Algebra, Physics and Satellites

Nearly every facet of modern life is touched by space, in particular, satellites orbiting the Earth. Communications, GPS, financial transactions and television are all dependant on satellites in orbit. But how do satellites get into orbit and how do they stay there? The answer lies in physics. This article will take a look at the physics behind orbiting satelllites and how humble algebra takes the physical laws and translate them into functional equations that allow us to find all sorts of answers.

So, what is an orbit? Well, it's simply a curved path a satellite takes around the Earth. Depending on how far the satellite is from the Earth, it moves faster or slower. At the heart of a satellite's orbit is the balance between two forces: centripetal and centrifugal forces. The centripetal force is the Earth’s gravity which pulls the satellite towards Earth, while the satellite's forward motion creates the centrifugal force that pulls the satellite away. When these forces balance perfectly, the satellite stays in a stable orbit (fig. 1).

To describe the motion of the satellite, we can use algebraic equations, we can use equations and Newton's laws of motion and universal gravitation.

Newton’s Law of Universal Gravitation states that every body in the known universe exerts a force of attraction on every other body, and can be described with the following equation:

Where:

• F is the gravitational force between Earth and the satellite.

• G is the gravitational constant, approximately 6.674×10−11 Nm2/kg.

• M is the mass of Earth, approximately 5.972×1024 kg.

• m is the mass of the satellite.

• r is the distance from the center of Earth to the satellite.

To calculate the centripetal force, we use the following relationship

Where:

• F is the centripetal force (which, in this case, equals the gravitational force).

• m is the mass of the satellite.

• v is the velocity of the satellite.

• r is the radius of the orbit (the distance from the center of Earth to the satellite).

Notice that these two relationships are simple inverse relationships between Force and radius, which implies that as radius increases, the force decreases. 

In this equation, n is an integer, and Force is inversely proportional to the radius between the objects. There are also direct relationships between Force and mass, which implies that the force increases or decrease with the mass of the object.

In this relationship, Force is directly proportional to mass. Now, to keep the satellite in orbit (prevent it from falling into Earth), the gravitational force and the centripetal force must be equal:

Simplifying the equality by cancelling like terms:

At this point, you may ask, what exactly keeps the satellite in space? Well, the mass stays the same regardless where it is in the universe, so radius from the Earth and the velocity must be contributing to keeping the satellite where it is. Since satellites are placed into different orbits based on what the satellite is used for, once the orbit is set, the radius is set. Thus, to keep the satellite in the set orbit, we calculate the velocity needed:

v is the velocity the satellite needs to stay in orbit at a distance r, from Earth. 

So, does this work? Let’s find out with a simple example. Let's say we want to find the orbital speed of a satellite at a height of 1000 km above Earth's surface. First, we need to calculate the total distance from the center of Earth. Using the Earth's radius as R and the satellite's height above the Earth as h, we substitute values R = 6378 km and h = 1000 km.

Now, using the equation for velocity and substituting known values:

• G = 6.674×10−11 Nm2/kg2

• M = 5.972×1024 kg

• r = 7.378×106 m

So, the satellite needs to travel at approximately 7.1 km/s to stay in an orbit 1000 km above Earth's surface. For reference, an F35 travels at 0.54 km/s (Mach 1.6). So, the satellite moves 13.15 times faster than an F35.

Another important aspect of satellite orbits is the orbital period; the time it takes for the satellite to complete one full revolution around Earth. This can be calculated using Kepler’s third law, which relates the orbital period T to the orbital radius r. The equation is:

Solving for T:

We can now calculate how long it will take for a satellite to complete one orbit around the Earth, depending on its distance from Earth. Note the relationship between Time and radius, and Time and Mass.

Thus, the period is directly proportional to the radius raised to the power of 1.5. Which gives you a quick way of understanding that the period increases as the radius increases.

On the other hand, the period is inversely proportional to the mass, thus as the mass gets bigger, the period gets smaller.

The idea of proportionality is fundamental and something we shall explore in later articles.

For fun, let’s try an example for the same satellite above that’s 1000 km above Earth, and using the same values for G, M, and r, we can plug them into the equation:

It would take the satellite 1 hour and 15 minutes to complete one complete orbit around Earth at 1000 km.

For reference, Low Earth Orbit (LEO) is 160 to 2000 km, with most satellites including the International Space Station (ISS), orbiting at 400 km. The ISS orbits in 1 hour 30 mins. Which makes sense; as we see from Kepler’s equation, period to radius are directly proportional.

So, what have we learned in this article? The study of algebra is key in space science. If you want to launch a satellite, Newton’s and Kepler’s laws apply, and algebra is the means of making these laws functional.