around a celestial body where its gravitational pull is stronger than that of a larger, nearby body (for example, Earth's gravity dominates over the Sun's at certain distances, and the Moon's gravity is dominated by Earth's within a certain region). ### Lunar-stationary Satellite A lunar-stationary satellite would require a period equal to the Moon's rotation period of 27.3 days. The calculated orbital radius would be 88,417 km. However, the Moon's Hill Sphere extends only up to approximately 60,000 km. Since the required radius (88,417 km) is outside the Moon's Hill Sphere, Earth's gravity would destabilize the orbit, preventing a stable lunar-stationary satellite. ### Synchronous Satellite for Venus A synchronous satellite for Venus would require a period equal to Venus's rotation period of 243 days. The calculated orbital radius is $1.53 \times 10^{9}\ \mathrm{m}$ or 1,530,000 km. Due to Venus's lower mass and slow rotation, its Hill Sphere is significantly smaller than this calculated radius. At such a distance, the satellite would be outside Venus's Hill Sphere and primarily influenced by the Sun's gravity, leading to an unstable orbit. Thus, Venus cannot have a feasible synchronous satellite. ## Possible Questions and Answers **Q:** What is the orbital radius from the center of Earth for a geostationary satellite? **A:** $4.23 \times 10^{7}\ \mathrm{m}$ or 42,300 km. **Q:** What is the orbital speed of a geostationary satellite around Earth? **A:** $3.07 \times 10^{3}\ \mathrm{m\ s^{-1}}$. **Q:** Example 15.6: Venus has a mass of $4.867 \times 10^{24}\ \mathrm{kg}$ and a period of 243 days. Calculate the theoretical radius of a synchronous satellite for Venus. **A:** Given $M_V = 4.867 \times 10^{24}\ \mathrm{kg}$, $T = 243\ \text{days} = 2.093 \times 10^{7}\ \mathrm{s}$. Using $r = \left(\frac{G M_{v} T^{2}}{4 \pi^{2}}\right)^{1 / 3}$: $$r = \left(\frac{6.67 \times 10^{-11} \times 4.867 \times 10^{24} \times (2.093 \times 10^{7})^{2}}{4 \times (3.14)^{2}}\right)^{1 / 3}$$ $$r = 1.53 \times 10^{9}\ \mathrm{m} = 1.53 \times 10^{6}\ \mathrm{km}$$ **Q:** Explain why a synchronous satellite around Venus, despite mathematical calculation, is not feasible. **A:** The calculated orbital radius of $1.53 \times 10^{6}\ \mathrm{km}$ for a synchronous satellite around Venus is far beyond Venus's Hill Sphere. Due to Venus's lower mass and incredibly slow rotation, its gravitational influence is weak at such a distance, and the satellite would be primarily affected by the Sun's gravity, making its orbit unstable. **Q:** Assignment 15.5: Calculate the height (from the surface of Earth) of a satellite in geostationary orbit. **A:** The orbital radius from Earth's center is $r = 4.23 \times 10^{7}\ \mathrm{m}$. Assuming Earth's radius $R_E \approx 6.37 \times 10^{6}\ \mathrm{m}$ (or 6370 km), Height $h = r - R_E = 4.23 \times 10^{7}\ \mathrm{m} - 6.37 \times 10^{6}\ \mathrm{m} = 3.593 \times 10^{7}\ \mathrm{m}$ or approximately 35,930 km. - Geostationary satellites orbit Earth with a 24-hour period, appearing fixed above a point on the surface. - Their orbital velocity is derived from balancing gravitational and centripetal forces. - The orbital radius for a geostationary satellite is calculated using Earth's mass, the gravitational constant, and the 24-hour period. - For Earth, the geostationary orbital radius from the center is approximately 42,300 km (or 35,930 km above the surface), with an orbital speed of approximately $3.07 \times 10^{3}\ \mathrm{m\ s^{-1}}$. - The feasibility of a synchronous orbit depends on the celestial body's Hill Sphere, which defines the region where its gravity dominates. For the Moon and Venus, synchronous satellites are not stable due to interference from larger gravitational bodies like Earth or the Sun, as their calculated orbital radii fall outside their respective Hill Spheres. | Concept | Formula | Value (for Earth's Geostationary Orbit) | | :----------------- | :--------------------------------------------------------- | :--------------------------------------------------------------- | | Orbital Velocity | $v_{0} = \sqrt{\frac{G M_{E}}{r}}$ or $v_{0} = \frac{2\pi r}{T}$ | $3.07 \times 10^{3}\ \mathrm{m\ s^{-1}}$ | | Orbital Radius | $r = \left(\frac{G M_{E} T^{2}}{4 \pi^{2}}\right)^{1/3}$ | $4.23 \times 10^{7}\ \mathrm{m}$ (42,300 km from center) | | Gravitational Constant | $G$ | $6.67 \times 10^{-11}\ \mathrm{N\ m^{2}\ kg^{-2}}$ | | Earth's Mass | $M_E$ | $6 \times 10^{24}\ \mathrm{kg}$ | | Period | $T$ | 86400 s (24 hours) | Geostationary satellites are crucial for telecommunications, weather monitoring, and broadcasting, providing continuous coverage to large areas of Earth.