Question

A synchronous Earth satellite is one which is placed in a west-to-east orbit over the Equator at such an. altitude that its period of revolution about Earth is 24 hours, the time for one rotation of Earth on its axis. Thus the orbital motion of the satellite is synchronized with Earth's rotation, and the satellite appears from Earth to remain stationary over a point on Earth's surface below. Such communication satellites as Syncom, Early Bird, Intelsat, and ATS are in synchronous orbits. Find the altitude for a synchronous Earth satellite.

Short answer

The altitude of a synchronous Earth satellite is approximately \(3.58 \times 10^7\) meters above the Earth's surface.

Step-by-step solution

Write down the known quantities

We have the following known quantities: 1. Period of revolution T = 24 hours 2. Earth's gravitational constant G = \(6.674 \times 10^{-11} Nm^{2}kg^{-2}\) 3. Mass of Earth M = \(5.972 \times 10^{24} kg\) 4. Earth's radius R = 6371 km

Convert the period of revolution to seconds

The period of revolution is given in hours, convert it into seconds. \(T = 24 \times 60 \times 60 = 86400\) seconds

Set up the equation for period and orbit radius

The equation relating the period (T), the gravitational constant (G), the mass of Earth (M), and the distance from the center of Earth to the satellite (r) is given by Kepler's Third Law: \(T^2 = \dfrac{4\pi^2 r^3}{GM}\) We need to solve for r.

Rearrange the equation and solve for r

Rearrange the equation to solve for r: \(r^3 = \dfrac{GM T^2}{4\pi^2}\) Substitute the known values for G, M, and T into the equation: \(r^3 = \dfrac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(86400)^2}{4\pi^2}\) Now, calculate r: \(r \approx 4.216 \times 10^7 m\)

Calculate the altitude of the satellite

To calculate the altitude (h) of the satellite, we need to subtract the Earth's radius from the distance r: \(h = r - R\) Convert Earth's radius R to meters: \(R = 6371 \times 10^3 m\) Substitute the values for r and R into the equation: \(h \approx (4.216 \times 10^7) - (6.371 \times 10^{6})\) Solve for h: \(h \approx 3.58 \times 10^7 m\) The altitude of a synchronous Earth satellite is approximately \(3.58 \times 10^7\) meters above the Earth's surface.

Key concepts

Kepler's Third Law

Johannes Kepler's contributions to astronomy are the foundation for understanding planetary motion. His Third Law, also known as the Law of Harmonies, beautifully connects the period of orbit to the radius of an object's orbit. In essence, it says that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit. Mathematically, it translates to the equation \(T^2 = \frac{4\pi^2 r^3}{GM}\), where G represents the gravitational constant and M is the mass of the more massive body, in this case, Earth.

For a synchronous Earth satellite, this law is pivotal in determining the correct altitude at which the satellite must orbit to remain geostationary. Because such a satellite must have an orbital period that matches Earth's rotational period, Kepler's Third Law allows us to calculate the exact distance from the center of Earth where this condition is fulfilled. It's a powerful example of how centuries-old principles are still being applied to modern technology.

Gravitational Constant

The gravitational constant, symbolized by G, is a key player in the universal law of gravitation and celestial mechanics. Its numerical value, \(6.674 \times 10^{-11} Nm^{2}kg^{-2}\), describes the strength of gravity between two objects. Sir Isaac Newton introduced us to this concept, which helps us understand that gravitational force is proportional to the masses of interacting bodies and inversely proportional to the square of the distance between their centers.

When calculating satellite orbits, G helps us predict the behavior of satellites in space. Its unchanging and universal value ensures that no matter where we are in the cosmos, we can rely on it as a constant of nature to guide our calculations and the design of satellite systems. Therefore, it is an indispensable part of satellite orbit calculations and understanding gravitational interactions on a cosmic scale.

Orbital Period

The orbital period of a satellite is the time it takes to complete one full orbit around a body, such as Earth. This period is especially critical for synchronous satellites that must match the rotation period of the planet. For Earth, a synchronous orbit means having an orbital period of exactly 24 hours, ensuring that the satellite stays over the same geographical point on the planet's surface.

A satellite's orbital period is tied to its speed and altitude. It's vital for satellite communication systems, as this synchronization allows for stable communication links. Understanding orbital periods is also essential for other types of orbits, like low Earth or polar orbits, which have different requirements and serve diverse purposes in satellite technology.

Satellite Orbit Calculations

Calculating a satellite orbit involves physics and mathematics to predict a satellite's path around a celestial body. It requires a blend of Newtonian mechanics and Kepler's Laws. When computing the altitude for a synchronous Earth satellite, one must account for the gravitational pull of Earth, ensuring that the satellite's velocity is neatly balanced to maintain a stable orbit.

As demonstrated in the provided exercise, the calculations begin by identifying known quantities, such as the gravitational constant (G), Earth's mass (M), and the desired orbital period (T). With the rearranged form of Kepler's Third Law, one can solve for the radius (r) and, by extension, the altitude (h) by subtracting Earth's radius. These calculations are not just theoretical; they are practical and vital for the deployment and maintenance of satellite systems on which global communication networks rely.