We are searching data for your request:

**Forums and discussions:**

**Manuals and reference books:**

**Data from registers:**

**Wait the end of the search in all databases.**

Upon completion, a link will appear to access the found materials.

Upon completion, a link will appear to access the found materials.

A satellite which is moving around a planet of mass $M$ in a circular orbit of radius $R$ is given a sudden thrust into the center of the planet, so that it deviates an angle $a$.

Find the latus-rectum $l$ and the eccentricity $e$ of the new orbit.

As a hint: a force directed inward does not change the angular momentum, and therefore does not change the semi-major axis nor the period. It's not clear which angle you're talking about, but knowing that the semi-major axis did not change should let you set the problem up as a simple geometry problem.

As mentioned in a comment, there's not enough information in the question (version 1.0) to solve the problem. To provide enough information, one needs to know uniquely specify where the two points in question lie on the orbit.

The standard way to do this is to use the true anomaly, the angular displacement between the point in question and the periapsis point, with the angle measured through the focus at which the central mass lies, and with angle measured positively in the direction of the orbital travel.

Before I get to the standard way, I'll look at the short and easy way. Suppose you know the mean anomaly for those two points, $M_1$ (initial) and $M_2$ (final). Then the time of flight from point 1 to point 2 is $frac

All you need to do then is to know how to transform those true anomaly values to mean anomaly values. This is done in two stages for each true anomaly value. First compute the eccentric anomaly via $ an frac E2 = sqrt

There are some gotchas in the above, which is that the expression $ an frac E2 = sqrt

Suppose, for sake of simplicity, that $mu$ is one length unit 3 per time unit 2 , and the semi-major axis is one length unit, and the eccentricity is 1/3. The mean motion is thus $sqrt

## Asteroid Orbit Determination

Near-earth asteroids, also known as “minor planets,” orbit the sun at roughly the same distance as Earth. Collisions with them are catastrophic – just ask the dinosaurs – so being able to predict their future position is an important application of vector calculus.

On the first day, participants learn celestial coordinates, and how to interpret an ephemeris to select a near-earth asteroid to study. Each team of three then writes an “observing proposal” similar to what an astronomer would submit to an observatory. On the third night, teams begin observing runs at the telescope.

After each run, the team locates the asteroid’s faint dot among the background stars (not always easy to do), then precisely measures its position relative to surrounding stars. Once they have at least three or four observations taken on different nights, they write software in Python to calculate the asteroid’s position and velocity vectors, then transform them into the six orbital elements that characterize the asteroid’s orbital ellipse, using numerical differentiation.

Each team performs every step themselves: choosing their asteroid, pointing the telescope, taking images, reducing the data, calculating the orbit. Some go on to improve the accuracy of their calculated orbits using additional observations to make differential corrections. Another option is to use Visual Python to make an animation of their asteroid orbiting the sun!

Each team’s observations are submitted to the Minor Planet Center of the International Astronomical Union, and used to improve future predictions of the asteroid’s position.

## Question: calculating orbit using another orbit - Astronomy

The Shenzhou-12 manned spaceship atop its Long March-2F carrier rocket is transferred to the launching area of Jiuquan Satellite Launch Center in northwest China, June 9, 2021. /Xinhua

**Editor's note:** Zhang Fan, PhD, is an associate professor of Astronomy at Beijing Normal University, where he serves as deputy director of the Gravitational Wave and Cosmology Laboratory. The article reflects the author's opinions and not necessarily the views of CGTN.

China's Shenzhou-12 mission has taken three astronauts up to join the Tiangong space station. They will serve as the first construction crew, and will carry out spacewalks to put the finishing touches on the exterior of the station and do some interior decorating. Such tasks have to be done in space, since the core module cannot be fully completed on the ground, or else it won't fit into the rocket, odds and ends inside might also become shaken loose during launch.

The construction schedule is intense, with the entire 90-tonne multi-chamber station slated for completion next year, to be achieved via 11 launch missions from two sites in China. Such an ambitious plan is made possible by the automations facilitated by the ingeniously designed robotic arms, one of which can crawl across the outside of the station and combine with another arm, substantially extending its reach. Newly-developed human hand mimicking robots may also find uses, offering the dexterity that may help reduce the need for dangerous and time-consuming space walks. Those experiences learned and disseminated by the International Space Station also play no small part, so its decommissioning in the near future would be a sad event for all, and hopefully avoided.

The reason we want to maintain a crewed permanent space facility is because there are things we can do there that we simply cannot do on the ground. At the moment, that mostly involves scientific experiments and observations. For example, because the Earth's atmosphere is not transparent to most frequency bands of light, we will have to go to space in order to see in X-ray, gamma-ray and many other colors of rays.

This undated photo shows taikonauts Nie Haisheng (C), Liu Boming (R) and Tang Hongbo, the crew for the Shenzhou-12 manned spaceflight mission. /Xinhua

China's Tiangong space station in particular, will be joined by a telescope that flies in rough formation, and approaches the station occasionally for re-fueling and repairs. There could even be a phase two, where a swarm of large mirrors use equipment onboard the station as focal point, to form an extremely large composite telescope, that would allow us to peer into the abyss of deeper space.

In the near future, microgravity manufacturing will likely be all the rage. Without pesky gravity constantly trying to pull things down and splatter liquids, materials cure and settle rather differently, offering up miracle properties that we crave. Also, while the human body still works in space, it won't work exactly the same way – just extrapolate from reversing the effect of gravity by hanging upside down on a tree.

Microgravity consequently offers exciting therapeutical possibilities. The transportation cost to and from space is the main obstacle here, but maybe not for much longer, once we have reusable rockets powered by hydrogen and oxygen split from water using solar power. It is therefore profitable that the current space station design be modular, and expandable into larger facilities suitable for scaled-up productions.

In the far future, once we manage to capture asteroids and guide them in Earth's orbit, we will be able to refit the station into a mining rig and use the resources extracted to carry out large scale construction in orbit, to build spaceships that don't need to be launched from the ground, so can in principle be of arbitrary sizes. These prospects may seem fantastical, but we live in interesting times, when incremental advancements and breakthroughs in multiple disciplines converge to make the stuff of science fiction plausible.

It is now an opportune time to anticipate and prepare for giant leaps in our technological capabilities, least to fend off potential adverse effects and pave the way for reaping maximal benefits. Take for example a possible new problem, that if a private commercial entity buys the decommissioned International Space Station and transfers its headquarters into something else there, renders its services through the internet beamed from space, and receives bitcoin as payment, thereby bypassing all the traditional physical and intangible infrastructures of Earthbound countries, then what exactly is its legal status? People may need to figure that out sooner than they think.

## Collaboration

JPL is a federally funded research center in Pasadena that carries out robotic space and earth-science missions and implements programs in planetary exploration, earth science, space-based astronomy, and technology development. JPL is managed for NASA by Caltech. The outdoor installation is the brainchild of Dan Goods and David Delgado, visual strategists at NASA’s Jet Propulsion Laboratory who worked in collaboration with composer Shane Myrbeck and architect Jason Klimoski of StudioKCA to produce the innovative “soundscape” experience.

The origin of a collaboration between The Huntington and NASA's Jet Propulsion Laboratory (JPL) stretches back almost a century. In the early 1900s, a handful of visionaries imagined Southern California as a global hub of learning across the humanities, arts, and sciences. One was Henry E. Huntington, who was developing his exceptional library. Another was solar astronomer George Ellery Hale, who played a central role in creating the California Institute of Technology. Caltech's pioneering work in rocket science would eventually lead to the creation of JPL. Hale convinced Huntington to envision his library as a nexus of research in history, literature, and art.

The Huntington Library today holds one of the largest history of science collections in North America. Highlights include Galileo's Sidereus Nuncius, which featured the first engravings of the moon peered through the newfangled telescope, and a 1923 logbook from astronomer Edwin Hubble on his observations from a telescope atop Mt. Wilson. These objects and others are on view in the permanent exhibition "Beautiful Science: Ideas That Changed the World" in The Huntington's Dibner Hall of the History of Science. The "Orbit Pavilion" at The Huntington celebrates the relationship between these two research-based institutions, the dynamic fulfillment of a vision first proposed almost 100 years ago.

## 1 Answer 1

### Rotation and Units

Rotation in Three.JS is measured in *radians*. For those that are completely unfamiliar with radians (a small excerpt from an old paper of mine):

Like the mathematical constant

Pi, aradian(roughly 57.3 degrees) is derived from the relationship between a circle's radius (or diameter) and its circumference. One radian istheangle which will always span an arc on the circumference of a circle which is equal in length to the radius of that same circle (true for any circle, regardless of size). Similarly, Pi is the ratio of circumference over diameter, such that the circumference of the unit circle is precisely Pi. Radians and degrees are not actuallytrueunits, in fact angles are in general dimensionless (like percentages and fractions, we do not use actual units to describe them).However, unlike the degree, the radian was

notdefined arbitrarily, making it the more natural choice in most cases often times being much easier and much more elegant, clear, and concise than using degrees in mathematical formulae. The Babylonians probably gave us the degree, dividing their circle into 6 equal sections (using the angle of an equilateral triangle). each of these 6 sections were probably further subdivided into 60 equal parts given theirsexagesimal(base 60) number system. This would also allow them to use such a system for astronomy because the estimated number of days in a year was much less accurate during their time and was often considered 360.

### Basic Rotation in Three.JS

So now, knowing you're working in radians, if you increment using the **first** of the following statements **once** in your anim function (callback to requestAnimFrame ), you will be incrementing the rotation of mesh on the x-axis by one radian

As the *last* two of the above statements show we can use Math.PI / 180 to easily convert a value in degrees into radians before the assignment if we wish to use degrees instead.

### Taking Framerate Into Account

In your case, you need to take into consideration how much time passes with each frame. This is your *delta*. You have to think of it like this: How many FPS are we running at? We'll declare a global clock variable which will store a THREE.Clock object which has an interface to the information we require. We need a global variable we'll call clock (needs to be accessible in other functions, specifically anim ):

Within init , create an instance of THREE.Clock storing it in the variable declared outside init (with a greater scope):

Then, in your anim function, you'll make two calls that will update two variables associated with clock :

- time (total elapsed time in milliseconds since the clock was instantiated)
- delta (time in milliseconds between each frame) in two other global variables:

Note that delta is *meant* to return the amount of time between each frame however, this will be true if and *only* if clock.getDelta is consistently called within anim / render

*Only*once each animation/render cycle- In the same place each animation cycle (beginning or end, which one shouldn't matter as far as I know)

The above conditions are a result of the THREE.Clock implementation. getDelta initially returns the amount of time since the clock was instantiated, afterwards the time it returns is simply the time since it was last called). If it somehow gets called mistakenly or inconsistently it's going to screw things up.

### Rotating With A Delta Factor

Now if your scene doesn't bog down the processor or GPU, Three.JS and it's included **requestAnimationFrame shim** will try (working with the available resources) to keep things running at a smooth 60 frames per second. This means ideally we will have approximately 1/60 = .016666 seconds between each frame, this is your delta value which you can read from the clock each frame and use it to normalize your speed based on the framerate by multiplying as shown below. This way you can get a value in terms of seconds regardless of small variations in the framerate which you can multiply each time in order to get a value in terms of seconds.

So, based on what we had at the beginning in your anim function you can use it like so:

### Rotational Speed and Units

Because our measures of angles, radians and degrees are not actually units then when we look at our units for angular velocity we will see that it is going to function of *only* time (rather than as a function of distance and time like you have in your code).

### Calculating Rotational Speeds Based On Time

As for your specific case, you don't need the radius to calculate the rotational speed (angular velocity), instead you can use the number of hours in a day (the amount of time it takes for a complete revolution, ie. 2 * Math.PI *radians* of rotation on it's axis). If you have a variable called revolutionTime then you can calculate it like so.

If you assume Earth has 24 hours = 24 * 60 * 60 = 86,400 in a day (it doesn't). Then we will get rotationalSpeed = 2 * PI / 86,400 , or roughly 0.0000727 *radians per second*. You should be able to find textbook values which may be more accurate than this (taking into account a more accurate measurement than our 24 hour flat figure for the amount of time it takes for Earth to complete a revolution.

### Considering Your Case Particularly

However, I wouldn't worry about making sure you have all of the angular velocities for the planets exactly correct. Instead, a better idea would be to work out what the ratios between each of the angular velocities (of the planets) are and use that. This will work better for a few reasons: You will likely want a faster rotational speed, this will allow you to use whatever rotational speed works well the important thing, as with any model (especially when it comes to astronomical models), is that you keep it to scale.

## Orbiter Space Flight Simulator 2016 Edition

Fed up with space games that insult your intelligence and violate every law of physics? Orbiter is a simulator that gives you an idea what space flight really feels like - today and in the not so distant future. And best of all: you can download it for free!

Launch the Space Shuttle from Kennedy Space Center and rendezvous with the International Space Station.

Recreate historic flights with addon spacecraft packages: Mercury, Gemini, Apollo, Vostok and more.

Plan interplanetary slingshots and tour the solar system with futuristic spacecraft.

Find and explore new worlds. Orbiter contains high-resolution models of many celestial bodies.

Design your own rockets, or download addons created by other users.

Learn about the concepts of space flight and orbital mechanics by playing and experimenting.

You are the commander of your spacecraft. Welcome to the flight deck!

Planetary bodies now support terrain elevation maps for modelling mountain ranges.

Write your own Orbiter plugin modules, and learn the basics of C++ programming along the way.

## Hyperbolic (and Parabolic?) Orbits

Imagining the satellite as a particle sliding around in a frictionless well representing the potential energy as pictured above, one can see how both circular and elliptical orbits might occur.

(*Optional*: More formally, we solved the equation of motion at the end of these earlier notes to find [ dfrac <1>

which is equivalent to the equation for an ellipse [ dfrac

However, that is not the whole story: what if a rogue planet comes flying towards the Solar System from outer space? What kind of orbit will it follow as it encounters the Sun&rsquos gravity? In fact, our analysis of the equations of motion is equally valid in this case, and the ( (r, , heta) ) equation is the same as that above! The new wrinkle is that *e*, which is always less than one for an ellipse, becomes greater than one, and this means that for some angles *r* can be infinite (the right-hand side of the above equation can be zero). The orbit is a *hyperbola*: the rogue comes in almost along a straight line at large distances, the Sun&rsquos gravity causes it to deviate, it swings around the Sun, then recedes tending to another straight line path as it leaves the System.

There is also the theoretical possibility of a parabolic orbit, going out to infinity but never approaching a straight line asymptote. However, this requires *exactly* the correct energy&mdashthe slightest difference would turn it into a very long ellipse or a hyperbola. In practice, of course, this delicate energy tuning would be upset by gravitational attraction from other planets.

## How do aphelion and perihelion compare to one another?

Aphelion is the point in the orbit of a Planet when it is farthest away from it's Star.

Perihelion is the point in a Planet's orbit when it is closest to it's Star.

The Average distance between the Earth and the Sun is about 149 million km.Average because the orbit of Earth around the Sun is not a perfect circle but an ellipse.

The Earth's Aphelion distance is about 152 million km with a Perihelion distance of about 146 million km.

You can calculate the Perihelion and Aphelion of any object if you know the eccentricity of the Object's orbit and the average distance between that object and the sun also known as the Semi-Major axis.

Aphelion and Perihelion distances can be easily calculated using the following formulas.

a is the Average distance between the star and the Object aka Semi-major axis.

## 3 Answers 3

I think these are two separate questions that should be approached separately.

1) "Why isn't $m$ in in the first equation?" The mass of a body does change the force acting on it. But the mass of a body also changes its acceleration. If you increase the mass of an object it feels a larger force, but it's also harder to move. The equation for gravitational force is $F = displaystyle GmM/

2) "Why is $m$ in the second equation?" Think about the moon and the earth. The earth is pulling on the moon, but the moon is also pulling on the earth! The two bodies actually orbit around their common center of mass. This is important for the relative velocity: we need to add how fast the earth is orbiting to how fast the moon is orbiting. That's why you have the $m$ term.