Keplerian Elements for the Moon and Earth around their Barycenter

Keplerian Elements for the Moon and Earth around their Barycenter

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So I'm looking to replicate the inner Solar System to a decent degree of accuracy. So far I've been utilizing the Keplerian Elements and Equations for the planets, as shown here. There is one issue with this though:

For the Earth and Moon, I only get the path of the Barycenter. This is great, but it means that the next thing I need is the Earth's and Moon's Keplerian Elements relative to their Barycenter. Now I can get the starting element values from JPL HORIZONS or the JPL Planetary Satellite Mean Orbital Parameters. However, I get a bit stuck on what the best way to evolve these values over time is.

I'm thinking that maybe the answer lies in utilizing the Longitude Rate (n) in the JPL Planetary Satellite Mean Orbital Parameters for approximate motion.

Does anyone have any solid resources I could utilize to determine a good solution to this problem (i.e. books/links)?

Also, just so I know, does a question like this actually belong here? Or is it more appropriate for Space Exploration?

The Keplerian Orbital Elements and Orbital Perturbations

In our last post, we constructed an abstract solar orbital hierarchy. This hierarchy contains no explicit orbital information: just the hierarchical structure of a solar system.

In order to ascribe actual orbital data to the various stars and barycenters of a solar system, we will need a way to describe orbits in PISES. In this post, we’ll get spun up on the 6 Keplerian Orbital Elements and what their limitations are.

Geocentrism: Failing Basic Physics.

John Martin attempts a 'Gish Gallop' (RationalWiki) of comments in response to this post (Geocentrism: Does NASA use Geocentrism?). These questions, and some additional, also appear to be available on John Martin's blog. Most of the comments demonstrate a very poor level of understanding of topics ranging from basic geometry to physics. This commonly occurs if someone bases their scientific understanding on interpretations of press releases with an incorrect or incomplete understanding of the more fundamental physical concepts.

It's not clear how many of these were Mr. Martin's own thoughts or some regurgitation of material from another source. However, considering how many of the answers I readily found online, it is clear that Mr. Martin was incapable of doing even the most basic of research.

Let me get this straight, you want to do all the calculations in the appropriate (heliocentric) frames, then use a coordinate transformation (a technique valid between ANY locations) to transform to a geocentric frame. So far that is okay.

But if you want use this technique to claim that it supports the Earth as a favored position in space, that would be a lie. This technique will work for any object, whether it be a planet or spacecraft.

  • This type of ploy also brings to light the fact that the heliocentrists can work in ANY convenient coordinate system, but the Geocentrists cannot do squat without the heliocentric techniques. Mr. Martin has basically admitted that Geocentrism claims are useless for any practical problems.
  • This tactic is also used by Geocentrists and other pseudoscientists to evade doing any actual work themselves that would be required of real science. They want to get by with just saying “We use the heliocentric results done by others and then transform the results to an Earth-centered system and this makes the Earth a favored position“. This is equivalent to stealing the work of others and trying to re-label it as your own.

Many of the questions below erroneously invoke Kepler's Laws. This is disturbing since Mr. Martin claims a degree in engineering yet apparently could not comprehend how many of these issues were covered in the previous post, Geocentrism: Does NASA use Geocentrism? Some of these errors are equivalent to using idealizations of a round earth and invoking the fact that the Earth actually deviates from perfectly round as a basis for claiming that the Earth is flat, or trying to claim the Gas Laws (wikipedia) are incorrect by applying them to the thermal expansion of liquids and solids.

Kepler's Laws (Wikipedia) are actually a special case of Newton's laws and gravitation LIMITED TO THE CASE OF ONLY TWO BODIES (Wikipedia). The instant the question involves more than two bodies, Kepler's Laws become, at best, an approximation. In the case where the system is dominated by a large mass with additional bodies of much smaller masses (such as the Sun being the dominant mass of the solar system), it is still possible to develop perturbation expansions where the orbit is close to Keplerian, but the Keplerian orbital elements (wikipedia) undergo a slow variation with time. This is the method used in algorithms such as VSOP87 (Wikipedia)

Q1 - according to Kepler’s first law, See more. “The orbit of every planet is an ellipse with the Sun at one of the two foci.“, yet the center of mass of the sun is always moving around the solar system barycenter. Therefore, as the foci of the planets ellipse moves, does this therefore invalidated the first law, or does the entire ellipse move with the moving foci at the center of the sun?

Q2- If the planet actually orbits the solar system barycenter, why then does Kepler’s first law say otherwise?

Q6: Trajectories look different in different coordinate systems, especially when those coordinate systems vary in time. LRO in Earth Centered and Moon Centered Coordinates

Q7- How are these relative prograde and retrograde motions of the earth on a monthly basis taken into account in the flight path calculations?

Q8- How are the calculations consistent with Kepler’s laws, when the earths flight path through space is not an ellipse, but a complex flower shape?

Rotation and translation of the Moon

The moon makes a rotation movement around its own axis in approximately 27.32 days. This period of rotation is called sidereal month.According to Cassini's first law, this is also the time it takes for the Moon to orbit Earth.

Synchronous rotation is responsible for Earth observers always seeing the same side of the Moon.

For his part, synodic month is the time that elapses between two identical and successive lunar phases.

The synodic month lasts 29.53 days and is due to the fact that the Earth is not still while the Moon orbits around it. For the relative Earth-Sun-Moon positions to be the same again, the Earth must advance 27º in its translational motion around the Sun.

The Moon also travels around the Earth following an elliptical orbit with a very small eccentricity. The eccentricity of an ellipse is a measure of its flattening. This small value means that the Moon's path is almost circular, which it travels at a rate of 1 km / s.

The orbits of the Earth and the Moon intersect at the points called nodes, which make eclipses possible, since seen from Earth the apparent sizes of the Sun and the Moon are very similar.

Converting State Vectors to Keplerian Orbital Elements for Binary Objects

Homework Statement:: I'm working on a personal project to convert objects from a simulation using state vectors for position and velocity to Keplerian orbital elements (semimajor axis, eccentricity, argument of periapsis, etc.). However, the equations I am using do not calculate the eccentricity correctly for binary objects.
Relevant Equations:: mu = gravitation parameter * mass of parent body
momentum = position x velocity
eccentricity = ((velocity x momentum) / mu) - (position / positionMagnitude)
where velocity, mommentum, and position are vectors.

Now, I've been working on this on and off for a while, so thank you for taking the time to read through this. Hopefully someone can give me some insight as to what the problem is, because I'm a computer science major, not a physics major.

The program I am working on converts objects from a program called Universe Sandbox (US) into objects for another program called Space Engine (SE). More specifically, this is for solar systems, real or imagined, and deals with stars, planets, moons, and any celestial objects. US is a simulator and stores information about objects using cartesian state vectors, which can be exported to represent a snapshot of the simulation. SE is an emulator and only uses simple Keplerian elements.

The paper I linked above has worked near-perfect so far and I have had few problems. But eccentricity is not calculated correctly for any binary relationships, and I'm not sure why. Here is the relevant code:

Let me give a little context for this code: What happens here is that every object in a solar system is connected through vectors of "children" and every object has a parent. For instance, a star would be at the top of the hierarchy and a moon would be at the bottom. Objects are sent through this CalcOrbit function from the bottom of the hierarchy (this is why the parent's position/velocity are subtracted from the object's), and their Keplerian elements are calculated as necessary.

Just to be clear, this equation works fine for most objects. But when an object passes through that is in a binary relationship, its "parent" is the barycenter between it and its partner. The barycenter has a total mass of the combined mass of the two objects. When eccentricity is calculated, it often comes out wildly incorrect (often around 0.9).

So I have tried to use other formulas, but had no success (I may have implemented them wrong, though). I have also tried changing the value of mu to its partner's mass, the barycenter's parent's mass, and other values, but none have worked. Unfortunately, I think I don't have the background in physics needed to know what the actual problem is, which is why I'm posting here. Is this problem even possible? Or are binary relationships too complicated to derive an eccentricity from a single snapshot of position/velocity?

If anyone has any ideas about why this doesn't work, or if I haven't provided enough context, please let me know.

The full code can be found here:
But be warned that most of the program is dedicated to identifying objects and their relationships, and much of it I didn't write myself. The relevant part will be the full CalcOrbit function around line 353.

Also, I apologize if this is not the correct forum to post this in. I was afraid there was too much code to post in the regular physics forum. Mods, please feel free to move this to the correct forum or to contact me about where would be the correct place for this post.

1 Answer 1

The answer to this question can be directly extracted from the definition of an osculating orbit.

the osculating orbit of an object in space at a given moment in time is the gravitational Kepler orbit (i.e. an elliptic or other conic one) that it would have around its central body if perturbations were absent.

The previous question that was linked here, How to find instantaneous positions of a central body of specified mass that results in an osculating orbit? , already has an answer explaining that keplerian elements are meaningless unless you know what central body they are referring to.

To calculate the osculating elements, the first thing you therefore need to do is selecting the central body. In the two different sets of element in your question the central bodies selected are thus the SSB and the Sun. Then using the position and mass of these central bodies, you use the position and velocity of the body of interest relative to the central body to solve the IVP for a kepler orbit.

If you check the two sets for the mass of the central body using:

you get respectively 1.9963e30 kg and 1.9984e30 kg. This corresponds pretty much to Sun and Solar system mass.

As you apparently discovered yourself, you can calculate osculating elements with respect to any arbitrary body, but of course it makes zero sense to express the state of Voyager 1 as a Kepler orbit around Earth. In a mathematical sense any arbitrary state can be expressed as if it were a Kepler orbit around any arbitrary central body but we all know that Voyager is not actually orbiting the Earth in some strange hyperbolic orbit.

When it comes to determining the mass of the central body without the provided period, that is not possible. You need either of both to fully define the system.

You do not need something that converts JPL Horizons ephemerides to cartesian coordinates because cartesian coordinates is the system in which the Horizons systems naturally works. Requesting Keplerian elements forces the Horizons system to use a conversion algorithm. That algorithm doesn't quite make sense for the Moon due to the fact that the Moon's mass is over 1% of the Earth's mass.

The easiest way to programmatically obtain cartesian coordinates of a natural solar system object is to use JPL's SPICE Toolkit, which is freely available.

This is an alternate answer, a list of other possibilities or options. @DavidHammen's answer is the best answer to your question.

  1. Caveat, check the jpl-horizions tag. is right. That's the gold. The source. The horses mouth. is a python package that gives you some python-wrapped access to many of those same features and math, and is easy to use.
  2. You can also just choose to get the state vectors (rather than the orbital elements) directly from Horizons like this:

To get the basic position and velocity (state vector) without any light time corrections, make sure it looks like this:

To get units of kilometers and kilometers per second, and ecliptic coordinates, choose these:

But if you want Coordinates compatible with RA and dec, use Earth's Mean Equator (z axis points towards declination of +90 degrees):

OK, so for the moon's position and velocity in ecliptic coordinates with the solar system barycenter as the origin (a good place to start) and to download a csv directly to your computer, make it look like this (note that z is smaller than x or y):

And 04-Jun-2017 00:00 will be (JD, date, x, y, z, vx, vy, vz):

If you chose a Geocentric origin, it will look like this (note x, y, z are all of order 1E+05):

And 04-Jun-2017 00:00 will be (JD, date, x, y, z, vx, vy, vz):

And if you chose the Earth's equator for a reference plane solar system barycenter for origin, it will look like this (note that z is now large):

Natural motion around the Martian moon Phobos: the dynamical substitutes of the Libration Point Orbits in an elliptic three-body problem with gravity harmonics

The Martian moon Phobos is becoming an appealing destination for future scientific missions. The orbital dynamics around this planetary satellite is particularly complex due to the unique combination of both small mass-ratio and length-scale of the Mars–Phobos couple: the resulting sphere of influence of the moon is very close to its surface, therefore both the classical two-body problem and circular restricted three-body problem (CR3BP) do not provide an accurate approximation to describe the spacecraft’s dynamics in the vicinity of Phobos. The aim of this paper is to extend the model of the CR3BP to consider the orbital eccentricity and the highly-inhomogeneous gravity field of Phobos, by incorporating the gravity harmonics series expansion into an elliptic R3BP, named ER3BP-GH. Following this, the dynamical substitutes of the Libration Point Orbits (LPOs) are computed in this more realistic model of the relative dynamics around Phobos, combining methodologies from dynamical systems theory and numerical continuation techniques. Results obtained show that the structure of the periodic and quasi-periodic LPOs differs substantially from the classical case without harmonics. Several potential applications of these natural orbits are presented to enable unique low-cost operations in the proximity of Phobos, such as close-range observation, communication, and passive radiation shielding for human spaceflight. Furthermore, their invariant manifolds are demonstrated to provide high-performance natural landing and take-off pathways to and from Phobos’ surface, and transfers from and to Martian orbits. These orbits could be exploited in upcoming and future space missions targeting the exploration of this Martian moon.

This is a preview of subscription content, access via your institution.

Tidal Energy

Simon P. Neill , M. Reza Hashemi , in Fundamentals of Ocean Renewable Energy , 2018

3.1 Tide Generating Forces

Newton’s Law of Universal Gravitation explains how every object in the universe attracts every other object with a force that is proportional to the product of their masses, and inversely proportional to the square of their distance apart

where F is the force between two masses m1 and m2, G is the gravitational constant (6.674 × 10 −11 m 3 /kg per s 2 ), and r is the distance between the centres of the masses ( Fig. 3.1 ).

Fig. 3.1 . Newton’s Law of Universal Gravitation.

In the case of the Earth-Moon system , the Earth and Moon orbit each other about their common centre of gravity, and so the gravitational attraction is balanced by the outward directed centrifugal force 1 ( Fig. 3.2 ). Because the mass of the Earth is two orders of magnitude greater than the mass of the Moon, the centre of gravity of the Earth-Moon system actually lies within the body of the Earth indeed, it is around 1700 km below the surface of the Earth. The Earth-Moon system turns about this centre of gravity once per month, and so every object on Earth turns with it. The centrifugal force (the dashed arrow directed outward from the Earth on Fig. 3.2 ) is the same everywhere, because all points on Earth experience the same orbital motion. However, the gravitational attraction of the Moon is progressively weaker as we move away from the Moon, as stated by Newton’s Law of Universal Gravitation ( Eq. 3.1 ). The result is that the centrifugal force and the Moon’s gravity cannot balance each other everywhere. The slight imbalances that occur are what cause the tides in the ocean. These imbalances are called the tide generating forces.

Fig. 3.2 . Earth-Moon system. The Earth and Moon attract each other due to Newton’s Law of Universal Gravitation (solid arrows), but this is balanced by the outward directed centrifugal force (dashed arrows).

At the centre of the Earth ( Fig. 3.3 ), the centrifugal force (dashed arrow) is equal and opposite to the Moon’s gravitational pull (solid arrow). At point A (position on the Earth’s surface closest to the Moon), the Moon’s gravity exceeds the centrifugal force, and this results in a net force towards the Moon (block arrow at A). At point B (position on the Earth’s surface furthest from the Moon), the Moon’s gravity is weaker than the centrifugal force. This results in a net force directed away from the moon at B (block arrow at B).

Fig. 3.3 . Net forces in an Earth-Moon system. See text for explanation.

The tide generating forces lead to a deformation in the shape of the ocean, stretching it out in both directions along the Earth-Moon axis ( Fig. 3.4 ). Because the Earth rotates within this ellipsoid, two high waters (and two low waters) are experienced per day. This is known as the equilibrium tidal model. However, there are three main problems with this equilibrium tidal model:

Fig. 3.4 . Vectors on the Earth’s surface indicate the difference between the gravitational force the Moon exerts at a given point on the Earth’s surface, and the force it would exert at the Earth’s centre. These resultant force vectors move water towards the Earth-Moon orbital plane, creating two bulges on opposite sides of the Earth.

the tidal wave cannot move fast enough to keep up with the Earth’s spin and maintain ‘equilibrium’

the equilibrium tide does not account for continents and

the equilibrium tide does not account for Earth’s rotation.

As can be seen in Fig. 3.4 , the equilibrium tide is actually a wave that is stretched around the circumference of the Earth. The wave has two crests: one directly under the Moon, and the other on the opposite side of the Earth. Between these two crests are two troughs. As the Earth spins, the equilibrium tidal model requires the wave to maintain itself with one crest directly under the moon. Tidal waves behave as shallow water waves, with speed c given as

where h is water depth. The mean depth of the world’s oceans is around 4000 m, and so c = 200 m/s. At the equator, the Earth turns at around 460 m/s. Therefore, tidal waves cannot move as fast as the Earth spins.

The equilibrium tide does not account for continents, which hinder the progress of the tidal wave. In addition, water that moves over a rotating Earth is deflected by Coriolis forces, which arise as a result of the Earth’s spin (see Section 3.6 ).

Wave speed is controlled by the local water depth ( Eq. 3.2 ). For a typical shelf sea water depth of h = 200 m, the tidal wave propagates at 44 m/s. Given a tidal period T of 12.42 h (see Section 3.8 ), the wavelength L = cT is around 2000 km.

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Watch the video: Earths motion around the Sun, not as simple as I thought (February 2023).