Astronomy

How to calculate center (not focus) of planetary orbit?

How to calculate center (not focus) of planetary orbit?


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I'm developing a website showing the movement of Solar System planets according to the passage of time. I'm using this reference to calculate all the necessary orbital parameters.

These are two images some days apart.

(Only planets up to Mars are shown here. The moon orbit has been highly exaggerated to become visible. Celestial bodies radii are not to scale either, except among themselves-except the Sun.)

The images show the high eccentricity of Mercury's orbit, causing it to be very far from the Sun in the first image, and very close to it in the second.

I want to draw the correct ellipse, instead of circles that constantly change radius. I have all the “orbital elements” given in the link above (N, i, w, a, e, M, r, v, x, y, z, lon, lat), but don't know how to extract an ellipse from them. I'm drawing the planets using x, y heliocentric coordinates. Even if I had the ellipse foci, JavaScript needs the geometric center of the ellipse to draw it, and I don't know either how to convert the former into the latter.

How to calculate the ellipse center from the parameters given above?


I assume that the Sun is at the origin: (0,0). Then we can look at this image of an ellipse:

The foci are at +-c, where c = ea, by the definition of eccentricity. To put the perigee on the negative y-axis, we can then say the center of the ellipse is at (0,ea). Then, to account for the argument of periapsis, w, we need to rotate the ellipse (and ellipse center) counter-clockwise around the origin to get the coordinates for the ellipse center, (x,y) =(-easin(w),eacos(w)). A better approximation would include the inclination, i, for a 2-D projection, but the inclination is small for Mercury so it may not matter for the purposes of your animation. This diagram for the orbital elements should also help:


I have calculated the perihelion and aphelion distances,

q = a*(1-e); // perihelion distance Q = a*(1+e); // aphelion distance

the perihelion and aphelion longitudes,

angPeri = (lon-v)*Math.PI/180; angAph = (lon-v+180)*Math.PI/180;

the perihelion and aphelion coordinates,

xPeri = q*Math.cos(angPeri); yPeri = q*Math.sin(angPeri); xAph = Q*Math.cos(angAph); yAph = Q*Math.sin(angAph);

then the center of the ellipse,

xEll = (xPeri+xAph)/2; yEll = (yPeri+yAph)/2;

calculatedaproportional to the canvas size,

aPx = a*maxPx/maxAU;

maxPxis the radius of the Mars orbit in pixels,maxAUis the radius of the Mars orbit in Astronomic Units. Then I've finally drawn the ellipse,

ctx.ellipse(canvasWidth/2+xEll,canvasHeight/2-yEll,aPx,aPx*Math.sqrt(1-e*e),angAph,0,2*Math.PI);

canvasWidth/2andcanvasHeight/2mean that the Sun is positioned in the center of the canvas.


From the Sun's center, always. When you deduce the equations of motion of planets, you're always calculating from the center. Plus, the results don't change when the Sun blows up as a red giant, or collapses as a dwarf.

But even if you measure from the surface, in most cases it won't make a huge difference. In Earth's case, it's a 0.5% error. It would be a larger error for internal planets, smaller error for external planets.

EDIT: Incorrect. Planets (and the Sun, too) orbit the common barycenter of the Solar System.

It's how the maths/physics works - barring miniscule alterations due to non-uniform distribution of mass, the centre of mass of a planet orbits around the centre of mass of the Sun.

And the centre of mass of the Solar_System (which is very near the centre of the Sun) orbits the centre of mass of the galaxy, etc.

Positions and distances are calculated with respect to the center of mass of a body, not with respect to the geometric center, or surface. The center of the Earth (and therefore its surface) moves with respect to its center of mass by about a centimeter. Mercury's center of mass is offset from its center of figure (geometric center) by 640 meters.

Radar ranging directly measures the surface to surface distance, but these distances are converted to center-of-mass distances in order to calculate orbits.


Some Properties of Ellipses

1. For an ellipse there are two points called foci (singular: focus) such that the sum of the distances to the foci from any point on the ellipse is a constant. In terms of the diagram shown to the left, with "x" marking the location of the foci, we have the equation

that defines the ellipse in terms of the distances a and b.

2. The amount of "flattening" of the ellipse is termed the eccentricity. Thus, in the following figure the ellipses become more eccentric from left to right. A circle may be viewed as a special case of an ellipse with zero eccentricity, while as the ellipse becomes more flattened the eccentricity approaches one. Thus, all ellipses have eccentricities lying between zero and one. The orbits of the planets are ellipses but the eccentricities are so small for most of the planets that they look circular at first glance. For most of the planets one must measure the geometry carefully to determine that they are not circles, but ellipses of small eccentricity. Pluto and Mercury are exceptions: their orbits are sufficiently eccentric that they can be seen by inspection to not be circles.

3. The long axis of the ellipse is called the major axis, while the short axis is called the minor axis (adjacent figure). Half of the major axis is termed a semimajor axis. The length of a semimajor axis is often termed the size of the ellipse. It can be shown that the average separation of a planet from the Sun as it goes around its elliptical orbit is equal to the length of the semimajor axis. Thus, by the "radius" of a planet's orbit one usually means the length of the semimajor axis.


3: Orbits and Gravity

  • Contributed by Andrew Fraknoi, David Morrison, & Wolff et al.
  • Sourced from OpenStax

How would you find a new planet at the outskirts of our solar system that is too dim to be seen with the unaided eye and is so far away that it moves very slowly among the stars? This was the problem confronting astronomers during the nineteenth century as they tried to pin down a full inventory of our solar system.

If we could look down on the solar system from somewhere out in space, interpreting planetary motions would be much simpler. But the fact is, we must observe the positions of all the other planets from our own moving planet. Scientists of the Renaissance did not know the details of Earth&rsquos motions any better than the motions of the other planets. Their problem, as we saw in Observing the Sky: The Birth of Astronomy, was that they had to deduce the nature of all planetary motion using only their earthbound observations of the other planets&rsquo positions in the sky. To solve this complex problem more fully, better observations and better models of the planetary system were needed.

  • 3.1: The Laws of Planetary Motion Tycho Brahe&rsquos accurate observations of planetary positions provided the data used by Johannes Kepler to derive his three fundamental laws of planetary motion. Kepler&rsquos laws describe the behavior of planets in their orbits as follows: (1) planetary orbits are ellipses with the Sun at one focus (2) in equal intervals, a planet&rsquos orbit sweeps out equal areas and (3) the relationship between the orbital period (P) and the semimajor axis (a) of an orbit is given by (P^2 = a^3) (when a is in units
  • 3.2: Newton&rsquos Great Synthesis In his Principia, Isaac Newton established the three laws that govern the motion of objects: (1) objects continue to be at rest or move with a constant velocity unless acted upon by an outside force (2) an outside force causes an acceleration (and changes the momentum) for an object and (3) for every action there is an equal and opposite reaction. Momentum is a measure of the motion of an object and depends on both its mass and its velocity.
  • 3.3: Newton&rsquos Universal Law of Gravitation Gravity, the attractive force between all masses, is what keeps the planets in orbit. Newton&rsquos universal law of gravitation relates the gravitational force to mass and distance. The force of gravity is what gives us our sense of weight. Unlike mass, which is constant, weight can vary depending on the force of gravity (or acceleration) you feel. When Kepler&rsquos laws are reexamined in the light of Newton&rsquos gravitational law, it becomes clear that the masses of both objects are important for the thir
  • 3.4: Orbits in the Solar System The closest point in a satellite orbit around Earth is its perigee, and the farthest point is its apogee (corresponding to perihelion and aphelion for an orbit around the Sun). The planets follow orbits around the Sun that are nearly circular and in the same plane. Most asteroids are found between Mars and Jupiter in the asteroid belt, whereas comets generally follow orbits of high eccentricity.
  • 3.5: Motions of Satellites and Spacecraft The orbit of an artificial satellite depends on the circumstances of its launch. The circular satellite velocity needed to orbit Earth&rsquos surface is 8 kilometers per second, and the escape speed from our planet is 11 kilometers per second. There are many possible interplanetary trajectories, including those that use gravity-assisted flybys of one object to redirect the spacecraft toward its next target.
  • 3.6: Gravity with More Than Two Bodies Calculating the gravitational interaction of more than two objects is complicated and requires large computers. If one object (like the Sun in our solar system) dominates gravitationally, it is possible to calculate the effects of a second object in terms of small perturbations. This approach was used by John Couch Adams and Urbain Le Verrier to predict the position of Neptune from its perturbations of the orbit of Uranus and thus discover a new planet mathematically.
  • 3.E: Orbits and Gravity (Exercises)

Thumbnail: This space habitat and laboratory orbits Earth once every 90 minutes. (credit: modification of work by NASA)


Calculating the Positions of Planets

How to Calculate the Positions of the Planets: An Overview

This brief explanation describes the methods used in the simulation above to determine the Right Ascension (RA) and Declination (Dec) of the Sun, Mercury, Venus, Mars, Jupiter, and Saturn. It assumes some background in astronomy. However, there is a reasonably verbose glossary linked to what may be unfamiliar terms. In truth, this explanation is more for those wishing to know exactly what assumptions underlie the above simulation. However, this explanation does serve as a good case study in how to determine the planets' positions.

The simulation overlays the RA and Dec of the five planets visible to the naked eye and the sun atop an cylindrical projection of the Earth's sky. The name planet roughly translates to "wanderer." Knowing the geocentric model held for so long, many of my students simply assume the planets move about the sky in much the same manner as the moon or sun--steadily creeping eastward. They do not, and I developed this simulation to illustrate their true wandering nature.

Solving Kepler's Equation & Calculating Ephemerides

When first approaching this problem, I knew that I would need to solve Kepler's Equation and have a little fun with reference frames. I was surprised by the explanations available on the web. Most of them fell short of what I was looking for. They were either strictly qualitative or, if quantitative, unnecessarily opaque. This resulted mostly from having too few or poorly executed diagrams and illustrations. There were a few notable exceptions, and I have cited them in the references section. As I never formally studied celestial mechanics, these sources were my teachers and greatly helped acquaint me with the problem and its solutions. What I've attempted to do here is to weave together a relatively cogent "How to" on the solving of Kepler's Equation and the calculation of planetary positions in the Earth's sky.

Johannes Kepler (1571-1630) was a mathematician, astronomer, and Copernican. He believed that the Sun, not the Earth, lay at the center of the universe. He refined Copernicus's view of a heliocentric (Sun-centered) universe, making it into more than simply a competing theory for the geocentric (Earth-centered) model. Under Kepler it would become the superior predictive model. In his work Kepler formulated three laws of planetary motion first set down together in Harmonice Mundi (Harmonies of the World), 1618, and here they are.

1. The planets orbit about the Sun in elliptical orbits with the sun centered at one of the ellipse's two foci (Figure 1).

2. An imaginary line connecting the sun and a planet sweeps out equal areas in equal times as the planet moves through its orbit. A consequence of this is that a planet moves fastest when closest to the Sun. Newton will have something to say about this.

3. The square of the period of a planet's orbit is proportional to its distance from the Sun cubed. When the units used for distance are Astronomical Units (AU) and time is measured in years, this relationship can be written explicitly as an equation relating the planet's period P and the semi-major axis of its orbit a (eq.1).


Kepler's Laws meant that given only a handful of orbital parameters, one could say where a planet had been and would be. To state this explicitly, astronomers make use of Kepler's Equation (eq.2).


Kepler's equation is a transcendental equation. This means there is no general solution. So to find the location of a planet at a time t, we must solve for that time using some numerical method. First let us work with what we have. NOTE: You may find it helpful to reference Figure 2 (pop-up) to help visualize some of the variables referenced here.

Only e is time independent. So we consult our orbital parameters for its value and then solve for the mean anomaly (eq.3), M in Kepler's Equation (eq.2). The mean anomaly is just the angle with the perihelion that the planet would have if the orbit was an ellipse with eccentricity = 0, i.e., a circle. We call the imaginary planet moving along such an orbit the mean planet. In such a case the planet would move with a velocity V = (2*PI)/Period .


We can now find the eccentric anomaly using some numerical method. This simulation makes use of successive approximation. Once we have a value for E with which we are happy, we can find the true anomaly (eq.4). The true anomaly is the ACTUAL angle between the perihelion and the planet.

From here it is a simple matter to find the planet's radial distance (eq.5) from the sun.

We now have the planet's polar coordinates (r, v) within the plane of its orbit such that the X axis points from the Sun towards the Perihelion, point P.

Now we find the Heliocentric Ecliptic coordinates (x, y, z) for the planet by converting from polar to cartesian coordinates and rotating the frame such that the X axis points towards the first point of Aries.

We then rotate the coordinates into Heliocentric Equatorial coordinates (X, Y, Z), making use the matrix below.

However, our display shows the positions of the planets from the Earth. So we need to switch our vantage point to that of a geocentric system. To do this we first repeat the above process, solving for the Earth's Heliocentric Equatorial coordinates. We want to know the Sun's Geocentric coordinates. So here we will approximate this as the inverse of Earth's heliocentric coordinates. This is the same method used to find the Sun's location for display. It is important to note, however, that this is just an approximation, as what we really find is not the location of the Earth but rather that of the Earth-Moon system's barycenter. This simplification is responsible for limiting the simulation's accuracy. Note: This is not an issue for the Build Your Own Solar System simulation for teachers as the hypothetical "Earth" has no moon in that simulation. We then add the Sun's geocentric coordinates to those of the heliocentric coordinates of our planet. This shifts the coordinates, giving us the Geocentric equatorial coordinates (xp, yp, zp) for the planet.

Having the planet's Geocentric coordinates, it is a simple matter to convert them into Right Ascension and Declination. Note: Watch you signs here if you're not careful, it WILL get messy.

That's it. We can now solve for many discreet times and collect the data into tables to construct ephemerides. If you are interested in finding more accurate calculations for the planets' positions, consider buying a copy of the Astronomical Almanac from the US Naval Observatory or making use of JPL's Horizons system.


References & Further Reading:

1. To anyone interested in why it is the orbits of the planets are elliptical, I suggest finding a copy of D. & J. Goodstein's Feynman's Lost Lecture: The Motion of Planets Around the Sun. W. W. Norton & Company. New York, NY. 1996.

2. A copy of Kepler's Harmonice Mundi (Harmonies of the World) as well as many other ground breaking texts in astronomy have been compiled into one tome: Stephen Hawking's On The Shoulders of Giants: The Great Works of Physics and Astronomy. Running Press. Philadelphia. 2002.

3. For what I found to be the most rigorous on-line handling of this material, try Dr. J. B. Tatum's Celestial Mechanics: http://orca.phys.uvic.ca/

tatum/celmechs.html (Link current as of April 2004).

4. The orbital parameters used here came from the JPL Solar System Dynamics Group's "Planetary Orbital Elements," JPL Solar System Dynamics: http://ssd.jpl.nasa.gov/elem_planets.html. (Link current as of April 2004).

Ascending node: The point of intersection between a planet's orbit and the plane of the Sun's equator, where the planet is moving northward ("upward") across the plane of the Sun's equator.

Astronomical Units (AU): A measure of distance where one AU is just about equal to the average distance of the Earth from the Sun, 1.49597870691 x 108 (± 3) kilometers.

Barycenter: The center of mass for a multi-body system of mutually orbiting bodies. The system orbits about the barycenter.

Celestial sphere: A gigantic imaginary sphere surrounding a stationary Earth upon which the stars are affixed. It was once believed that the celestial sphere was real. However, it is now regarded solely as a convenient descriptive tool.

Celestial equator: The projection of the Earth's equator onto the celestial sphere.

Copernican: One who subscribes to the Copernican world view of a heliocentric universe, i.e., one who believes that the Earth orbits around a fixed Sun.

Declination (DEC): A heavenly object's position in the sky as measured along a meridian in degrees (0 to 90 degrees) north (+) or south (-) from the equator.

Descending node: The point of intersection between a planet's orbit and the plane of the Sun's equator, where the planet is moving southward ("downward") across the plane of the Sun's equator.

Eccentric Anomaly: a time dependent term in Kepler's equation which must be solved for in order to calculate a planet's position on its orbit.

Eccentricity: A measure of how "elliptical" an eclipse is (measured from 0 to 1). For example, a circle has an eccentricity of zero, not very elliptical. A relationship can be stated mathematically between the semi-major axis a, the semi-minor axis b and the eccentricity e where:

Above are four ellipses with varying eccentricities. The first is a circle.

Ecliptic: As seen from the Earth the ecliptic is the Sun's annual path across the sky.

Ephemerides: plural of ephemeris. Tables containing the calculated positions (usually RA and DEC) of celestial objects for different times, usually at regular intervals.

Ellipse: One of the conic sections, those shapes which are the intersection of a cone and plane. The ellipse is a geometric shape that looks like a squashed circle. You can easily make an ellipse with two thumb tacks and a loop of string. Place the two tacks into a paper and loop the string around them. Place a pencil in the loop of string and move it outwards until the loop becomes taut. Move the pencil around the tacks always keeping the slack out of the loop. The figure drawn is an ellipse. The points where the thumbtacks lie are the foci of the ellipse (singular focus).

Elliptical: Shaped like an ellipse.

First point of Aries: The position against the background stars of the Earth's descending node as seen from the Sun.

Foci: Plural of focus. See ellipse.

Geocentric: Earth centered.

Geocentric equatorial coordinates: An X,Y,Z coordinate system centered on the Earth in which the Earth's equator lies in the X-Y plane.

Heliocentric: Sun centered.

Heliocentric ecliptic cartesian coordinates: An X,Y,Z coordinate system centered on the Sun in which the ecliptic lies in the X-Y plane.

Heliocentric equatorial cartesian coordinates: An X,Y,Z coordinate system centered on the Sun in which the Sun's equator lies in the X-Y plane.

Kepler's Equation: An equation derived from Kepler's Laws whose solution can specify the position of a planet in its orbit for a specified time given a set of orbital parameters.

Mean anomaly: The angle between the perihelion and the mean planet as measured in the plane of its orbit.

Mean planet: An imaginary planet which moves at a constant velocity around a circular orbit with a radius equal to the semi-major axis of the actual planet's orbit.

Numerical method: A method for solving mathematics problems, usually by computer, through the repeated use of simple arithmetic operations.

Orbital parameters: A set of physical parameters for the orbit of a planet sufficient to predict the position of the planet at a given time t. The orbital parameters used in the simulation above can be found at: http://ssd.jpl.nasa.gov/elem_planets.html (valid as of June 2004).

Perihelion: The closest point on a planet's orbit to the Sun.

Period (of a planet): The length of time it takes a planet to return to the same place in its orbit.

PI (): The ratio of a circle's circumference C to its diameter D.

Polar Coordinates: A means of denoting a point's location by use of its radial distance from the origin and the angle it is from the x axis.

Radial distance: How far something is from the coordinate axes as measured directly out from the axes.

Retrograde Motion: The westward motion of the planets against the background stars. In order to maintain the Earth's central location and a commitment to perfect circular motion, geocentrists devised a set of epicycles (orbits within orbits) upon which the planets would rotate. The motion of the planet about its epicycle allowed for the presence of retrograde motion. However, heliocentrists' Sun-centered model had no need for epicycles as retrograde motion could bee seen as one planet simply overtaking another as they raced about the Sun. See Animation.

Right Ascension (RA): A heavenly object's position in the sky as measured in hours:minutes:seconds east (+) or west (-) from the vernal equinox.

Semi-major axis: See ellipse.

Semi-minor axis: See ellipse.

Successive approximation: A numerical method by which a solution is found to an equation by substituting in guesses for the answer on both sides of the equation. The sides are evaluated and the first guess that produces a difference between the sides of less than a pre-defined tolerance is taken to be the answer.

Transcendental equation: An equation for which a general solution cannot be found algebraically as it contains transcendental (non-algebraic) functions.

True Anomaly: The angle between the perihelion and the planet as measured in the plane of its orbit.

Vector: A quantity consisting of both direction and magnitude (e.g., velocity).

Velocity: A measure of an object's motion that includes both the object's speed and direction.

Vernal equinox: Both the date (around March 21st) on which the Sun crosses the celestial equator moving northward and the point against the background stars where this occurs.


Introduction

First, it is important to have all materials prepared and ready for this laboratory activity. (Note: usually, I have a set student who has the classroom job of passing out all necessary materials to students during the 'Warm Up' activity.)

Materials: (each student receives one of each)

  • 1 sheet of blank paper (8.5 x 11 inches)
  • Pencil (students should not use pens, as they often don't work well when using the drawing compass)
  • Scissors
  • Approximately 22 cm of string (have these already cut!)
  • A thick piece of cardboard (pieces of cardboard can be purchased, but it's usually just as easy to cut up pieces of cardboard boxes)
  • A drawing compass
  • 2 push pins
  • Metric ruler
  • Calculator

Introduction: [Note: Please refer to resource below for student note sheets, as they will be periodically writing down information during this section.]

I usually introduce the activity by having all of the laboratory materials ready under a document camera. I then set up one push pin in the center of my piece of paper, loop the string around it, and draw a circle. I then very directly ask students, "What shape did I just draw?"

Generally, they're all able to correctly identify that I drew a circle.

Then, I repeat the procedure, but this time, I use two push pins in the center of the piece of paper, drawing an ellipse. After constructing it, I ask students the same question: "What shape did I just draw?"

Here, responses tend to be mixed. Often, students will guess things like a circle or oval. Then I introduce the concept of an ellipse by explaining that this shape, while nearly circular, is not actually a circle. It is what scientists call an ellipse. Then, I pose the question: "What makes a circle different from an ellipse?"

Again, answers tend to vary, but students generally are able to identify that the circle has one center point (focus), while the constructed ellipse has two (foci). I then clarify that these "center points" are actually called foci.

After defining foci, I then introduce the relationship between ellipses and orbital shapes. I state that: "All of the objects that orbit the Sun may appear circular to us, but they actually all have elliptical orbits. All of these objects that orbit the Sun, including the Earth, orbit in an elliptical shape, with the Sun at one of the foci. [Note: This is Kepler's First Law, although this will not be introduced until a later lesson]

I then ask, "If the Sun it at one of the foci, what is usually at the other foci?"

I select a variety of responses, but usually students are quick to understand that the other orbital foci is just a point in space - there is nothing there.

I then point them to a sample image (refer to Laboratory Introduction resource) showing an illustration of an elliptical orbit, having students note the two foci. Then, I point them to the image below (refer to Laboratory Introduction resource), which has an elliptical orbit with two additional parts labeled, the "distance between foci" and the "length of major axis". We then collectively define the major axis as the "distance between orbits that pass through both foci."

"But the interesting part is, we can actually see how elliptical each planet's orbit is. By using the same algebraic equation that astronomers use, we can figure out the orbital eccentricity of each planet. The eccentricity is actually how elliptical, or circle-like, a planet's orbit is. We can calculate it using an equation. The equation itself is listed below (refer to Laboratory Introduction resource), but it is actually a ratio between the distance between the two foci and the length of the major axis (HSA-CED.A.2 HSA-CED.A.4)."

Then, we do a quick example:

"Let's look at the ellipse I drew for you earlier in class. I can use my metric ruler to measure the distance between foci, which is the numerator ("d") in my eccentricity equation. Then, I can also use my metric ruler to measure the length of the major axis, which is the denominator ("L") in my eccentricity equation. Then, I can use my calculator (Note: Have students round to the thousandths place when entering these numbers into the calculator) to determine "e", which is the orbital eccentricity.


The Laws of Planetary Motion

Kepler's First Law:

Kepler's First Law is illustrated in the image shown above. The Sun is not at the center of the ellipse, but is instead at one focus (generally there is nothing at the other focus of the ellipse). The planet then follows the ellipse in its orbit, which means that the Earth-Sun distance is constantly changing as the planet goes around its orbit. For purpose of illustration we have shown the orbit as rather eccentric remember that the actual orbits are much less eccentric than this.

Kepler's Second Law:

II. The line joining the planet to the Sun sweeps out equal areas in equal times as the planet travels around the ellipse.

Kepler's second law is illustrated in the preceding figure. The line joining the Sun and planet sweeps out equal areas in equal times, so the planet moves faster when it is nearer the Sun. Thus, a planet executes elliptical motion with constantly changing angular speed as it moves about its orbit. The point of nearest approach of the planet to the Sun is termed perihelion the point of greatest separation is termed aphelion. Hence, by Kepler's second law, the planet moves fastest when it is near perihelion and slowest when it is near aphelion.

Kepler's Third Law:

In this equation P represents the period of revolution (orbit) for a planet around the sun and R represents the length of its semimajor axis. The subscripts "1" and "2" distinguish quantities for planet 1 and 2 respectively. The periods for the two planets are assumed to be in the same time units and the lengths of the semimajor axes for the two planets are assumed to be in the same distance units.

Here is a java applet allowing you to investigate Kepler's Laws, and Here is an animation illustrating the actual relative periods of the inner planets.


Activity Details

  • Subjects:MATHEMATICS
  • Types:PROBLEM SET, TEACHABLE MOMENTS
  • Grade Levels:9 - 12
  • Primary Topic:PROBLEM SOLVING
  • Additional Topics:
    ALGEBRA
    GEOMETRY
    MOTION AND FORCES
    PHYSICAL SCIENCES
    TEACHABLE MOMENTS
  • Time Required: 30 mins - 1 hr
  • Next Generation Science Standards (Website)

Use mathematical or computational representations to predict the motion of orbiting objects in the solar system

(+) Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant.


2 Answers 2

How long time does it take before three planets achieve the same relative position?

The answer is never, except for the case when their orbital periods can be expressed with low integers, like the 4:2:1 resonance of Io, Europa and Ganymede

However, what you are asking about is when they are going to be in almost the same position again, a quazi-period.

To find those periods, we are pretty much only left by brute forcing as our method. A nice little detail about the case with three planets is that the inner planet is always aligned with one of the other ones at the closest three-planet alignments. That allows us to calculate accurate solutions. In the cases of four or even five planets I simply give up.

To check all the possibilities, we can use a program. Here is an example of a function in JavaScript returning a list of quazi-periods and alignment error:

For Jupiter, Saturn and Uranus, I get the following output:

13.81170069444156. 0.30449020900657225
39.71676854387252. 0.12441143762575813
41.43510208332468. 0.08652937298028318
139.00868990355383. 0.06455996830984656
138.1170069444156. 0.04490209006572288
179.55210902774027. 0.041627282914560304
317.6691159721559. 0.0032748071511630172
3991.581500693611. 0.002329597100612091
4309.250616665767. 0.0009452100505313865

The first of this periods is of no use, as the error in alignment is almost a third of an orbit. Note that the one you found (that is really impressive you did,actually) gives an error in the alignment of less than a percent. We have to look at periods more than a thousand years long to find any better alignment.


Kepler&rsquos Second Law

Kepler&rsquos Second Law is based on the speed of the object as it orbits.

  • In the Earth-Sun example shown in Figure 2, the Earth will travel faster and faster as it gets closer to the sun.
  • As the Earth moves away from the sun, it will move slower and slower.
  • It&rsquos almost like the Earth is being &ldquoslingshot&rdquo around the sun very quickly as it passes near it.

Kepler didn&rsquot talk about speed when he wrote out his second law . Instead, he looked at a mathematical detail that pops out because we are talking about ellipses.

&ldquoAn imaginary line from the sun to the planet
sweeps out equal areas in equal times.&rdquo

If we were to look at the area the Earth sweeps out in a 15 day period, first when close to the sun (Figure 3) and then when far away (Figure 4 ), we would get diagrams that look like this.

Notice how in Figure 3 we have a stubby, fat, (basically) triangular area that was swept out by the line, but in Figure 4 we have a tall, thin, (basically) triangular area swept out.

  • If we calculate the area that I have (more or less) shaded in as triangles, you would find that they are equal.
  • This just shows that the planet is moving a lot faster when it is closer to the sun, since you can see that it traveled a greater distance along its orbit during that time.

How to calculate center (not focus) of planetary orbit? - Astronomy

When planning deep space missions it is obvious that accurate positions of the planets must be known to plot the interplanetary trajectories. The computational methods presented in this page are those described by Jean Meeus in his book Astronomical Formulae for Calculators, Fourth Edition, Willmann-Bell Inc., 1988.

Julian Date (JD) is the system of time measurement for scientific use by the astronomy community. It is the interval of time in days and fractions of a day since 4713 BC January-1 Greenwich noon. The Julian day begins at Greenwich mean noon, that is at 12:00 Universal Time (UT).

To convert a calendar date to Julian Date, perform the following steps:

Y = year
M = month
D = day (includes hours, minutes & seconds as fraction of day)

If the date is equal to or after 1582-Oct-15 (i.e. the date is in the Gregorian calendar), calculate

A = INT( Y / 100 )
B = 2 &ndash A + INT( A / 4 )

If the date is before 1582-Oct-15 (i.e. the date is in the Julian calendar), it is not necessary to calculate A and B.

JD = INT(365.25 × Y) + INT(30.6001 × (M + 1)) + D + 1720994.5 + B

where B is added only if the date is in the Gregorian calendar.

Calculate the JD corresponding to 1976-July-20, 12:00 UT.

A = INT(1976 / 100) = 19
B = 2 &ndash 19 + INT( 19 / 4 ) = &ndash13

The orbital elements of the major planets can be expressed as polynomials of the form

where T is the time measured in Julian centuries of 36525 ephemeris days from the epoch 1900 January 0.5 ET = JD 2415020.0. In other words,

This quantity is negative before the beginning of the year 1900, positive afterwards. Note that T is expressed in centuries, and thus should be taken with a sufficient number of decimals (at least nine).

L = mean longitude of the planet
a = semimajor axis of the orbit (this element is a constant for each planet)
e = eccentricity of the orbit
i = inclination on the plane of the ecliptic
w = argument of perihelion
W = longitude of ascending node

The longitude of the perihelion can be calculated from

and the planet's mean anomaly is

The perihelion and aphelion distances are

The quantities L and p are measured in two different planes, namely from the vernal equinox along the ecliptic to the orbit's ascending node, and then from this node along the orbit.

Table 1 gives the coefficients ai for the orbital elements of the major planets Mercury to Neptune. The elements for Earth are not given in Table 1.

TABLE 1
Elements for the mean equinox of the date
a0 a1 a2 a3
MERCURY
L 178.179 078 +149 474.070 78 +0.000 3011
a 0.387 0986
e 0.205 614 21 +0.000 020 46 &ndash0.000 000 030
i 7.002 881 +0.001 8608 &ndash0.000 0183
w 28.753 753 +0.370 2806 +0.000 1208
W 47.145 944 +1.185 2083 +0.000 1739
VENUS
L 342.767 053 +58 519.211 91 +0.000 3097
a 0.723 3316
e 0.006 820 69 &ndash0.000 047 74 +0.000 000 091
i 3.393 631 +0.001 0058 &ndash0.000 0010
w 54.384 186 +0.508 1861 &ndash0.001 3864
W 75.779 647 +0.899 8500 +0.000 4100
MARS
L 293.737 334 +19 141.695 51 +0.000 3107
a 1.523 6883
e 0.093 312 90 +0.000 092 064 &ndash0.000 000 077
i 1.850 333 &ndash0.000 6750 +0.000 0126
w 285.431 761 +1.069 7667 +0.000 1313 +0.000 004 14
W 48.786 442 +0.770 9917 &ndash0.000 0014 &ndash0.000 005 33
JUPITER
L 238.049 257 +3036.301 986 +0.000 3347 &ndash0.000 001 65
a 5.202 561
e 0.048 334 75 +0.000 164 180 &ndash0.000 000 4676 &ndash0.000 000 0017
i 1.308 736 &ndash0.005 6961 +0.000 0039
w 273.277 558 +0.559 4317 +0.000 704 05 +0.000 005 08
W 99.443 414 +1.010 5300 +0.000 352 22 &ndash0.000 008 51
SATURN
L 266.564 377 +1223.509 884 +0.000 3245 &ndash0.000 0058
a 9.554 747
e 0.055 892 32 &ndash0.000 345 50 &ndash0.000 000 728 +0.000 000 000 74
i 2.492 519 &ndash0.003 9189 &ndash0.000 015 49 +0.000 000 04
w 338.307 800 +1.085 2207 +0.000 978 54 +0.000 009 92
W 112.790 414 +0.873 1951 &ndash0.000 152 18 &ndash0.000 005 31
URANUS
L 244.197 470 +429.863 546 +0.000 3160 &ndash0.000 000 60
a 19.218 14
e 0.046 3444 &ndash0.000 026 58 +0.000 000 077
i 0.772 464 +0.000 6253 +0.000 0395
w 98.071 581 +0.985 7650 &ndash0.001 0745 &ndash0.000 000 61
W 73.477 111 +0.498 6678 +0.001 3117
NEPTUNE
L 84.457 994 +219.885 914 +0.000 3205 &ndash0.000 000 60
a 30.109 57
e 0.008 997 04 +0.000 006 330 &ndash0.000 000 002
i 1.779 242 &ndash0.009 5436 &ndash0.000 0091
w 276.045 975 +0.325 6394 +0.000 140 95 +0.000 004 113
W 130.681 389 +1.098 9350 +0.000 249 87 &ndash0.000 004 718

Calculate the orbital elements of Mars on 1976-July-20, 12:00 UT.

From our previous calculation we have JD 2442980.0 therefore,

T = (2442980.0 &ndash 2415020.0) / 36525 = 0.765503080

Consequently, from Table 1, we find

L = 293.737334 + (19141.69551 × 0.765503080) + (0.0003107 × 0.765503080 2 ) = 14946.764387 o = 186.764387 o

a = 1.5236883 AU
e = 0.093383330
i = 1.849824 o
w = 286.250750 o
W = 49.376635 o

The longitude of perihelion and the mean anomaly are,

p = 286.250750 + 49.376635 = 335.627385 o
M = 186.764387 &ndash 335.627385 = &ndash148.862998 = 211.137002 o

Since the plane of the ecliptic is the plane of Earth's orbit around the Sun, for Earth i = 0. The angles w and W are, therefore, not determined. The semimajor axis of Earth is a = 1.0000002 AU. The remaining orbital elements are calculated as follows:

L = 99.69668 + 36000.76892 T + 0.0003025 T 2

e = 0.01675104 &ndash 0.0000418 T &ndash 0.000000126 T 2

M = 358.47583 + 35999.04975 T &ndash 0.000150 T 2 &ndash 0.0000033 T 3

Calculate the orbital elements of Earth on 1976-July-20, 12:00 UT.

L = 99.69668 + (36000.76892 × 0.765503080) + (0.0003025 × 0.765503080 2 ) = 27658.396351 o = 298.396351 o

a = 1.0000002 AU
e = 0.016718968
M = 195.859204 o

p = 298.396351 &ndash 195.859204 = 102.537147 o

From the values of e and M, calculate the eccentric anomaly E from

The above is a transcendental equation in E that must be solved by iteration. It is also important that the angles M and E be expressed in radians. The equation can be solved in degrees if we replace e with eo = e × 180/ p , giving us

Then calculate the true anomaly n from

tan( n /2) = [ (1 + e) / (1 &ndash e) ] 1/2 × tan(E/2)

The radius vector of the planet can be calculated by one of the following

r = a × ( 1 &ndash e 2 ) / ( 1 + e × cos n )

The planet's argument of latitude is

The ecliptical longitude l can be deduced from ( l &ndash W ), which is given by

If i o , as for the major planets, ( l &ndash W ) and u must lie in the same quadrant. When a programmable calculator or computer is used, in order to avoid the use of other tests, the preceding formula can be better written as

tan( l &ndash W ) = cos i × sin u / cos u

and then the conversion from rectangular to polar coordinates should be applied to the numerator and the denominator of the fraction in the right-hand side. This will give ( l &ndash W ) directly in the correct quadrant.

The planet's ecliptic latitude b is given by

We have now obtained the heliocentric ecliptical coordinates l , b , and r of the planet for the given instant.

Calculate the heliocentric ecliptical coordinates of Mars on 1976-July-20, 12:00 UT.

As previously calculated, we have

L = 186.764387 o
a = 1.5236883 AU
e = 0.093383330
i = 1.849824 o
w = 286.250750 o
W = 49.376635 o
p = 335.627385 o
M = 211.137002 o

Following the steps outlined above, we have

eo = 0.093383330 × 180/ p = 5.3504707

E = 211.137002 + 5.3504707 × sin E
E = 208.577611 o

tan( n /2) = [ (1 + 0.093383330) / (1 &ndash 0.093383330) ] 1/2 × tan(208.577611 / 2)
n = 206.114239 o

r = 1.5236883 × ( 1 &ndash 0.093383330 × cos 208.577611 ) = 1.648641 AU

u = 186.764387 + 206.114239 &ndash 211.137002 &ndash 49.376635 = 132.364988 o

tan( l &ndash 49.376635) = cos 1.849824 × sin 132.364988 / cos 132.364988
l = 181.756494 o

sin b = sin 132.364988 × sin 1.849824
b = 1.366666 o

A perturbation is a disturbance in the regular and usually elliptical course of motion of a celestial body that is produced by some force additional to that that causes its regular motion. The most important perturbations in the motions of the major planets are caused by the gravitational influence of other planets. These perturbations must be accounted for if better accuracy is needed than that attainable using the above data alone. The perturbations in the motions of the giant planets are particularly important in longitude, they can be larger than 0.3 degree for Jupiter, and larger than 1.0 degree for Saturn.

Since the purpose of this web site is to provide only a basic understanding of interplanetary space flight, the extra accuracy attained by taking into account the principle perturbations is unnecessary. The planetary positions derived by the methods described above are adequate for illustrative and education purposes.

In order to calculate an accurate position of the Moon, it is necessary to take into account hundreds of periodic terms in the Moon's longitude, latitude and parallax. In his book, Jean Meeus presents a rigorous method for calculating the Moon's position, however he limits his solution to about 100 of the most important periodic terms, being satisfied with a small inaccuracy. However, even this less accurate solution is too cumbersome to present here. Fortunately, Meeus provides suggestions to simplify the method to produce a low accuracy solution. The accuracy appears to be about ± 0.3 o in longitude, ± 0.1 o in latitude, and ± 0.01 o in parallax.

Using the method described below, one obtains the geocentric longitude l and the geocentric latitude b of the center of the Moon, referred to the mean equinox of the date. If necessary, l and b can be converted to right ascension a and declination d using the following formulae.

tan a = ( sin l cos e &ndash tan b sin e ) / cos l

sin d = sin b cos e + cos b sin e sin l

where e , the obliquity of the ecliptic, is
e = 23.452294 &ndash 0.0130125 T &ndash 0.00000164 T 2 + 0.000000503 T 3

The equatorial horizontal parallax p of the Moon too is obtained. When the parallax p is known, the distance between the centers of Earth and Moon, in kilometers, can be found from

For the given instant, calculate the JD and then T by means of the formula previous presented. Then calculate the angles L', M, M', D and F by means of the following formulae, in which the various constants are expressed in degrees and decimals.

Moon's mean longitude:
L' = 270.434164 + 481267.8831 × T

Sun's mean anomaly:
M = 358.475833 + 35999.0498 × T

Moon's mean anomaly:
M' = 296.104608 + 477198.8491 × T

Moon's mean elongation:
D = 350.737486 + 445267.1142 × T

Mean distance of Moon from its ascending node:
F = 11.250889 + 483202.0251 × T

With the values of L', M, M', D and F calculated, l , b and p can be obtained by means of the following expressions where, again, all the coefficients are given in degrees and decimals.

l = L' + 6.288750 sin M'
+ 1.274018 sin(2D&ndashM')
+ 0.658309 sin 2D
+ 0.213616 sin 2M'
&ndash 0.185596 sin M
&ndash 0.114336 sin 2F

b = 5.128189 sin F
+ 0.280606 sin(M'+F)
+ 0.277693 sin(M'&ndashF)
+ 0.173238 sin(2D&ndashF)
+ 0.055413 sin(2D+F&ndashM')
+ 0.046272 sin(2D&ndashF&ndashM')

p = 0.950724
+ 0.051818 cos M'
+ 0.009531 cos(2D&ndashM')
+ 0.007843 cos 2D
+ 0.002824 cos 2M'
+ 0.000857 cos(2D+M')

Calculate the geocentric coordinates and parallax of the Moon on 1968-December-24, 10:00 UT.

Using the previously demonstrated method,

JD = 2440214.9167, T = 0.689799224

Using the above equations for angles L', M, M', D and F, we have

L' = 328.646595 o
M = 350.592460 o
M' = 67.500542 o
D = 55.647457 o
F = 323.632971 o

The Moon's longitude is L' plus the sum of the periodic terms

l = 328.646595
+ 5.810070
+ 0.881713
+ 0.613362
+ 0.151046
+ 0.030337
+ 0.109184

b = &ndash2.480685 o
p = 0.9717311 o

The distance to the Moon is,

D = 6378.14 / sin 0.9717311 = 376,090 km

Converting geocentric ecliptical coordinates to right ascension and declination, we have

e = 23.443317 o
a = 331.29323 o = 22 h 5 m 10.4 s
d = &ndash14.41295 o

The angular separation distance d between two celestial bodies, whose longitudes and latitudes are known, is given by the formula

In a previous example, we found that the coordinates of Mars on 1976-July-20 12:00 UT are

l = 181.756494 o , b = 1.366666 o

We're given that on the same date and time the coordinates of Earth are

The angular separation between the planets is

cos d = sin 1.366666 × sin 0 + cos 1.366666 × cos 0 × cos(181.756494 &ndash 297.883130)
d = 116.118642 o