Satellite orbits can be any of the four conic sections. This page deals mostly with elliptical orbits, though we conclude with an examination of the hyperbolic orbit.

Orbital Elements

Figure 4.3

Figure 4.2

To mathematically describe an orbit one must define six quantities, called orbital elements. They are

• Semi-Major Axis, a
• Eccentricity, e
• Inclination, i
• Argument of Periapsis, $\omega$
• Time of Periapsis Passage, T
• Longitude of Ascending Node, $\Omega$

An orbiting satellite follows an oval shaped path known as an ellipse with the body being orbited, called the primary, located at one of two points called foci. An ellipse is defined to be a curve with the following property: for each point on an ellipse, the sum of its distances from two fixed points, called foci, is constant (see Figure 4.2). The longest and shortest lines that can be drawn through the center of an ellipse are called the major axis and minor axis, respectively.

The semi-major axis is one-half of the major axis and represents a satellite's mean distance from its primary.

Eccentricity is the distance between the foci divided by the length of the major axis and is a number between zero and one. An eccentricity of zero indicates a circle.

Inclination is the angular distance between a satellite's orbital plane and the equator of its primary (or the ecliptic plane in the case of heliocentric, or sun centered, orbits). An inclination of zero degrees indicates an orbit about the primary's equator in the same direction as the primary's rotation, a direction called prograde (or direct). An inclination of 90 degrees indicates a polar orbit. An inclination of 180 degrees indicates a retrograde equatorial orbit. A retrograde orbit is one in which a satellite moves in a direction opposite to the rotation of its primary.

Periapsis is the point in an orbit closest to the primary. The opposite of periapsis, the farthest point in an orbit, is called apoapsis. Periapsis and apoapsis are usually modified to apply to the body being orbited, such as perihelion and aphelion for the Sun, perigee and apogee for Earth, perijove and apojove for Jupiter, perilune and apolune for the Moon, etc. The argument of periapsis is the angular distance between the ascending node and the point of periapsis (see Figure 4.3). The time of periapsis passage is the time in which a satellite moves through its point of periapsis.

Nodes are the points where an orbit crosses a plane, such as a satellite crossing the Earth's equatorial plane. If the satellite crosses the plane going from south to north, the node is the ascending node; if moving from north to south, it is the descending node. The longitude of the ascending node is the node's celestial longitude. Celestial longitude is analogous to longitude on Earth and is measured in degrees counter-clockwise from zero with zero longitude being in the direction of the vernal equinox.

In general, three observations of an object in orbit are required to calculate the six orbital elements. Two other quantities often used to describe orbits are period and true anomaly. Period, $P$, is the length of time required for a satellite to complete one orbit. True anomaly, $\nu$, is the angular distance of a point in an orbit past the point of periapsis, measured in degrees.

Types Of Orbits

For a spacecraft to achieve Earth orbit, it must be launched to an elevation above the Earth's atmosphere and accelerated to orbital velocity. The most energy efficient orbit, that is one that requires the least amount of propellant, is a direct low inclination orbit. To achieve such an orbit, a spacecraft is launched in an eastward direction from a site near the Earth's equator. The advantage being that the rotational speed of the Earth contributes to the spacecraft's final orbital speed. At the United States' launch site in Cape Canaveral (28.5 degrees north latitude) a due east launch results in a "free ride" of 1,471 km/h (914 mph). Launching a spacecraft in a direction other than east, or from a site far from the equator, results in an orbit of higher inclination. High inclination orbits are less able to take advantage of the initial speed provided by the Earth's rotation, thus the launch vehicle must provide a greater part, or all, of the energy required to attain orbital velocity. Although high inclination orbits are less energy efficient, they do have advantages over equatorial orbits for certain applications. Below we describe several types of orbits and the advantages of each:

Geosynchronous orbits (Geo) are circular orbits around the Earth having a period of 24 hours. A geosynchronous orbit with an inclination of zero degrees is called a geostationary orbit. A spacecraft in a geostationary orbit appears to hang motionless above one position on the Earth's equator. For this reason, they are ideal for some types of communication and meteorological satellites. A spacecraft in an inclined geosynchronous orbit will appear to follow a regular figure-8 pattern in the sky once every orbit. To attain geosynchronous orbit, a spacecraft is first launched into an elliptical orbit with an apogee of 35,786 km (22,236 miles) called a geosynchronous transfer orbit (GTO). The orbit is then circularized by firing the spacecraft's engine at apogee.

Polar orbits (PO) are orbits with an inclination of 90 degrees. Polar orbits are useful for satellites that carry out mapping and/or surveillance operations because as the planet rotates the spacecraft has access to virtually every point on the planet's surface.

Walking orbits: An orbiting satellite is subjected to a great many gravitational influences. First, planets are not perfectly spherical and they have slightly uneven mass distribution. These fluctuations have an effect on a spacecraft's trajectory. Also the sun, moon, and planets contribute a gravitational influence on an orbiting satellite. With proper planning it is possible to design an orbit which takes advantage of these influences to induce a precession in the satellite's orbital plane. The resulting orbit is called a walking orbit, or a precessing orbit.

Sun synchronous orbits (SSO) are walking orbits whose orbital plane precesses with the same period as the planet's solar orbit period. In such an orbit, a satellite crosses periapsis at about the same local time every orbit. This is useful if a satellite is carrying instruments which depend on a certain angle of solar illumination on the planet's surface. In order to maintain an exact synchronous timing, it may be necessary to conduct occasional propulsive maneuvers to adjust the orbit.

Molniya orbits are highly eccentric Earth orbits with a period of approximately 12 hours (2 revolutions per day). The orbital inclination is chosen so the rate of change of perigee is zero, thus both apogee and perigee can be maintained over fixed latitudes. This condition occurs at inclinations of 63.4 and 116.6 degrees. For these orbits the argument of perigee is typically placed in the southern hemisphere, so the satellite remains above the northern hemisphere near apogee for approximately 11 hours per orbit. This orientation can provide good ground coverage at high northern latitudes.

Hohmann transfer orbits are interplanetary trajectories whose advantage is that they consume the least possible amount of propellant. A Hohmann transfer orbit to an outer planet, such as Mars, is achieved by launching a spacecraft and accelerating it in the direction of Earth's revolution around the sun until it breaks free of the Earth's gravity and reaches a velocity which places it in a sun orbit with an aphelion equal to the orbit of the outer planet. Upon reaching its destination, the spacecraft must decelerate so that the planet's gravity can capture it into a planetary orbit

To send a spacecraft to an inner planet, such as Venus, the spacecraft is launched an accelerated in the direction opposite of Earth's revolution around the sun (i.e. decelerated) until it achieves a sun orbit with a perihelion equal to the orbit of the inner planet. It should be noted that the spacecraft continues to move in the same direction as Earth, only more slowly.

To reach a planet requires that the spacecraft be inserted into an interplanetary trajectory at the correct time so that the spacecraft arrives at the planet's orbit when the planet will be at the point where the spacecraft will intercept it. This task is comparable to a quarterback "leading" his receiver so that the football and receiver arrive at the same point at the same time. The interval of time in which a spacecraft must be launched in order to complete its mission is called a launch window.

Newton's Laws of Motion and Universal Gravitation

Newton's laws of motion describe the relationship between the motion of a particle and the forces acting on it.

The first law states that if no forces are acting, a body at rest will remain at rest, and a body in motion will remain in motion in a straight line. Thus, if no forces are acting, the velocity (both magnitude and direction) will remain constant.

The second law tells us that if a force is applied there will be a change in velocity, i.e. an acceleration, proportional to the magnitude of the force and in the direction in which the force is applied. This law may be summarized by the equation

(4.1) $F = ma$

where $F$ is the force, $m$ is the mass of the particle, and $a$ is the acceleration.

The third law states that if body 1 exerts a force on body 2, then body 2 will exert a force of equal strength, but opposite in direction, on body 1. This law is commonly stated, "for every action there is an equal and opposite reaction".

In his law of universal gravitation, Newton states that two particles having masses $m_1$ and $m_2$ and separated by a distance $r$ are attracted to each other with equal and opposite forces directed along the line joining the particles. The common magnitude $F$ of the two forces is

(4.2) $F = G \left ( \frac{m_1m_2}{r^2} \right )$

where $G$ is an universal constant, called the constant of gravitation, and has the value $6.67259x10^{-11} \frac{Nm^2}{kg^2}$.

Let's now look at the force that the Earth exerts on an object. If the object has a mass $m$, and the Earth has a mass $M$, and the object's distance from the center of the Earth is $r$, then the forst that the Earth exerts on the object is $\frac{GmM}{r^2}$. If we drop the object, the Earth's gravity will cause it to accelerate toward the center of the Earth. By Newton's second law ($F = ma$), this acceleration $g$ must equal $\frac{\frac{GmM}{r^2}}{m}$, or

(4.3) $g = \frac{GM}{r^2}$

At the surface of the earth, this acceleration has the value $9.80665 \frac{m}{s^2}$.

Many of the upcoming computations will be somewhat simplified if we express the product $GM$ as a constant, which for Earth has the value $3.986005x10^{14}\frac{m^3}{s^2}$. The product GM is often represented by the Greek letter $\mu$.

For additional useful constants please see the appendix Basic Constants.

For a refresher on SI versus U.S. units see the appendix Weights & Measures.

Uniform Circular Motion

In the simple case of free fall, a particle accelerates toward the center of the Earth while moving in a straight line. The velocity of the particle changes in magnitude, but not in direction. In the case of uniform circular motion a particle moves in a circle with constant speed. The velocity of the particle changes continuously in direction, but not in magnitude. From Newton's laws we see that since the direction of the velocity is changing, there is an acceleration. This acceleration, called centripetal acceleration is directed inward toward the center of the circle and is given by

(4.4) $a = \frac{v^2}{r}$

where $v$ is the speed of the particle and $r$ is the radius of the circle. Every accelerating particle must have a force acting on it, defined by Newton's second law ($F = ma$). Thus, a particle undergoing uniform circular motion is under the influence of a force, called centripetal force, whose magnitude is given by

(4.5) $F = \frac{mv^2}{r}$

The direction of $F$ at any instant must be in the direction of $a$ at the same instant, that is radially inward.

A satellite in orbit is acted on only by the forces of gravity. The inward acceleration which causes the satellite to move in a circular orbit is the gravitational acceleration caused by the body around which the satellite orbits. Hence, the satellite's centripetal acceleration is $g$, that is $g = \frac{v^2} {r}$. From Newton's law of universal gravitation we know that $g = \frac{GM} {r^2}$. Therefore, by setting these equations equal to one another we find that, for a circular orbit,

$\frac{v^2}{r} = \frac{GM}{r^2}$, or

(4.6) $v = \sqrt{\frac{GM}{r}}$

var GM = 3.986005e14;
function v(r) {
  return Math.sqrt(GM/r);
}
var R_earth = 6378.14 * 1000; // meters
var altitude_satellite = 200 * 1000; // meters
var R_satellite = R_earth + altitude_satellite;
emit("velocity: %d m/s", v(R_satellite));

Motions of Planets and Satellites

Through a lifelong study of the motions of bodies in the solar system, Johannes Kepler (1571-1630) was able to derive three basic laws known as Kepler's laws of planetary motion. Using the data compiled by his mentor Tycho Brahe (1546-1601), Kepler found the following regularities after years of laborious calculations:

1. All planets move in elliptical orbits with the sun at one focus.
2. A line joining any planet to the sun sweeps out equal areas in equal times.
3. The square of the period of any planet about the sun is proportional to the

cube of the planet's mean distance from the sun.

These laws can be deduced from Newton's laws of motion and law of universal gravitation. Indeed, Newton used Kepler's work as basic information in the formulation of his gravitational theory.

As Kepler pointed out, all planets move in elliptical orbits, however, we can learn much about planetary motion by considering the special case of circular orbits. We shall neglect the forces between planets, considering only a planet's interaction with the sun. These consideration apply equally well to the motion of a satellite about a planet.

Let's examine the case of two bodies of masses $M$ and $m$ moving in circular orbits under the influence of each other's gravitational attraction. The center of mass of this system of two bodies lies along the line joining them at a point $C$ such that $mr = MR$. The large body of mass $M$ moves in an orbit of constant radius $R$ and the small body of mass $m$ in an orbit of constant radius $r$, both having the same angular velocity $\omega$. For this to happen, the gravitational force acting on each body must provide the necessary centripetal acceleration. Since these gravitational forces are a simple action-reaction pair, the centripetal forces must be equal but opposite in direction. That is, $m\omega^2r$ must equal $M\omega^2R$. The specific requirement then, is that the gravitational force acting on either body must equal the centripetal force needed to keep it moving in its circular orbit, that is

(4.7) $\frac{GMm}{(R+r)^2} = m \omega^2 r$

If one body has a much greater mass than the other, as is the case of the sun and a planet or the Earth and a satellite, its distance from the center of mass is much smaller than that other body. If we assume that $m$ is negligible compared to $M$, then $R$ is negligible compared to $r$. Thus, equation (4.7) becomes

(4.8) $GM = \omega^2 r^3$

If we express the angular velocity in terms of the period of revolution, $\omega = \frac{2\pi}{P}$, we obtain

$GM = \frac{4\pi^2r^3}{p^2}$, or

(4.9) $p^2 = \frac{4\pi^2r^3}{GM}$

where $P$ is the period of revolution. This is a basic equation of planetary and satellite motion. It also holds for elliptical orbits if we define $r$ to be the semi-major axis ($a$) of the orbit.

A significant consequence of this equation is that it predicts Kepler's third law of planetary motion, that is $p^2 \sim r^3$.

function p(r) {
  return Math.sqrt((4 * Math.pow(Math.PI, 2) * Math.pow(r, 3)) / GM);
}
emit("period: %d seconds", p(R_satellite));

function r(p) {
  return Math.cbrt((Math.pow(p, 2) * GM) / (4 * Math.pow(Math.PI, 2)));
}
emit("radius: %d meters", r(86164.1));

Kepler's second law of planetary motion must, of course, hold true for circular orbits. In such orbits both $\omega$ and $r$ are constant so that equal areas are swept out in equal times by the line joining a planet and the sun. For elliptical orbits, however, both $\omega$ and $r$ will vary with time. Let's now consider this case.

Figure 4.5 shows a particle revolving around $C$ along some arbitrary path. The area swept out by the radius vector in a short time interval $\Delta t$ is shown shaded. This area, neglecting the small triangular region at the end, is one-half the base times the height or approximately $\frac{r(r\omega\Delta t)} {2}$. This expression becomes more exact as $\Delta t$ approaches zero, i.e. the small triangle goes to zero more rapidly than the large one. The rate at which area is being swept out instantaneously is therefore

(4.10) $\lim_{t \to 0} \left [ \frac{r(r \omega \Delta t)}{2} \right ] = \frac {\omega r^2} {2}$

For any given body moving under the influence of a central force, the value $\omega r^2$ is constant.

Let's now consider two points, $P_1$ and $P_2$ in an orbit with radii $r_1$ and $r_2$, and velocities $v_1$ and $v_2$. Since the velocity is always tangent to the path, it can be seen that if $\gamma$ is the angle between $r$ and $v$, then

(4.11) $v\sin \gamma = \omega r$

where $v\sin \gamma$ is the transverse component of $v$. Multiplying through by $r$, we have

(4.12) $rv\sin\gamma = \omega r^2 = \text{Constant}$

or, for two points $P_1$ and $P_1$ on the orbital path

(4.13) $r_1v_1\sin\gamma_1 = r_2v_2\sin\gamma_2$

Note that at periapsis and apoapsis, $\gamma = \text{90 degrees}$. Thus, letting $P_1$ and $P_2$ be these two points we get

(4.14) $R_pV_p = R_aV_a$

Let's now look at the energy of the above particle at points $P_1$ and $P_2$. Conservation of energy states that the sum of the kinetic energy and the potential energy of a particle remains constant. The kinetic energy $T$ of a particle is given by $\frac{mv^2}{2}$ while the potential energy of gravity $V$ is calculated by the equation $-\frac{GMm}{r}$. Applying the conservation of energy we have

$T_1 + V_1 = T_2 + V_2$, or

$\frac{mv_1^2}{2} - \frac{GMm}{r_1} = \frac{mv_2^2}{2} - \frac{GMm}{r_2}$, or

(4.15) $v_2^2 - v_1^2 = 2GM\left(\frac{1}{r_2} - \frac{1}{r_1}\right)$

From equations (4.14) and (4.15) we obtain

(4.16) $V_p = \sqrt{\frac{2GMR_a}{R_p(R_a+R_p)}}$, and

(4.17) $V_a = \sqrt{\frac{2GMR*p}{R_a(R_a+R_p)}}$

Rearranging terms we get

(4.18) $R_a = \frac{R_p}{\left ( \frac{2GM}{R_pV_p^2} - 1 \right )}$, and

(4.19) $R_p = \frac{R_a}{\left ( \frac{2GM}{R_aV_a^2} - 1 \right )}$

function V_p(R_a, R_p) {
  return Math.sqrt((2 * GM * R_a) / (R_p * (R_a + R_p)));
};
function V_a(R_a, R_p) {
  return Math.sqrt((2 * GM * R_p) / (R_a * (R_a + R_p)));
};
R_a = 500*1000 + R_earth;
R_p = 250*1000 + R_earth;
emit("V_p: %d m/s", V_p(R_a, R_p));
emit("V_a: %d m/s", V_a(R_a, R_p));

function R_a(R_p, V_p) {
  return R_p / (((2*GM)/(R_p*Math.pow(V_p, 2))) - 1);
}
emit(
  "Altitude @ apogee = %.1f km",
  (R_a(R_earth + 200*1000, 7850) - R_earth) / 1000);

The eccentricity $e$ of an orbit is given by

(4.20) $e = \frac{R_pV_p^2}{GM} - 1$

function e(R_p, V_p) {
  return ((R_p * Math.pow(V_p, 2)) / GM) - 1;
}
emit("e = %.5f", e(R_earth + 200*1000, 7850));

If the semi-major axis $a$ and the eccentricity $e$ of an orbit are known, then the periapsis and apoapsis distances can be calculated by

(4.21) $R_p = a(1-e)$, and

(4.22) $R_a = a(1+e)$

also note $R_p+R_a = 2a$

a = 6700 * 1000; // meters
e = 0.01;
R_p = a*(1 - e);
R_a = a*(1 + e);
emit("Altitude @ perigee = %.1f km", (R_p - R_earth) / 1000);
emit("Altitude @ apogee = %.1f km", (R_a - R_earth) / 1000);

Launch of a Space Vehicle

The launch of a satellite or space vehicle consists of a period of powered flight during which the vehicle is lifted above the Earth's atmosphere and accelerated to orbital velocity by a rocket, or launch vehicle. Powered flight concludes at burnout of the rocket's last stage at which time the vehicle begins its free flight. During free flight the space vehicle is assumed to be subjected only to the gravitational pull of the Earth. If the vehicle moves far from the Earth, its trajectory may be affected by the gravitational influence of the sun, moon, or another planet.

Figure 4.7

A space vehicle's orbit may be determined from the position and the velocity of the vehicle at the beginning of its free flight. A vehicle's position and velocity can be described by the variables $r$, $v$, and, $\gamma$ where $r$ is the vehicle's distance from the center of the Earth, $v$ is its velocity, and $\gamma$ is the angle between the position and the velocity vectors, called the zenith angle (see Figure 4.7). If we let $r_1$, $v_1$, and $\gamma_1$ be the initial (launch) values of $r$, $v$, and $\gamma$, then we may consider these as given quantities. If we let point $P_2$ represent the perigee, then equation (4.13) becomes

(4.23) $v_2 = V_p = \frac{r_1v_1\sin{\gamma_1}}{R_p}$

Substituting equation (4.23) into (4.15), we can obtain an equation for the perigee radius $R_p$.

(4.24) $\frac{r_1^2 v_1^2 \sin^2 \gamma_1}{R_p^2} - v_1^2 = 2GM \left( \frac{1} {R_p} - \frac{1}{r_1} \right)$

Multiplying through by $\frac{-R_p^2}{r_1^2v_1^2}$ and rearranging, we get

(4.25) $\left(\frac{R_p}{r_1}\right)^2(1-C) + \left(\frac{R_p}{r_1}\right)C - \sin^2\gamma_1 = 0$

where $C = \frac{2GM}{r_1v_1^2}$

Note that this is a simple quadratic equation in the ratio $\frac{R_p }{r_1}$ and that $\frac{2GM}{r_1v_1^2}$ is a nondimensional parameter of the orbit.

Solving for $\frac{R_p}{r_1}$ gives

(4.26) $\left(\dfrac{R_p}{r_1}\right)_{1,2} = \dfrac{-C \pm \sqrt{C^2 - 4(1-C) (-\sin^2\gamma_1)}} {2 (1-C)}$

Like any quadratic, the above equation yields two answers. The smaller of the two answers corresponds to $R_p$, the periapsis radius. The other root corresponds to the apoapsis radius, $R_a$.

Please note that in practice spacecraft launches are usually terminated at either perigee or apogee, i.e. $\gamma = 90$. This condition results in the minimum use of propellant.

var r1 = R_earth + (250*1000);
var v1 = 7900;
var g1 = 89 * (Math.PI / 180);
var C = (2*GM) / (r1 * Math.pow(v1, 2));
var root = Math.pow(C, 2) - 4*(1 - C) * -Math.pow(Math.sin(g1), 2);
var R_p = r1 * ((-C + Math.sqrt(root)) / (2*(1-C)));
var R_a = r1 * ((-C - Math.sqrt(root)) / (2*(1-C)));
emit("Altitude @ perigee: %.1f km", (R_p - R_earth) / 1000);
emit("Altitude @ apogee: %.1f km", (R_a - R_earth) / 1000);

Equation (4.26) gives the values of $R_p$ and $R_a$ from which the eccentricity of the orbit can be calculated, however, it may be simpler to calculate the eccentricity $e$ directly from the equation

(4.27) $e = \sqrt{\left(\frac{r_1v_1^2}{GM}-1\right)^2\sin^2\gamma_1+\cos^2\gamma_1}$

var r1 = R_earth + (250*1000);
var v1 = 7900;
var g1 = 89 * (Math.PI / 180);
function e(r1, v1, g1) {
  return Math.sqrt(
    Math.pow(((r1*Math.pow(v1, 2)) / GM) - 1, 2) * Math.pow(Math.sin(g1), 2)
    + Math.pow(Math.cos(g1), 2));
};
emit("e = %.7f", e(r1, v1, g1));

To pin down a satellite's orbit in space, we need to know the angle $\nu$, the true anomaly, from the periapsis point to the launch point. This angle is given by

(4.28) $\tan\nu = \dfrac{\left(\frac{r_1v_1^2}{GM}\right)\sin\gamma_1\cos\gamma_1}{\left(\frac{r_1v_1^2}{GM}\right)\sin^2\gamma_1 - 1}$

var rvGM = (r1*Math.pow(v1, 2))/GM;
var v = Math.atan((rvGM*Math.sin(g1)*Math.cos(g1))/((rvGM*Math.pow(Math.sin(g1), 2)) - 1));
emit("true anomaly = %.3f°", v / (Math.PI / 180));

In most calculations, the complement of the zenith angle is used, denoted by $\Phi$. This angle is called the flight-path angle, and is positive when the velocity vector is directed away from the primary as shown in Figure 4.8. When flight-path angle is used, equations (4.26) through (4.28) are rewritten as follows:

(4.29) $\left(\dfrac{R_p}{r_1}\right)_{1,2} = \dfrac{-C \pm \sqrt{C^2 - 4(1-C) (-\cos^2\Phi})} {2 (1-C)}$

where $C = \frac{GM}{rv^2}$

(4.30) $e = \sqrt{\left(\frac{rv^2}{GM}-1\right)^2\cos^2\Phi + \sin^2\Phi}$

(4.31) $\tan\nu = \dfrac{\left(\frac{rv^2}{GM}\right)\cos\Phi\sin\Phi}{\left (\frac{rv^2}{GM}\right)\cos^2\Phi-1}$

The semi-major axis is, of course, equal to $\frac{R_p+R_a}{2}$, though it may be easier to calculate it directly as follows:

(4.32) $a = \dfrac{1}{\left(\frac{2}{r} - \frac{v^2}{GM}\right)}$

function a(r, v) {
  return 1 / ((2 / r) - (Math.pow(v, 2) / GM));
}
emit("%d m", a(r1, v1));

If $e$ is solved for directly using equation (4.27) or (4.30), and $a$ is solved for using equation (4.32), $R_p$ and $R_a$ can be solved for simply using equations (4.21) and (4.22).

The End.

The next bit requires a notion of "sidereal time", which has to do with the position of the Earth relative to some fixed stars. This starts going into some very heavy math involving real-world circumstances and I was never able to find or write a Javascript library to compute the "local apparent sidereal time" - required for the problem sets. If anyone reads this and feels the urge to fill me in, feel free to drop me a line.