Humans were curious enough to learn how to predict and accurately simulate the movement of the Earth around the Sun. Little did we know about the consequences of our curiosity, as years later we are using the same equations to put rockets and satellites in space. In this article, we will learn about the orbit equation and the different types of orbital trajectories.
Difference Between Trajectories and Orbits
An orbit is a type of trajectory, but a trajectory is not an orbit.
A trajectory is a path followed by a moving body.
An orbit is a trajectory that repeats periodically.
An orbit is also called an orbital trajectory. The word trajectory is used in relation to projectiles, while the word orbit is used in connection with celestial bodies or artificial satellites. For example, the path followed by a satellite around a planet is an orbit as it occurs repeatedly. The path followed by a launching rocket is a trajectory, as it only occurs once.
Orbital Trajectories: Meaning and Properties
To understand orbital trajectories, we need to consider a body moving towards a central body influenced by a gravitational force, or in other words a two-body system. To make the math simpler and reduce the problem to a one-body problem, we study the orbital trajectory in the Center of Mass frame of reference. We look at the orbital trajectory of the reduced mass \(\mu\) about the origin, where we locate the total mass \(M\) at rest:
\[\begin{align*}M&=m_1+m_2,\\ \mu&=\frac{m_1m_2}{m_1+m_2},\\\mu&=\frac{m_1m_2}M.\end{align*}\]
In the equation above, \(m_1\) is the mass of the central body in kilograms, \(m_2\) is the mass of the orbiting body in kilograms, \(M\) is the total mass in kilograms, and \(\mu\) is the reduced mass in kilograms.
Fig. 1 - Visualization of the reduced mass. A two-body problem is turned into a one-body problem by using the Center of Mass frame and the reduced mass.
Now that we know this, we can express the orbit equation in polar coordinates, which reduces the problem to two variables.
The polar coordinate system is a two-dimensional coordinate system in which a point in the plane can be located by measuring the distance from a reference point and an angle of rotation around the origin.
Orbital Trajectory: Equation
The orbit equation is now given by
$$r=\frac{l^2}{{\mu}Gm_1m_2}\frac1{1-e\cos\left(\theta\right)},$$
where \(G\) is the gravitational constant \(6.67\times10^{-11}\;\frac{\mathrm N\;\mathrm m^2}{\mathrm{kg}^2}\), \(r\) is the distance between the bodies in \(\mathrm m\), \(\theta\) is the angle between \(r\) and the longest axis of the orbit in degrees or radians, \(^\circ\) or \(\mathrm{rad}\), \(e\) is the eccentricity of the orbital trajectory, \(l\) is the angular momentum in \(\frac{\mathrm{kg}\;\mathrm m^2}{\mathrm s}\), \(m\) is the mass of the orbiting body in \(\mathrm{kg}\), and \(\mu\) is the reduced mass in \(\mathrm{kg}\). This equation has been used experimentally to launch satellites and rockets in orbit.
The periapsis is located in the horizontal axis of the polar coordinate system, and it represents the \(r_\max\), the longest distance between the orbiting body and the central body.
Angular momentum is a property of mass that is rotating along a fixed axis. It is the rotational version of linear momentum.
Fig. 2 - Representation of our reduced mass system. This is an elliptical orbit using the reduced mass. The maximum and minimum distances from the central body are located on the horizontal axis.
To derive the orbit equation we must first begin by describing the energy of the reduced mass system orbiting the system's center of mass. We know that the kinetic energy and the gravitational potential of the system, so we can express the total energy:
$$E=\frac{1}{2}\mu v^2-\frac{Gm_1m_2}{r},$$
where \mu is the reduced mass in \(\text{kg}\) and \(v\) is the relative velocity between the two bodies.
The velocity can be defined in polar coordinates as,
\begin{align*}\vec{v}&=v_\text{r} \hat{r} +v_\theta \hat{\theta},\\v&=|\vec{v}|,\\v&=|\frac{\text{d}\vec{r}}{\text{d}t}|,\end{align*}
where \(v_\text{r}=\frac{\text{d}r}{\text{d}t}\) and \(v_\theta=r\left(\frac{\text{d}\theta}{\text{d}t}\right)\).
The energy of the system is now:
$$E=\frac{1}{2}\mu\left[\left(\frac{\text{d}r}{\text{d}t}\right)^2+\left(r\frac{\text{d}\theta}{\text{d}t}\right)^2\right]-\frac{Gm_1m_2}{r}.$$
The angular momentum is given by
$$\begin{align*}\vec{l}&=\vec{r}\times\mu\vec{v},\\\vec{l}&=r \hat{r}\times\mu\left(v_\text{r} \hat{r}+v_\theta \hat{\theta}\right),\\\vec{l}&=r\mu v_\theta \hat{k},\\\vec{l}&=\mu r^2\frac{\text{d}\theta}{\text{d}t} \hat{k},\\l&=\mu r^2\frac{\text{d}\theta}{\text{d}t},\\\frac{\text{d}\theta}{\text{d}t}&=\frac{l}{\mu r^2}.\end{align*}$$
Now we rewrite the total energy of the system,
$$E=\frac{1}{2}\mu\left(\frac{\text{d}r}{\text{d}t}\right)^2+\frac{1}{2}\frac{l^2}{\mu r^2}-\frac{Gm_1m_2}{r}.$$
We can write it in terms of \(\frac{\text{d}r}{\text{d}t}\):
$$\frac{\text{d}r}{\text{d}t}=\sqrt{\frac{2}{\mu}}\left(E-\frac{1}{2}\frac{l^2}{\mu r^2}+\frac{Gm_1m_2}{r}\right)^{\frac{1}{2}}$$
Now we divide \(\frac{\text{d}\theta}{\text{d}t}\) by \(\frac{\text{d}r}{\text{d}t}\) to obtain \(\frac{\text{d}\theta}{\text{d}r}\), an expression that relates the distance \(r\) to the angle \(\theta\). It gives the orbit equation in differential form:
\begin{align*}\frac{\text{d}\theta}{\text{d}r}&=\frac{\frac{\text{d}\theta}{\text{d}t}}{\frac{\text{d}r}{\text{d}t}},\\\frac{\text{d}\theta}{\text{d}r}&=\frac{l}{\sqrt{2\mu}}\frac{\left(\frac{1}{r^2}\right)}{\left(E-\frac{l^2}{2\mu r^2}+\frac{Gm_1m_2}{r}\right)^{\frac{1}{2}}},\\\text{d}\theta&=\frac{l}{\sqrt{2\mu}}\frac{\left(\frac{1}{r^2}\right)}{\left(E-\frac{l^2}{2\mu r^2}+\frac{Gm_1m_2}{r}\right)^{\frac{1}{2}}}\text{d}r.\end{align*}
Before we do any integration, we need to do some substitutions. We do the substitution \(u=\frac{1}{r}\), \(\text{d}u=-\left(\frac{1}{r^2}\right)\text{d}r\,\):
$$\text{d}\theta=-\frac{l}{\sqrt{2\mu}}\frac{\text{d}u}{\left(E-\frac{l^2}{2\mu}u^2+Gm_1m_2u\right)^{\frac{1}{2}}}.$$
Next, we multiply and divide the right hand side of the equation by \(\frac{\sqrt{2\mu}}{l}\):
$$\begin{align*}\text{d}\theta&=-\frac{\text{d}u}{\left(\frac{2\mu E}{l^2}-u^2+2\left(\frac{\mu Gm_1m_2}{l^2}\right)u\right)^{\frac{1}{2}}},\\\text{d}\theta&=-\frac{\text{d}u}{\left(\frac{2\mu E}{l^2}-u^2+\frac{2u}{r_0}\right)^{\frac{1}{2}}},\end{align*}$$
where we have defined \(r_0=\frac{l^2}{\mu Gm_1m_2}\).
Now we add and subtract \(\frac{1}{{r_0}^2}\) inside the parenthesis with the square root:
\begin{align*}\text{d}\theta&=-\frac{\text{d}u}{\left(\frac{2\mu E}{l^2}+\frac{1}{{r_0}^2}-u^2+\frac{2u}{r_0}-\frac{1}{{r_0}^2}\right)^{\frac{1}{2}}},\\\text{d}\theta&=-\frac{\text{d}u}{\left(\frac{2\mu E}{l^2}+\frac{1}{{r_0}^2}-\left(u-\frac{1}{r_0}\right)^2\right)^{\frac{1}{2}}},\\\text{d}\theta&=-\frac{r_0\text{d}u}{\left(\frac{2\mu E{r_0}^2}{l^2}+1-\left(r_0 u-1\right)^{2}\right)^{\frac{1}{2}}},\end{align*}
where we define the eccentricity as \(\epsilon=\sqrt{\frac{2\mu E{r_0}^2}{l^2}+1}\). This dimensionless quantity is responsible for the shape of the orbit. This is better discussed in the article Orbital Trajectories.
We rewrite our equation in terms of the eccentricity:
$$\text{d}\theta=-\frac{r_0\text{d}u}{\left({\epsilon}^2-\left(r_0 u-1\right)^{2}\right)^{\frac{1}{2}}}.$$
Finally, we do the last substitution before solving the integral, \(r_0u-1=\epsilon \cos{\alpha}\) and \(r_0 \text{d}u=-\epsilon\sin{\alpha}\text{d}\alpha\):
\begin{align*}\text{d}\theta&=-\frac{-\epsilon\sin{\alpha}\text{d}\alpha}{\left({\epsilon}^2-{\epsilon}^2{\cos^2{\alpha}}\right)^{\frac{1}{2}}},\\\text{d}\theta&=-\frac{-\bcancel{\epsilon}\sin{\alpha}\text{d}\alpha}{\bcancel{\epsilon}\left(1-\cos^2{\alpha}\right)^{\frac{1}{2}}},\\\text{d}\theta&=\frac{\sin{\alpha}\text{d}\alpha}{\left(sin^2{\alpha}\right)^{\frac{1}{2}}},\\\text{d}\theta&=\frac{\bcancel{\sin{\alpha}}\text{d}\alpha}{\bcancel{sin{\alpha}}}\,\\\theta&=\int{\text{d}\alpha},\\\theta&=\alpha + \text{constant}.\end{align*}
We already know that \(r=\frac{1}{u}\). If we choose the constant to be zero, we find that:
\begin{align*}r_0u-1&=\epsilon\cos{\alpha},\\r_0u-1&=\epsilon\cos{\theta},\\u&=\frac{1-\epsilon\cos{\theta}}{r_0}.\end{align*}
Our final expression for the orbit equation is:
\begin{align*}r&=\frac{1}{u},\\r&=\frac{r_0}{1+\epsilon\cos{\theta}}.\end{align*}
Alternately, if we choose the constant to be \(\pi\) we get the same orbit equation but with the vertical axis reflected,
$$r=\frac{r_0}{1-\epsilon\cos{\theta}}.$$
We can immediately determine \(r_\min\) and \(r_\max\), the minimum and maximum distances between the orbiting body and the central body. To get \(r_\max\), \(\theta\) must be \(0^\circ\). To get \(r_\min\), \(\theta\) must be \(180^\circ\). If we substitute these values in the above orbit equation for \(\theta\), we get the equations for \(r_\min\) and \(r_\max\):
\[\begin{align*}r_\max&=\frac{l^2}{{\mu}Gm_1m_2}\frac1{1-e\cos\left(0^\circ\right)}=\frac{l^2}{{\mu}Gm_1m_2}\frac1{1-e},\\ r_\min&=\frac{l^2}{{\mu}Gm_1m_2}\frac1{1-e\cos\left(180^\circ\right)}=\frac{l^2}{{\mu}Gm_1m_2}\frac1{1+e}.\end{align*}\]
As we can see from the figure below, the relation between \(r\) and \(\theta\) describes a conic section.
Fig. 3 - Visualization of different conic sections.
Diagram for Orbital Trajectories
The eccentricity \(e\) of the orbit is the parameter that will determine the shape of the orbital trajectory and it is given by
$$e=\sqrt{1+\frac{2El^2}{\mu\left(Gm_1m_2\right)^2}},$$
where we introduce the parameter \(E\), which is the total energy of the system in joules, \(\mathrm J\).
Fig. 4 - Different types of orbital trajectories depend on the eccentricity: in blue we have a circular orbit, in green we have an elliptical orbit, in red we have a parabolic orbit, and in purple we have a hyperbolic orbit.
Estimation of the Shape of the Orbital Trajectory
There are 4 possible cases that represent different types of orbital trajectories.
\(E=0\)
If \(E=0\) then \(e=1\), which means that this will not be an orbit, but a trajectory in the form of a parabola.
So you can understand why this case is a trajectory and not an orbit, we can solve the expression for \(r_\max\) with \(e=1\):
$$r_\max=\frac{l^2}{{\mu}Gm_1m_2}\frac1{1-e}=\frac{l^2}{{\mu}Gm_1m_2}\frac1{1-1}=\frac{l^2}{{\mu}Gm_1m_2}\frac10.$$
We see that te maximum distance is undefined, so it does not exist. That means that this cannot be an orbit because for orbits we always have a maximum distance.
\(E>0\)
If \(E>0\) then \(e>1\). This corresponds to a trajectory in the form of a hyperbola. This is also not an orbit.
\(E<0\)
If \(E<0\) then \(0< e<1\). This corresponds to an orbit in the form of an ellipse. Earth's orbit around the Sun is an ellipse with an eccentricity of \(0.017\). As the distance and velocities of this orbit are different at various moments the total energy of the system and the angular momentum are conserved, but the gravitational potential and kinetic energy are not conserved.
\(E=-\frac{\mu\left(Gm_1m_2\right)^2}{2l^2}\)
If \(E=-\frac{\mu\left(Gm_1m_2\right)^2}{2l^2}\) then \(e=0\). This corresponds to an orbit in the form of a circle. As the distance and velocities of this orbit are the same at any moment the total energy of the system, the angular momentum, the gravitational potential, and kinetic energy are conserved.
If we solve the orbit equation for \(e=0\), we find that the distance \(r\) from the central body is constant, so we have a circular orbit:
$$r=\frac{l^2}{{\mu}Gm_1m_2}\frac1{1-e\cos\left(\theta\right)}=\frac{l^2}{{\mu}Gm_1m_2}\frac1{1-0}=\frac{l^2}{{\mu}Gm_1m_2}.$$
Calculation of the Motion of a Satellite in an Orbital Trajectory: Circular Orbit
For a circular orbit, we know the only force acting is the gravitational force. Because we know it is circular motion we can substitute the acceleration term with the centripetal force term as follows:
$$\frac{m_2v^2}r=\frac{Gm_1m_2}{r^2}$$
$$m_2v^2=\frac{Gm_1m_2}r$$
If we multiply both sides of the equations by \(\frac12\), we can express this equation in terms of the Kinetic and Potential energies, and define the total energy of the system for a circular orbit. We know that the gravitational potential energy for a circular orbit can be expressed as
$$U=-\frac{Gm_1m_2}r,$$
so we can express the kinetic energy and the total energy of the system as follows:
\[\begin{align*}K&=-\frac{U}{2},\\ E&=K+U,\\ E&=-\frac{U}{2}+U,\\ E&=\frac{U}{2},\\ E&=-\frac{Gm_1m_2}{2r}.\end{align*}\]
Q: A satellite is orbiting the Earth with mass \(m_\oplus\). Assume the satellite has mass \(m\) and that the type of orbit is circular with an orbital distance \(r\). What is the total mechanical energy for this system?
A: The total mechanical energy is
$$E=-\frac{Gmm_\oplus}{2r}.$$
The Trajectory of the Earth
The only orbital trajectories that correspond to bound orbits are the last two cases. So, planets can only have either circular or elliptical orbits around the Sun, confirming Kepler's First Law. All planets' orbits have different eccentricities. For example, Venus has an almost circular orbit and Mercury has the most elliptical orbit of all planetary orbits in the Solar System. The Earth's orbit has an eccentricity of \(0.017\). One complete orbit of the Earth around the Sun takes 365.256 days and a distance of 940 million kilometers.
Planet | Eccentricity |
Mercury | \(0.206\) |
Venus | \(0.007\) |
Earth | \(0.017\) |
Mars | \(0.094\) |
Jupiter | \(0.049\) |
Saturn | \(0.052\) |
Neptune | \(0.010\) |
We need two coordinates to track the position of an orbiting satellite in the sky, the azimuth and elevation angles. The azimuth angle is an angle of \(360^\circ\) degrees of rotation horizontally, while the elevation angle is an angle of \(90^\circ\) degrees of rotation vertically. These angles are measured from the antenna on Earth. We present a diagram below to help you visualize these angles.
Fig. 5 - Diagram to visualize the azimuth and elevation angles and be able to track the position of a satellite in orbit.
The equation to calculate the azimuth and elevation angles varies with the type of orbit. Here we will discuss the equation for satellites in a geostationary orbit. If we want the more general equation, it gets a little bit more complicated as we have to add some factors like the satellite's height above sea level and a time-dependent equation, as not every orbit has the same period as the Earth.
We can express the equation for the azimuth and elevation angles of geostationary satellites as follows:
$$A=180^\circ+\;\tan^{-1}\left(\frac{\tan\left(S-N\right)}{\tan\left(L\right)}\right)$$
$$E=\tan^{-1}\left(\frac{\cos\left(S-N\right)\cos\left(L\right)-0.1512}{\sqrt{1-\cos^2\left(S-N\right)\cos^2\left(L\right)}}\right),$$
where \(S\) is the satellite's longitude in degrees, \({}^\circ\), \(N\) is the longitude of the antenna site in degrees, \({}^\circ\), and \(L\) is the latitude of the antenna site in degrees, \({}^\circ\).
Orbital Trajectory - Key takeaways
- A trajectory is a path followed by a moving body.
- An orbit is a trajectory that repeats periodically.
- An orbit is a type of trajectory, but a trajectory is not an orbit.
- The polar coordinate system is a two-dimensional coordinate system \((r,\theta)\) in which a point in the plane can be located by measuring the distance from a reference point and an angle of rotation around the origin.
- The relation between \(r\) and \(\theta\) describes a conic section.
- The eccentricity of the orbit is the parameter that will determine the shape of the orbital trajectory.
- \(e=1\), means that this will not be an orbit, but a trajectory in the form of a parabola.
- \(e>1\), corresponds to a trajectory in the form of a hyperbola.
- \(0<e<1\), corresponds to an orbit in the form of an ellipse.
- \(e=0\), corresponds to an orbit in the form of a circle.
- Planets can only have either circular or elliptical orbits around the Sun.
- We know that gravitational potential energy for a circular orbit can be expressed as,\(U=-\frac{Gm_1m_2}r\).
- So for a circular orbit, we can express the kinetic energy and the total energy of the system as,\(K=-\frac U2\) and \(E=\frac U2\).
References
- Fig. 1 - Visualization of the reduced mass. A two-body problem is turned into a one-body problem by using the Center of Mass frame and the reduced mass, StudySmarter Originals
- Fig. 2 - Representation of our reduced mass system. This is an elliptical orbit using the reduced mass. The maximum and minimum distances from the central body are located in the horizontal axis, StudySmarter Originals
- Fig. 3 - Visualization of different conic sections, StudySmarter Originals
- Fig. 4 - Different types of orbital trajectories depend on the eccentricity: in blue we have a circular orbit, in green we have an elliptical orbit, in red we have a parabolic orbit, and in purple we have a hyperbolic orbit, StudySmarter Originals
- Fig. 5 - Diagram to visualize the azimuth and elevation angles and be able to track the position of a satellite in orbit, StudySmarter Originals
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