Astronomy

Circular orbits

Circular orbits


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First of all, I'm studying orbits for a hobby: world building. Unfortunately, my mathematical abilities approach a ridiculous low threshold, which means I am stuck with reading the simplest explanations, which in turn leave me asking tons of fairly basic questions.

Allow me to start with a simple point. I know that Kepler's Laws state that planetary orbits must always be elliptical. I also know that Earth's orbit varies from more elliptical to less elliptical, and that its less elliptical stage is nearly circular.

So… what would happen if Earth did have a circular orbit? Why is it impossible for any planet (or moon, by the way) to orbit another body in a perfectly circular path?


You've been given an answer, and it's perfectly valid, but here's something from a different perspective (less strict).

A circle is really just a particular case of an ellipse. Take an ellipse, and change it, by moving its focal points closer together. When those two points coincide, what you get is a circle. It's still an ellipse, technically - one that happens to have both focal points in the same place, is all.

So yes, you can actually have planetary orbits, or any orbits, circular. There's nothing forbidding that. It's just pretty unlikely that this will occur via a natural process.

As indicated elsewhere, in the real world, all orbits and trajectories are a bit imperfect due to perturbations - whether they be elliptical, circular, parabolic or hyperbolic, they are always a bit perturbed by external factors. In many cases, perturbations are so tiny that you can ignore them.

When a planet is orbiting the Sun, and the orbit is elliptical, the Sun will be in one of those two focal points; the other point has no particular signification. If you could circularize that orbit, then the Sun would be in the center of the circle, of course.

Kepler's laws remain valid for a circular orbit:

  1. The orbit of every planet is an ellipse with the Sun at one of the two foci.

Still true. A circle is an ellipse where the foci coincide.

  1. A line joining a planet and the Sun sweeps out equal areas during equal intervals of time.

Still true. On a circular orbit, the planet moves at constant speed, so the swept area remains constant per time.

  1. The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Still true. The semi-major axis becomes the radius of the circle.


You must understand that Kepler's laws now have more of a historic interest. They are not exactly at the bleeding edge of science anymore. During Kepler's time, it seemed reasonable to state that all orbits must be elliptical (in the strict sense of the term), but now we know that trajectories (including orbits, or closed trajectories) can be circular, elliptical, parabolic or hyperbolic, depending on a few factors.

We also know that perturbations actually deflect all these trajectories a little bit from ideal shapes (but it's usually a very tiny effect).

We also know that relativity makes all "elliptical" orbits more complex - they remain close to elliptic, but the whole ellipse keeps turning around the central star very slowly.

All this stuff was not known during Kepler's time, so take his laws for what they are - a snapshot of the development of our understanding in time.


Why is it impossible for any planet (or moon, by the way) to orbit another body in a perfectly circular path?

One way to look at it is from the perspective of probability and statistics. Think of position and velocity as random variables drawn from some continuous probability distributions. Given some position vector, the velocity vector has to have a very specific value to yield a circular orbit. The probability of drawing a specific value from a well-behaved continuous probability distribution is identically zero.

An even better way to look at it: Even perfectly elliptical orbits aren't possible. Kepler's laws are an approximation that result from assuming a universe that obeys Newtonian mechanics and comprises but two point masses. Newtonian mechanics is only approximately valid in the real universe, objects are lumpy and can only approximately be treated as point masses, and there are a lot more than two objects in the universe.

Suppose that by some fluke chance, an object appears to have a perfectly circular orbit at some point in time (to within measurement error). The non-Newtonian nature of the universe, the non-spherical mass distributions of the objects, and the multiplicity of objects means that a moment later the object will no longer appear to have a perfectly circular orbit.


A circle is an ellipse with eccentricity zero. And in fact tidal evolution can drive orbital eccentricity to values negligibly close to zero. See Regarding the Putative Eccentricity of Charon's Orbit.

From observations we already know Pluto and Charon move about each other in very nearly circular orbits. I expect when New Horizons flies by the system in July, 2015, we will know their orbits more precisely.


The ellipse family includes perfect circles. An ellipse is a dilated circle. Draw a circle on a rubber sheet and stretch it to get an ellipse, pretty easy. Draw an ellipse on a stretched rubber sheet and relax it to see if you get a circle, much harder right. Now lets take the earth's orbit around the sun. The earth isn't perfect sphere, that rotates around a tilted axis and the center of gravity varies depending on the orbital position (season) with respect to the sun. The polar ice caps, the tidal distributions, the clouds all play apart. The moon and other planets also have some effects that vary with position. The moon and sun have their own similar idiosyncrasies. You can probably name many more.
So for most planetary orbits the bottom line is that there are too many variables that must line up exactly all the time to result in a perfect circular orbit and an elliptical orbit is the result.


Orbits

In the vast frictionless environment of space, gravitational attraction can cause two massive bodies to orbit each other.

If the relative velocity between two bodies is too low they will collide. If the velocity is too high they will move away from each other. In between these extremes exist a range of stable orbits that can repeat with almost no change.

Yet, over time orbits degrade. The Moon's orbit gets 38 mm farther away every year because its kinetic energy changes into tidal friction.


Angular momentum and torque

A particle of mass m and velocity v has linear momentum p = mv. The particle may also have angular momentum L with respect to a given point in space. If r is the vector from the point to the particle, then

Notice that angular momentum is always a vector perpendicular to the plane defined by the vectors r and p (or v). For example, if the particle (or a planet) is in a circular orbit, its angular momentum with respect to the centre of the circle is perpendicular to the plane of the orbit and in the direction given by the vector cross product right-hand rule, as shown in Figure 10 . Moreover, since in the case of a circular orbit, r is perpendicular to p (or v), the magnitude of L is simply

The significance of angular momentum arises from its derivative with respect to time, where p has been replaced by mv and the constant m has been factored out. Using the product rule of differential calculus, />

In the first term on the right-hand side of equation ( 46 ), dr/dt is simply the velocity v, leaving v × v. Since the cross product of any vector with itself is always zero, that term drops out, leaving

Here, dv/dt is the acceleration a of the particle. Thus, if equation ( 47 ) is multiplied by m, the left-hand side becomes dL/dt, as in equation ( 45 ), and the right-hand side may be written r × ma. Since, according to Newton’s second law, ma is equal to F, the net force acting on the particle, the result is

Equation ( 48 ) means that any change in the angular momentum of a particle must be produced by a force that is not acting along the same direction as r. One particularly important application is the solar system. Each planet is held in its orbit by its gravitational attraction to the Sun, a force that acts along the vector from the Sun to the planet. Thus, the force of gravity cannot change the angular momentum of any planet with respect to the Sun. Therefore, each planet has constant angular momentum with respect to the Sun. This conclusion is correct even though the real orbits of the planets are not circles but ellipses.

The quantity r × F is called the torque τ. Torque may be thought of as a kind of twisting force, the kind needed to tighten a bolt or to set a body into rotation. Using this definition, equation ( 48 ) may be rewritten

Equation ( 49 ) means that if there is no torque acting on a particle, its angular momentum is constant, or conserved. Suppose, however, that some agent applies a force Fa to the particle resulting in a torque equal to r × Fa. According to Newton’s third law, the particle must apply a force −Fa to the agent. Thus, there is a torque equal to −r × Fa acting on the agent. The torque on the particle causes its angular momentum to change at a rate given by dL/dt = r × Fa. However, the angular momentum La of the agent is changing at the rate dLa/dt = −r × Fa. Therefore, dL/dt + dLa/dt = 0, meaning that the total angular momentum of particle plus agent is constant, or conserved. This principle may be generalized to include all interactions between bodies of any kind, acting by way of forces of any kind. Total angular momentum is always conserved. The law of conservation of angular momentum is one of the most important principles in all of physics.


Researchers identify circular orbits for 74 small exoplanets

The system Kepler-444 formed when the Milky Way galaxy was a youthful two billion years old. The planets were detected from the dimming that occurs when they transit the disc of their parent star, as shown in this artist's conception. Credit: NASA

Viewed from above, our solar system's planetary orbits around the sun resemble rings around a bulls-eye. Each planet, including Earth, keeps to a roughly circular path, always maintaining the same distance from the sun.

For decades, astronomers have wondered whether the solar system's circular orbits might be a rarity in our universe. Now a new analysis suggests that such orbital regularity is instead the norm, at least for systems with planets as small as Earth.

In a paper published in the Astrophysical Journal, researchers from MIT and Aarhus University in Denmark report that 74 exoplanets, located hundreds of light-years away, orbit their respective stars in circular patterns, much like the planets of our solar system.

These 74 exoplanets, which orbit 28 stars, are about the size of Earth, and their circular trajectories stand in stark contrast to those of more massive exoplanets, some of which come extremely close to their stars before hurtling far out in highly eccentric, elongated orbits.

"Twenty years ago, we only knew about our solar system, and everything was circular and so everyone expected circular orbits everywhere," says Vincent Van Eylen, a visiting graduate student in MIT's Department of Physics. "Then we started finding giant exoplanets, and we found suddenly a whole range of eccentricities, so there was an open question about whether this would also hold for smaller planets. We find that for small planets, circular is probably the norm."

Ultimately, Van Eylen says that's good news in the search for life elsewhere. Among other requirements, for a planet to be habitable, it would have to be about the size of Earth—small and compact enough to be made of rock, not gas. If a small planet also maintained a circular orbit, it would be even more hospitable to life, as it would support a stable climate year-round. (In contrast, a planet with a more eccentric orbit might experience dramatic swings in climate as it orbited close in, then far out from its star.)

"If eccentric orbits are common for habitable planets, that would be quite a worry for life, because they would have such a large range of climate properties," Van Eylen says. "But what we find is, probably we don't have to worry too much because circular cases are fairly common."

In the past, researchers have calculated the orbital eccentricities of large, "gas giant" exoplanets using radial velocity—a technique that measures a star's movement. As a planet orbits a star, its gravitational force will tug on the star, causing it to move in a pattern that reflects the planet's orbit. However, the technique is most successful for larger planets, as they exert enough gravitational pull to influence their stars.

Researchers commonly find smaller planets by using a transit-detecting method, in which they study the light given off by a star, in search of dips in starlight that signify when a planet crosses, or "transits," in front of that star, momentarily diminishing its light. Ordinarily, this method only illuminates a planet's existence, not its orbit. But Van Eylen and his colleague Simon Albrecht, of Aarhus University, devised a way to glean orbital information from stellar transit data.

They first reasoned that if they knew the mass and radius of a planet's star, they could calculate how long a planet would take to orbit that star, if its orbit were circular. The mass and radius of a star determines its gravitational pull, which in turn influences how fast a planet travels around the star.

By calculating a planet's orbital velocity in a circular orbit, they could then estimate a transit's duration—how long a planet would take to cross in front of a star. If the calculated transit matched an actual transit, the researchers reasoned that the planet's orbit must be circular. If the transit were longer or shorter, the orbit must be more elongated, or eccentric.

To obtain actual transit data, the team looked through data collected over the past four years by NASA's Kepler telescope—a space observatory that surveys a slice of the sky in search of habitable planets. The telescope has monitored the brightness of over 145,000 stars, only a fraction of which have been characterized in any detail.

The team chose to concentrate on 28 stars for which mass and radius have previously been measured, using asteroseismology—a technique that measures stellar pulsations, which reflect a star's mass and radius.

These 28 stars host multiplanet systems—74 exoplanets in all. The researchers obtained Kepler data for each exoplanet, looking not only for the occurrence of transits, but also their duration. Given the mass and radius of the host stars, the team calculated each planet's transit duration if its orbit were circular, then compared the estimated transit durations with actual transit durations from Kepler data.

Across the board, Van Eylen and Albrecht found the calculated and actual transit durations matched, suggesting that all 74 exoplanets maintain circular, not eccentric, orbits.

"We found that most of them matched pretty closely, which means they're pretty close to being circular," Van Eylen says. "We are very certain that if very high eccentricities were common, we would've seen that, which we don't."

Van Eylen says the orbital results for these smaller planets may eventually help to explain why larger planets have more extreme orbits.

"We want to understand why some exoplanets have extremely eccentric orbits, while in other cases, such as the solar system, planets orbit mostly circularly," Van Eylen says. "This is one of the first times we've reliably measured the eccentricities of small planets, and it's exciting to see they are different from the giant planets, but similar to the solar system."

David Kipping, an astronomer at the Harvard-Smithsonian Center for Astrophysics, notes that Van Eylen's sample of 74 exoplanets is a relatively small slice, considering the hundreds of thousands of stars in the sky.

"I think that the evidence for smaller planets having more circular orbits is presently tentative," says Kipping, who was not involved in the research. "It prompts us to investigate this question in more detail and see whether this is indeed a universal trend, or a feature of the small sample considered."

In regard to our own solar system, Kipping speculates that with a larger sample of planetary systems, "one might investigate eccentricity as a function of multiplicity, and see whether the solar system's eight planets are typical or not."


Are circular orbits normal

The objects cannot be stationary relative to each other (zero tangential velocity) and in orbit around each other (non-zero tangential velocity) at the same time.

The ellipse is the general case, a circle is a special ellipse. In practice an orbit will never be an exact circle, but it can be a good approximation to it.

Only to a good approximation. You won't swing it in a perfect circle, the rope is elastic and so on. That is the point. Sometimes the approximations are better, sometimes they are worse.

Unlike a rope, gravity has nothing that would prefer a fixed distance.

Let's try to go back to two-body problem, where second body has a negligible mass compared to the first one. Let's imagine that this system is somehow miraculously isolated from rest of the universe. Now you can imagine that the trajectory of the second body around the first one will depend on some initial conditions, namely the initial position and velocity of the second body. Let's consider the first body at rest all the time. Now just considering kinematics of uniform circular motion, for each orbit (distance from the center of the first body) exists only single value of speed of the second body which allows its circular motion. And more over the direction of its initial movement must by exactly tangential, i.e exactly at right angle with radial direction (toward the center of the first body). Whatever else initial velocity will not lead to a circular motion (ellipse, parabola and hyperbola are possible depending on the total energy of the system).

As you can see, we made some approximations and considered unrealistic isolated system and even though the probability of second body orbiting the first one in an exact circle is very low. Now you can ask yourself, how probable is this to happen when planets are being formed in any real stellar system, where all bodies and dust interact with each other.


(F) = centripetal force [N] vector
(a) = centripetal acceleration [m/s²] vector
(v) = tangential velocity [m/s] vector
(r) = radius of the circular path [m] vector
(m) = mass [kg]

The term centripetal means pointing at the center. In all circular motion force and acceleration always points at the center of the circle. This is often confused with centrifugal, which means away from the center.

Tangential means touching at only one point. Since the velocity is perpendicular to the centripetal force, it doesn't enter or exit the circle.

Yes, the ball is accelerating towards the center of the circle.

It's possible to produce an acceleration similar to gravity without a gravitational field. It's done by rotating a room to produce a centripetal acceleration.

Example: You are designing a rotating spaceship with artificial gravity. In your plans, the ship is shaped like a wheel with a diameter of 50.0 m. How fast does the outer edge have to move to produce a centripetal acceleration equal to the gravity felt on Earth's surface? solution $d = 2r quad quad r = 25,mathrm$ $a = frac>$ $sqrt = v$ $sqrt <(9.8)(25)>= v$ $15.7 ,mathrm< frac> =v$
Example: Use the circumference of a circle and the tangential velocity of the ship to calculate how long one rotation will take. hint $C = 2 pi r$ $v=frac$ $v=frac<2 pi r>$ solution

Your 50 meter spaceship design was built. It produced the illusion of 9.8 m/s² of gravity, but people are complaining they feel strange when standing up.

Example: Use the circumference of a circle at a person's head and the time for one rotation to calculate the velocity at the top of a 2 meter tall person's head. strategy

A person on the rotating spaceship will have their feet on the outer edge, and their heads pointed towards the center. This means the head will have a shorter radius and a slower speed.

$d = 2r quad d = 50 , mathrm$ $r = 25-2 = 23, mathrm$

We also know that the total time for one rotation must be the same at any distance from the center. We calculated the time for one rotation in the previous example.

We already calculated that one rotation takes 10 seconds.

$d = 2r quad d = 50 , mathrm$ $r = 25-2 = 23, mathrm$
$v=frac quad C = 2 pi r$ $v=frac<2 pi r>$ $v=frac<2 pi (23)><(10)>$ $v=14.45, mathrm< frac>$
Example: What acceleration would a 2 meter tall person feel at their head? Compare that to the acceleration felt at their feet. solution $a = frac>$ $a = frac<14.45^<2>><23>$ $a = 9.07,mathrm< frac>$
$frac<9.07> <9.8>= 0.92$

A person's head only feels 92% of the gravity at their feet.


Question: What changes could be made to the spaceship design to reduce the difference in acceleration between a person's head and feet? answer

The ship's radius could be increased, but that would increase the costs.

The ship could produce a lower acceleration, maybe half of Earth's gravity.


Circular orbits - Astronomy

The force needed by a body of mass m, to keep in a circular motion at a distance R, from the centre of a circle with velocity v, is the centripetal force Fc, where

The direction of the force is towards the centre of the circle of motion, and its magnitude and direction can both be derived from a consideration of Newton’s second law of motion.

In astronomy many stars, planets and disks of material move in circular orbits and require a force equivalent to the centripetal force to maintain their circular motion. This force is usually gravity. By balancing the gravitational and centripetal forces it is possible to obtain estimates of the mass within a given radius from the rotation curves of galaxies or accretion disks around supermassive black holes.

When we sit on a merry-go-round or in a car taking a corner we “feel” the centrifugal force pulling us away from the centre of our circular motion which has the same magnitude but opposite sign as the centripedal force. This “pseudo-force” should not be confused with the reality of the centripetal force, but arises because of Newton’s third law of motion: “For every action, there is an equal, but opposite, reaction”. The centrifugal force is a pseudo-force because if the centripetal force ceased for an object in circular motion, the centrifugal force the body is “feeling” would instantly disappear, and the object would travel tangentially to its line of motion. It only arises because the body is in a non-inertial frame of reference.

Study Astronomy Online at Swinburne University
All material is © Swinburne University of Technology except where indicated.


If you added up all the energy you would need for each piece, you would derive the Gravitational Binding Energy for the body: This depends on two quantities: the Mass (M) and the Radius (R) of the body.

The formula above is a "proportionality", it tells us how the binding energy scales with the mass and radius of the object. The constant out front we need depends on details of how matter is distributed in the object. For example, a sphere of uniform density would have a constant of 3/5.

For the Earth, the Gravitational Binding Energy is about 2x10 32 Joules, or about 12 days of the Sun's total energy output!


I was reading and came across an article which stated that all the planets orbitted the sun elliptically while the earth was the only planet to orbit the sun in a circular motion.


Is this true.
Also what technology did we use to figure this out??

All the planets, including the Earth, orbit in ellipses. Neptune and Venus actually orbit closer to a circle than the Earth does.

As to technology, telescopes.

No orbit, including the Earth's, is perfectly circular. The Earth's orbit is, in fact, an ellipse. The distance between the Earth and the Sun varies by about 5,000,000 km throughout the year.

The very concept of a "perfectly circular orbit" is flawed, since the circle is not a stable orbit. Stability implies self-correction: if you perturb a stable system, it naturally returns to the same stable state. A circular orbit is not stable, because any perturbation, even a passing comet, would irreversibly perturb the orbit into an ellipse.

Well, SpaceTiger, I was trying to make it easy to understand: if you start with a circle, any perturbation, no matter how small, leaves you with an ellipse (or other conic section, I suppose).

I read that earths orbit was significantly different from that of all the other planets..How true is this statement.

I read that earths orbit was significantly different from that of all the other planets..How true is this statement.

Not true at all. Earth's orbit is slightly elliptical, close enough to circular that you can't tell it's an ellipse unless you pay close attention. And that pretty much describes the orbits of the other (traditional) planets too. Some are a little more elliptical, others a little less, but there isn't any deep pattern, it bears all the earmarks of accidental initial conditions just randomly distribution a small varying eccentricity on the various planets. Oh, Jupiter and Saturn have a certain "tide locking", and the Earth and Moon show another, but that again is slight and historically contingent, not "deep".


Circular orbits - Astronomy

This is a beta 3.4a version of the Circular Orbit program written by TeJaun RiChard, a senior at Shaw High School in East Cleveland, Ohio, during a shadowing experience in July 2012 for Future Connections, Ian Breyfogle, a senior at Kent State University, and Tom Benson from NASA Glenn. You are invited to participate in the beta testing. If you find errors in the program or would like to suggest improvements, please send an e-mail to [email protected]

Due to IT security concerns, many users are currently experiencing problems running NASA Glenn educational applets. There are security settings that you can adjust that may correct this problem.

With this software you can investigate how a satellite orbits a planet by changing the values of different orbital parameters. The velocity needed to remain in a circular orbit depends on the altitude at which you orbit, and the gravitational pull of the planet that you are orbiting. Using this simulator, you can study these effects.

There are two versions of CircularOrbit which require different levels of experience with the package, knowledge of orbital mechanics, and computer technology. This web page contains the on-line student version of the program. It includes an on-line user's manual which describes the various options available in the program and includes hyperlinks to pages in the Beginner's Guide to Rockets describing the math and science of rockets. More experienced users can select a version of the program which does not include these instructions and loads faster on your computer. You can download these versions of the program to your computer by clicking on this yellow button:

The program is provided as Corbit.zip. You must save this file on your hard drive and "Extract" the necessary files from Corbit.zip. Click on "Corbit.html" to launch your browser and load the program. With the downloaded version, you can run the program off-line and do not have to be connected to the Internet.

If you see only a grey box at the top of this page, be sure that Java is enabled in your browser. If Java is enabled, and you are using the Windows XP operating system, you may need to get a newer version of Java. Go to this link: http://www.java.com/en/index.jsp, try the "Download It Now" button, and then select "Yes" when the download box from Sun pops up.

    Information is presented to you using labels. A label has a descriptive word displayed in a colored box. Some labels give instructions for the next phase of design and launch, some labels express the state of the calculations.

  1. The Blue "Compute" button causes the program to calculate the orbital velocity, altitude and period based on current input values.
  2. White buttons are possible input selections that you can choose. Clicking on a selection button causes the button to turn Yellow and enables the input slider and text box.
  1. A white box with black numbers is an input box and you can change the value of the number. To change the value in an input box, select the box by moving the cursor into the box and clicking the mouse, then backspace over the old number, enter a new number.
  2. A black box with colored numbers is an output box and the value is computed by the program.

The program screen is divided into two main parts:

  1. On the left of the screen are the control buttons, labels, input box that you use to change orbital parameters. Details of the Input Variables are given below.
  2. On the right of the screen is the graphics window in which you will see the satellite orbiting the planet. Details are given in Graphics.

You move the graphic within the view window by moving your cursor into the window, hold down the left mouse button and drag to a new location. You can change the size of the graphic by moving the "Zoom" widget in the same way. If you loose your picture, or want to return to the default settings, click on the "Find" button at the bottom of the view window.

The speed needed to orbit a planet depends on the altitude above the planet and on the gravitational acceleration produced by the planet. A formula describing this relation was developed by Johannes Kepler in the early 1600's:

where V is the velocity for a circular orbit, g0 is the surface gravitational constant of the Earth (32.2 ft/sec^2), Re is the mean Earth radius (3963 miles), and h is the height of the orbit in miles. If the rocket was launched from the Moon or Mars, the rocket would require a different orbital velocity because of the different planetary radius and gravitational constant. For a 100 mile high orbit around the Earth, the orbital velocity is 17,478 mph. Knowing the velocity and the radius of the circular orbit, we can also calculate the time needed to complete an orbit. This time is called the orbital period.

T^2 = (4 * pi^2 * (Re + h)^3) / (g0 * Re^2)

Looking at these equations, we see that as the height above the planet increases, the velocity needed to maintain an orbit decreases. A spacecraft flying at a lower orbit must travel faster than a spacecraft flying at a higher orbit.

Input variables are located on the left side of the screen. You can select the planet by using the choice button. Click on the menu and drag to select any planet in the solar system or the Earth's Moon.. The corresponding gravitational constant and planet radius is displayed below the choice buttons. Calculations and input can be entered in either English (Imperial) or Metric units by using the "Units" choice button. The period of the orbit can be expressed in "Minutes", "Hours", or "Days" by using the choice button located next to the "Compute" button. You then select the desired input variable: altitude, velocity, or time by using the choice button. Set the value for the desired input by using the input box and then push the "Compute" button which sends the information to the computer, performs the calculations and displays the results. You can also set the value of altitude, velocity, or time by using the sliders located next to the input boxes. When using the sliders, you do not have to press the "Compute" button. You can change the maximum altitude on the sliders by using the input box at the bottom of the panel. Changing the maximum altitude automatically changes the minimum velocity and maximum time.

We will continue to improve and update CircularOrbit based on your input. The history of changes is included here:

  1. On 22 Nov 13, version 3.4a was released. This version of the program allows the user to specify their own planet for orbiting. You can change the values of the planet radius and gravitational constant using either sliders or text boxes.
  2. On 6 Aug 12, version 3.3 was released. This version of the program includes photos of the planets of the solar system. It also corrects some small problems in the operation of the input boxes. Versions 3.0- 3.2 were development versions and were not released to the public.
  3. On 27 Jul 12, version 2.7 was released. This version of the program includes slider input for the altitude, velocity and time. Versions 2.0- 2.6 were development versions and were not released to the public
  4. On 20 Jul 12, version 1.8 was released. This version of the program includes the graphics output and includes all the planets of the solar system and the Earth's Moon. Input for this version was limited to input boxes. Versions 1.1- 1.7 were development versions and were not released to the public
  5. On 10 Nov 05, version 1.0 of Orbit Calculator was released. This version did not include graphics on only solved for the orbital velocity around the Earth, Moon, and Mars.

Notice that orbital flight is a combination of altitude and horizontal velocity. The recent Space Ship 1 flight acquired the necessary altitude to "go into space", but lacked the horizontal velocity needed to "go into orbit".


Watch the video: GCSE Science Revision Physics Orbital Motion Triple (June 2022).