Earth Satellites: Meaning with Key Principles and Types of Orbits

Earth satellites are objects that revolve around our planet. They are held in their path by the force of Earth's gravity. 

Some of the common Earth satellites range from the Moon (a natural one) to artificial ones that are used for communication, weather forecasting, navigation, and research. 

Considering the nature of the orbits of these Earth satellites, they are technically elliptical. But for simplifying calculations, we can often use them as circular orbits. This is one of the most essential sections of the Gravitation chapter. 

The next section will tell you the principles used in the world of Physics to define Earth satellites. 

The motion of a satellite is a delicate balance between its forward momentum and the constant pull of Earth's gravity. 

Tricks to approach the physics of Earth satellite orbits. 

• Be clear with the Law of Conservation of Momentum. 

You can tell that the satellite's initial forward momentum is imparted during launch by its rocket. That operates on the law of conservation of momentum by expelling propellant backward.

• Know about the role of centripetal force in circular motion. 

The core principle here is to understand that for a satellite in orbit, the gravity between Earth and the satellite creates the necessary centripetal force for its circular motion.

• Remember the signs for kinetic and potential energies. 

The convention of setting potential energy to zero at infinity can be counterintuitive, especially when you are used to positive potential energy values. In essence, mgh near Earth's surface. Just remember, a negative total energy signifies a bound system from which the satellite cannot escape.

With this in mind, let’s understand the following. 

Orbital Speed of an Earth Satellite: Why It Stays Aloft

For a satellite of mass $m$  in a circular orbit of radius $r$  around the Earth (mass $M$  ), the gravitational force provides the necessary centripetal force for circular motion.

The gravitational force is: $F=\frac{GMm}{r^2}$

The centripetal force required is:  $F_c=\frac{m{v}_o^2}{r}$

Equating these forces: $\frac{GMm}{r^2}=\frac{m{v}_o^2}{r}$

Then we cancel  $m$  (since $m\ne 0$ ) and simplify: $v_o^2=\frac{GM}{r}\Rightarrow v_o=\sqrt[]{\frac{GM}{r}}$

For a satellite close to Earth's surface ( $r\approx R$ , Earth's radius), and since $g=\frac{GM}{R^2}$ , we have $GM=g{R}^2$ . Thus: $v_o=\sqrt[]{\frac{g{R}^2}{R}}=\sqrt[]{gR}$

Using $g=9.8\mathrm{m}/{\mathrm{s}}^2$  and $R=6400\mathrm{k}\mathrm{m}=6.4\times {10}^6\mathrm{m}$  :

$v_o\approx \sqrt[]{9.8\times 6.4\times {10}^6}\approx 7.92\times {10}^3\mathrm{m}/\mathrm{s}\approx 7.9\mathrm{k}\mathrm{m}/\mathrm{s}$

Key Insights  

• Orbital speed decreases with increasing orbital radius $v_o\propto \frac{1}{\sqrt[]{r}}$  
• Orbital speed is independent of the satellite's mass.

Time Period of a Satellite: How Long it Takes Around the Orbit

The time period $T$ is the time taken for one complete orbit. For a circular orbit, the distance travelled is the circumference $2\pi r$ , and the speed is $v_o$ .

Thus: $T=\frac{\text{Distance}}{\text{Speed}}=\frac{2\pi r}{v_o}$

Substituting the expression for orbital speed
$v_o=\sqrt[]{\frac{GM}{r}}$ : $T=\frac{2\pi r}{\sqrt[]{\frac{GM}{r}}}=2\pi r\cdot \sqrt[]{\frac{r}{GM}}=2\pi \sqrt[]{\frac{r^3}{GM}}$

Since $GM=g{R}^2$  : $T=2\pi \sqrt[]{\frac{r^3}{g{R}^2}}$

For a satellite close to Earth's surface ( $r\approx R$ ): $T=2\pi \sqrt[]{\frac{R^3}{g{R}^2}}=2\pi \sqrt[]{\frac{R}{g}}$

Using $R=6.4\times {10}^6\mathrm{m},g=9.8\mathrm{m}/{\mathrm{s}}^2$ : $T\approx 2\pi \sqrt[]{\frac{6.4\times {10}^6}{9.8}}\approx 2\pi \sqrt[]{6.53\times {10}^5}\approx 5077\mathrm{s}\approx 84.6\mathrm{m}\mathrm{i}\mathrm{n}$

Key Insights

• This formula is a direct application of Kepler's Third Law of Planetary Motion. For elliptical orbits, the relationship shows that T² ∝ a³. That means the square of the time it takes to complete one orbit varies directly with the cube of the semi-major axis length.
• Satellites orbiting close to Earth (in low Earth orbit or LEO) typically complete one full orbit in approximately 85 to 90 minutes.

Energy of a Satellite: Considering a Bound System

A satellite's total mechanical energy in orbit consists of both its kinetic energy (K.E.) and gravitational potential energy (P.E.) added together.

Kinetic Energy

The kinetic energy $K$  is:

$K=\frac{1}{2}m{v}_o^2$

Substitute $v_o^2=\frac{GM}{r}$ : $K=\frac{1}{2}m\cdot \frac{GM}{r}=\frac{GMm}{2r}$

Potential Energy (Gravitational)

The gravitational potential energy $U$  is: $U=-\frac{GMm}{r}$

Total Energy

The total energy $E$ is: $E=K+U=\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r}$

Key Insights

• The negative total energy value demonstrates that the satellite is gravitationally bound to Earth and cannot break free without external energy input.
• For circular orbits, an important relationship exists where the kinetic energy is always half the magnitude of the potential energy (K.E. = -½ * P.E.).

 

Satellite missions usually require specific orbital configurations. Here we will talk about two types. Geostationary and polar. 

Characteristics	Geostationary (GEO)	Polar Orbits
Altitude/Path	~35,900 km above Earth's surface (orbital radius ~42,300 km) in the equatorial plane, moving west to east	Passes over/near Earth's poles, typically in low Earth orbit
Orbital Period	24 hours (matches Earth's rotational period of 86,400 seconds)	Varies (usually ~90 minutes for LEO)
Orbital Speed	~3.1 km/s	Varies with altitude (typically ~7.8 km/s for LEO)
Key Characteristics	Appears stationary relative to any point on Earth's surface due to synchronous rotation	Covers the entire planet surface, as the Earth rotates beneath the satellite
Calculation Method	Uses formula T = 2π√(r³/GM) where T = 24 hours, solving for orbital radius r	Period varies based on altitude, using the same orbital mechanics principles
Primary Applications	Communication satellites, weather monitoring, broadcasting	Earth observation, mapping, and polar research