Geostationary Orbit

There are many Orbit an artificial satellite may use. The orbit type is determined by what the satellite is used for. We will start with one of the simplest orbits, a Geostationary orbit.

A Geostationary Orbit is a type of orbit where a satellite appears to remain in the same location relative to an observer on earth. As the Earth rotates, the satellite moves along an orbit that keeps pace with the rotating earth. Both the earth and the satellite complete one rotation(or orbit) in 24 hours, but how? The question that we will answer is how high do we have to make our orbit in order to fulfill this condition?

Uses

The satellites that use this orbit are primarily Weather Predictions and Earth Observation or communication satellites. If you’ve ever seen a Dish or DIRECTV satellite antenna, they are pointing towards a geostationary communications satellite. The advantage of this orbit is that the satellite antennas on earth can be fixed and do not have to track a moving object. the GOES weather satellites, which take images of whole hemispheres of our planet, are also in a geostationary orbit.

Orbital Mechanics 101

If you’ve ever taken a physics class you know what this equation means, if not then let me introduce you to Newton’s Second Law of motion. It states that the Force acting on an object equals the time rate of change of it’s momentum. Momentum is simply the product of mass times Velocity and in this example our mass is not changing. Therefore we write the time rate of change of velocity as Acceleration \(a\). $$F=ma$$ There is another law that we need to learn in order to answer our question and that is Newton’s Law of Universal Gravitation. This law states that objects attract other objects with a force that is directly proportional to the product of the masses divided by the square of the distance between the two objects, multiplied by a gravitational constant. This relationship is also known as the Inverse Square Law $$F_g=G\frac{m_1m_2}{r^2}$$

Because we have a circular orbit, we can use the equations for uniform circular motion. The equation for centripetal force is: $$F_c=\frac{mv^2}{r}$$ The centripetal force points toward the center of the circle, and gravity pulls toward the center of the orbit so we can set these two equations equal to each other assuming that \(m_2\) is the mass of the satellite and \(m_1\) is the mass of the Earth and then solve for $$\frac{m_2v^2}{r}=G\frac{m_1m_2}{r^2}$$ We then simplify to solve for the orbital velocity \(v\). $$v^2=G\frac{m_2}{r}$$ Because our orbit is a circle we can find the orbital period by taking the distance around the orbit (the circumference of the circle) and dividing it by the orbital velocity. $$v=\frac{2\pi r}{P}$$ We then combine these two equations into a more useful form. $$G\frac{m_2}{r}=(\frac{2\pi r}{P})^2$$ $$G\frac{m_2}{r}=\frac{4\pi^2r^2}{P^2}$$ $$Gm_2=\frac{4\pi^2r^3}{P^2}$$

We have a special case here, \( G\) and \(m_2\) are fixed as long as the satellite remains in earths orbit. We combine these into a standard graviational parameter\(\mu\). We want to solve for the orbital radius to get us our final equation. $$r=\sqrt[3]{\frac{\mu P^2}{4\pi^2}}$$ Now we substitute our desired period into the equation \(P=86164s\) and \(\mu=3.986004418e+14\) and we get our final value for the orbit height of \(42164140.1m\). $$42164140.1m$$ To find the height above the earth’s surface we need to subtract the radius of the earth from this number to get our final orbital altitude.

$$ r – 6371000 = 35793140.1m$$ For our final altitude of \(35,793.14km\) above the surface of the earth. Finally the Geostationary orbit is aligned with the equator and therefore we say that has an inclination of zero degrees.

Conclusion

In this post we ran through some basic orbital mechanics and In future posts we’re going to look into more complicated orbits and why we use them.