Orbit equations in a plane

We are now in a position to find the most basic equations to do calculations on satellite orbits in the plane.

Figure 1: Circular satellite orbit. Illustrated by Andøya Space Education

First, let us imagine a satellite which is in a circular orbit around Earth, as seen in figure 1. There is only one force acting on the satellite, which is the gravitational force of the Earth. The force is given as

F_\text{G} = G \, \dfrac{M\, m}{r^2},

where M is the mass of the Earth, m is the mass of the satellite and G is the gravitational constant. We know that the centripetal force F_\text{C} equals the force of gravity F_\text{G}:

F_\text{C} = F_\text{G} \quad \Longleftrightarrow \quad m \, \dfrac{v^2}{r} = G \, \dfrac{M \, m}{r^2}.

Simplifying and solving for the velocity gives

v = \sqrt{\dfrac{\mu}{r}} = \sqrt{\dfrac{\mu}{R_\text{E} + h}},

where \mu = G M is called the gravitational parameter. For Earth it becomes \mu_\text{E} \approx 398.6 \times 10^{12} \,\text{ m}^3/\text{s}^2 = 398 \, 600 \,\text{km}^3/\text{s}^2. Here, R_\text{E} \approx 6371 \, \text{km} is the Earth mean radius and h is the altitude of the satellite above Earth’s (mean) surface. Then, the velocity above is that which a satellite must have to maintain a circular velocity around Earth at a given radius or altitude.

We know that distance equals mean velocity times the time spent moving that distance, or d = v \, t. Using the circumference of a circular orbit d = 2\pi r and orbital time period T, we have

2\pi r = v \, T = \sqrt{\dfrac{\mu}{r}} \ T.

Solving for time, we get

T = 2\pi \sqrt{\dfrac{r^3}{\mu}} = 2\pi \sqrt{\dfrac{(R_\text{E} + h)^3}{\mu}}.

For a general ellipse this expression is very similar:

T = 2\pi \sqrt{\dfrac{a^3}{\mu}},

where a is the semi-major axis.

As long as the satellite is not firing its motors, its mechanical energy will always be conserved (in a Keplerian orbit; relativistic effects could cause non-conservation). The mechanical energy is given by

\dfrac{1}{2}\, mv^2-m\,\dfrac{\mu}{r} = -m\, \dfrac{\mu}{2a} .

Note that the potential energy -m\mu /2a is a constant for a general gravitational field, given only by the semi-major axis for the orbit. This is valid for all ellipses. Usually we rewrite the equation to

v^2-2\,\dfrac{\mu}{r} = -\dfrac{\mu}{a}.

Enter the ideal rocket equation \Delta v = v_\text{e} \ln \left( m_\text{i}/m_\text{f} \right). The assumption for the equation is that the satellite burns its engine for a short time, which is usually true for a satellite with a chemical rocket engine. When planning a satellite orbit, it is important to have a so-called ‘delta-v budget’, which states exactly how much fuel is needed for the satellite to reach a given orbit and to stay there. As several \Delta v can be added together, the total mass needed for the total \Delta v_\text{TOT} can be easily calculated. We will come back to this shortly.

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This article is part of a pre-course program used by Andøya Space Education in Fly a Rocket! and similar programs.