If I have a binary object system of equal masses, why don't they always have the same circular orbit around their center of mass, like in the photo?
There's no reason why 2 bodies of equal mass couldn't have elliptical orbits around each other. There's an example of that here
The simple way to think about this is, if two bodies of similar mass approach each other, one of two things can happen, they either have sufficient velocity to pass each other with some hyperbolic curving of both objects or there's enough gravitational attraction that they capture each other in orbit, but the shape or eccentricity of the orbit depends on the ratio between the tangential velocity and the orbital speed at the closest pass. For a circular capture the ratio needs to be exactly 1, and that kind of exactness is rare, so we nearly always get an elliptical orbit with an orbital capture.
Solar-system planet or moon formation (not moon capture) tends to be much more circular, because elliptical orbits of objects tends to cancel out as large bodies coalesce, but you're never going to get 2 similar mass objects orbiting each other in a condensing gas cloud. Most of the mass inevitably collects in the gravitational center. There can be, however, a tidal influence that circularizes orbits over time, never reaching a full circle, cause that's impossible, but slowly becoming more circular.
Why don't the stars in a binary star system of equal masses always orbit their center of mass in a circular orbit? - Astronomy
I. Binary Stars -- Roughly half of the points of light we see in the night sky are actually two stars that orbit each other, too close together for our eyes to be able to resolve them separately -- Sometimes the two stars are close enough that they can interact with each other, usually by exchanging some mass, which alters their evolution as stars. These are called "close pairs". -- Other binaries are widely separated, so can be considered to be two independent stars that just happen to be orbiting each other. These are "wide pairs". -- There are three types of binaries: visual, which means you can actually see the two stars in a telescope (no orbiting binaries have a wide enough separation to be seen with the naked eye) spectroscopic, which means you can see the presence of the orbit due to the Doppler shifting of the stellar spectral lines and eclipsing, which means you know there is a binary there because one gets in the other's way periodically during its orbit. Surprisigly, it was this last method that was first used to suggest that binary stars were possible, but their existence was not immediately accepted.
II. Visual Binaries -- visual binaries include Mizar in the Big Dipper (though don't be confused by nearby Alcor-- you can see these two with the naked eye, so they are not orbiting each other), and Castor (alpha Geminorum). Castor had an important place in binary star discovery, because Herschel used it in 1801 to demonstrate the binary-star orbit and prove it was indeed a binary. -- the orbital period of visual binaries is very long (recall Kepler's law, and note that to be visual, the separation must be huge). For example, the two stars in Castor (called the primary and the secondary) orbit each other every 467 years. -- visual binaries are very important because they allow us to determine the mass of the system. Remember that the force of gravity depends on the masses of both stars, and by looking at the orbit, we can tell what the force of gravity has to be. It turns out that what can be derived is the sum of the two masses (not their product). Without this information, we would not have been able to determine the relationship between mass and luminosity for stars in the H-R diagram.
III. Spectroscopic and Eclipsing Binaries -- spectroscopic binaries are seen from their line spectrum, by following the Doppler shifts as the stars orbit each other. -- to see these, you need big velocities, not big separations. That means you need the stars to be close to each other, not far away. So these types of observations can detect exactly the stars that visual techniques cannot. -- Typical periods of spectroscopic binaries are measured in days, not years. The orbital speeds are tens of kilometers per second, similar to Earth's orbital speed around the Sun. -- eclipsing binaries can only be seen when we view the orbit edge-on, so that the stars will occasionally get in each other's way. So only a small fraction of binaries can be seen to eclipse. -- the first eclipsing binary observed was Algol (beta Persei). In 1783, John Goodricke found that Algol dimmed periodically to about a third of its normal brightness. -- the eclipse is very important for understanding the size of the stars, because big stars can block out a lot of light, and small stars only a fraction of the light. So eclipsing binaries also provide an important piece in the puzzle of understanding the attributes of stars.
IV. Formation of Binaries -- wide binaries: tidal capture or conucleation models have been suggested -- close binaries: fragmentation or fission (for very close binaries)
V. Close Binaries and Stellar Evolution -- Algol is an interesting system. There are actually more than just two stars, but we'll focus on the close binary (the third star is well separated from these two). Calculations show that the two stars are a hot dwarf and a cool giant. The hot dwarf looks just like a 3.7 solar-mass main sequence star, and the cool giant looks like a 0.8 solar-mass red giant. What's wrong with this picture? -- If both stars formed at the same time as a binary, and then evolved the way single stars do, by the time the low-mass star became a red giant, the hot high-mass star would no longer be on the main sequence: in fact, it would have gone supernova. -- How can you have a high-mass main sequence star that is the same age as the low-mass red giant? You need a different evolutionary path. The evolution is affected by mass exchange in the binary. Thus the higher mass star used to be the lower mass star, so is taking longer to evolve, but then when the originally lower mass star becomes a red giant, it loses most of its mass to the other star. In this way, the star that ends up with more mass is actually less evolved.
VI. Mass Exchange and the Roche Lobe -- How does a star lose mass to a binary companion? As we move farther from one star and closer to the other, the force of gravity shifts from being mostly toward the first star to being mostly toward the second. Thus gas that is far away from the center of one star can end up being pulled onto the other star. -- The region around each star inside which the star will be able to pull gas in is called the Roche Lobe. It looks like a figure 8. -- Originally, both stars will be well within their Roche Lobes, so will evolve normally as if they were alone. But when the higher-mass star evolves faster into its red giant phase, its radius increases so much that it fills up its Roche Lobe. As it continues to expand in size, the mass that leaves the Roche Lobe gets pulled over onto the companion star. This gives the exchange of mass that completely changes the evolution.
VII. General Consequences of Mass Exchange -- Binary stars revolve about their center of mass, and have a fixed angular momentum which is conserved as mass is exchanged. If mass moves from the lower mass star to the higher mass star, it moves from being farther from the center of mass to being closer. But to conserve angular momentum, this means that the rest of the lower mass star has to move farther away from the center of mass. So the distance between the stars must increase. If instead mass moves from the higher mass star to the lower, then it is moving farther from the center of mass. To conserve angular momentum, the lower mass star has to be pulled in closer to the center of mass, so the distance between stars would decrease. -- Thus mass transfer affects the binary separation, and the two stars are always at their closest when their masses are equal. Normally, what happens is the high-mass star evolves faster, so expands into a red giant. But before it gets to a red giant, it fills its Roche lobe and begins to transfer mass to the other star. Since this brings the two masses closer to each other, it also brings the two stars closer together, which assists the mass transfer and makes it occur very fast. Eventually the first star has transferred so much of its mass to the other star that the distance between the stars increases again, and the mass transfer stops. The less massive star now was once the more massive! -- The star that was stripped of mass is still on its way to becoming a white dwarf, neutron star, or black hole, depending on the mass of its core as usual. If it goes supernova, the binary may actually survive, and create exotic systems like pulsars or black holes orbiting a main-sequence star. -- A key point to remember in all this is that close binaries alter stellar evolution by changing the mass of the stars, but otherwise, once you know the mass, the individual stars appear much as they would if they were simple isolated stars. So the basic evolutionary scheme proceeds normally before and after mass transfer.
VIII. Type Ia Supernovae -- Of particular interest is when you have a close binary with stars with masses not too much more than the Sun, and one star evolves into a white dwarf. Then when the other puffs out into a red giant for the first or second time, it can pass some of it mass over to the white dwarf. But remember that if a white dwarf has its mass go above 1.4 solar masses, say by mass transfer from its companion, then its electrons will go relativistic (move at close to the speed of light), and under those conditions, degeneracy pressure is not as efficient and cannot hold up against gravity. So the white dwarf, when it gets to pretty much exactly 1.4 solar masses, will collapse under its own gravity. However, the white dwarf is probably mostly carbon at this point, so it has lots of fuel that has not yet undergone nuclear burning. When it suddenly contracts, instead of a core-collapse supernova (where the energy comes from gravity), the rapid fusion of carbon will generate enough energy to completely explode the entire star-- there is no core left to collapse. We say this type of supernova is a thermonuclear runaway, much like a huge fusion bomb. It's jargony name is a type Ia supernova, and the reason it is so important is that it tends to have very consistent properties-- and is bright enough to be seen at huge distances. Thus it makes a good "standard candle", which is a constant search in astronomy, because it makes it very good at determining the distance to the galaxy in which the supernova occured. This will have cosmological importance, as we shall see. Week 11 notes
Why does my binary star simulation only work for equal masses and initial speeds?
if i place two objects with certain masses into the simulation with no initial speeds, then as expected they accelerate towards each other. Something weird happens at the end, but that is just because i haven't told the simulation to anything if they get too close, so as the distance between the two objects get very small the accelerations blow up. When I give one of the objects an initial velocity perpendiculor to the distance between them, I get an elliptical orbit as expected. However, I am having trouble getting binary stars to work. Most of the time, I get something that looks two elliptical orbits about a common centre of mass, as expected, but each consecutive orbit is shifted down by a constant amount. See the image below.
I got the above image when i tried two objects with different masses and equal but opposite velocities. At first I though maybe this could be explained by the binary star system 'moving through space', since the relative positions of the stars are indeed two ellipcital orbits, but relative to the space around them they are also moving downwards.
However, I think there is some other issue with my code, because if I only give one of the stars an initial velocity, this happens:
In this picture, only the object represented by the green trajectory had an initial velocity. Somehow the fact that it had no initial velocity means it only does half an ellipse?? Indeed, if I give if even just a small velocity in the opposite direction, it does complete an orbit:
3 Answers 3
The centre of mass of the binary system cannot move because there are no external forces acting.
The line joining the two stars must always pass through the centre of mass, because by definition the centre of mass lies on the line between the two stars.
That means the two stars must orbit with the same period. If their periods weren't the same they could not remain on opposite sides of the COM.
Kepler's third law is irrelevant here. It applies to many (small) planets orbiting a (large) central star, not to a binary star system.
If the stars have masses $m_1$ and $m_2$ and the radii from the COM are $r_1$ and $r_2$, then from the definition of the COM, $m_1r_1 = m_2r_2$.
The (gravitational) central force $F$ acting on each star is the same, but the central accelerations are different because the masses are different. If the radial accelerations are $a_1$ and $a_2$, then $a_1 = F/m_1$ and $a_2 = F/m_2$.
If the angular velocities are $omega_1$ and $omega_2$, for circular orbits we have $a_1 = r_1 omega_1^2$ and $a_2 = r_2 omega_2^2$.
In non-circumbinary planets, if a planet's distance to its primary exceeds about one fifth of the closest approach of the other star, orbital stability is not guaranteed.  Whether planets might form in binaries at all had long been unclear, given that gravitational forces might interfere with planet formation. Theoretical work by Alan Boss at the Carnegie Institution has shown that gas giants can form around stars in binary systems much as they do around solitary stars. 
Studies of Alpha Centauri, the nearest star system to the Sun, suggested that binaries need not be discounted in the search for habitable planets. Centauri A and B have an 11 au distance at closest approach (23 au mean), and both have stable habitable zones.   A study of long-term orbital stability for simulated planets within the system shows that planets within approximately three au of either star may remain stable (i.e. the semi-major axis deviating by less than 5%). The habitable zone for Alpha Centauri A extends, conservatively estimated, from 1.37 to 1.76 au  and that of Alpha Centauri B from 0.77 to 1.14 au  —well within the stable region in both cases. 
For a circumbinary planet, orbital stability is guaranteed only if the planet's distance from the stars is significantly greater than star-to-star distance.
The minimum stable star-to-circumbinary-planet separation is about 2–4 times the binary star separation, or orbital period about 3–8 times the binary period. The innermost planets in all the Kepler circumbinary systems have been found orbiting close to this radius. The planets have semi-major axes that lie between 1.09 and 1.46 times this critical radius. The reason could be that migration might become inefficient near the critical radius, leaving planets just outside this radius. 
For example, Kepler-47c is a gas giant in the circumbinary habitable zone of the Kepler-47 system.
If Earth-like planets form in or migrate into the circumbinary habitable zone they are capable of sustaining liquid water on their surface in spite of the dynamical and radiative interaction with the binary star. 
The limits of stability for S-type and P-type orbits within binary as well as triple stellar systems have been established as a function of the orbital characteristics of the stars, for both prograde and retrograde motions of stars and planets. 
Measuring distance between two stars in a binary system
I am trying to teach myself some basic maths for astronomy from a book, namely trying to calculate the distance between two stars in a binary system.
One thing i am confused with is what angular separation means and how it can be translated to true physical distance between them using trigonometry. I am trying to visualise it but its a bit confusing at the moment. The book did not explain it as it seems to have presumed i already understood this.
Firstly what does it mean for each star to have an angular separation from centre of mass, eg if star A has 5 arcsec and star B has 10 arcsec what are these angles relative to? And if you know the distance D of the system from Earth, i presume trigonometry can be done to solve it but i am struggling to visualise how to draw it out to do the trigonometry for it at the moment.
I made a drawing to show the setup of what i think it might mean:
When they say angular separation from COM is this correct thinking? If so how are they defining the angle? What constitutes the 0 arcsec line, what counts as the positive x axis in space i guess is what im asking.
Secondly from that i do not understand how you might calculate the physical separation distance between the two stars (white line magnitude). Mainly because i am confused by the angle situation.
That 8-Star System in Star Trek: Picard Really Could Exist
To revist this article, visit My Profile, then View saved stories.
To revist this article, visit My Profile, then View saved stories.
You've heard of binary star systems, right? It's where there are two stars close to each other, and they both orbit around a common center of mass. Sure, you saw binary stars in the original movie, where Luke Skywalker was on the desert planet Tatooine. Oh wait, that's Star Wars and this is Star Trek. I'm kidding, I know the difference. But binary stars are real.
So … what about an octonary system, one with eight stars, all gravitationally interacting with each other? That's what we get in Star Trek: Picard. In this case it’s actually an artificially created system set up by an alien race long ago as a warning sign for future civilizations—uh, long story. We'll know more after watching the season finale, which comes out today.
But you’re thinking the same thing I am: Could an eight-star system exist in the real universe? And if it did, how could the stars be arranged so the system was stable? How would it all move? As Enoch, the navigation hologram, says in the show, “The gravitational mechanics would have to be … incredibly complex.” In other words, we should try to model this thing!
I should mention that there’s a little physics backstory here—a famous situation called the three-body problem. See, if you have two objects that are gravitationally interacting with each other, like, say, the Earth and the sun, that’s a solvable problem. With a bit of math you can turn it into an equivalent one-dimensional, one object problem. It's complicated, but also seemingly magical. You can get an equation that determines the future position and velocity of both objects for all time.
But it turns out that with three (or more) bodies, there’s no way to derive an equation of motion. To model such a system you have to use a numerical calculation. That’s where you break the trajectories into small time intervals. At each step, you calculate where each object will be at the end of the interval, based on the forces at work, and you just keep doing that till you map out the whole system.
So with three objects, we’d have to calculate the net gravitational force on each object. Remember that the gravitational force is an attractive interaction between two objects with mass. Its magnitude depends on the product of the two masses (let's call them mA and mB), and is inversely proportional to the square of the distance (r) between their centers:
Why don't the stars in a binary star system of equal masses always orbit their center of mass in a circular orbit? - Astronomy
Stars spend most of their lives on the main sequence, in the core-hydrogen-burning phase of stellar evolution, stably fusing hydrogen into helium at their centers. Stars leave the main sequence when the hydrogen in their cores is exhausted. For the Sun, which is about halfway through its main-sequence lifetime, this stage will occur about 5 billion years from now. Low-mass stars evolve much more slowly than the Sun, and high-mass stars evolve much faster.
When the central nuclear fires in the interior of a solar-mass star cease, the helium in the star's core is still too cool to fuse into anything heavier. With no internal energy source, the helium core is unable to support itself against its own gravity and begins to shrink. The star at this stage is in the hydrogen-shell-burning phase, in which the nonburning helium at the center is surrounded by a layer of burning hydrogen. The energy released by the contracting helium core heats the hydrogen-burning shell, greatly increasing the nuclear reaction rates there. As a result, the star becomes much brighter while the envelope expands and cools. A low-mass star like the Sun moves off the main sequence on the HR diagram first along the subgiant branch, then almost vertically up the red-giant branch.
As the helium core contracts, it heats up. Eventually, it reaches the point at which helium begins to fuse into carbon. The net effect of the fusion reactions is that three helium nuclei (or alpha particles) combine to form a nucleus of carbon in the triple-alpha process.
In a star like the Sun, conditions at the onset of helium burning are such that the electrons in the core have become degeneratethey can be thought of as tiny, hard spheres that once brought into contact, present stiff resistance to being compressed any further. This electron degeneracy pressure makes the core unable to "react" to the new energy source, and helium burning begins explosively, in the helium flash. The flash expands the core and reduces the star's luminosity, sending it onto the horizontal branch of the HR diagram. The star now has a core of burning helium surrounded by a shell of burning hydrogen.
As helium burns in the core it forms an inner core of nonburning carbon. The carbon core shrinks and heats the overlying burning layers, and the star once again becomes a red giant. It reenters the red-giant region of the HR diagram along the asymptotic giant branch, becoming an extremely luminous red supergiant star.
The core of a low-mass star never becomes hot enough to fuse carbon. Such a star continues to ascend the asymptotic giant branch until its envelope is ejected into space as a planetary nebula. At that point the core becomes visible as a hot, faint, and extremely dense white dwarf. The planetary nebula diffuses into space, carrying helium and carbon into the interstellar medium. The white dwarf cools and fades, eventually becoming a cold black dwarf.
Evolutionary changes happen more rapidly for high-mass stars than for low-mass stars because larger mass results in higher central temperatures. High-mass stars do not experience a helium flash and do attain central temperatures high enough to fuse carbon. They form heavier and heavier elements in their cores, at a more and more rapid pace, and eventually die explosively.
The theory of stellar evolution can be tested by observing star clusters, all of whose stars formed at the same time. As time goes by, the most massive stars evolve off the main sequence first, then the intermediate-mass stars, and so on. At any instant, no stars with masses above the cluster's main-sequence turnoff mass remain on the main sequence. Stars below this mass have not yet evolved into giants and so still lie on the main sequence. By comparing a particular cluster's main-sequence turnoff mass with theoretical predictions, astronomers can measure the age of the cluster.
Stars in binary systems can evolve quite differently from isolated stars because of interactions with their companions. Each star is surrounded by a teardrop-shaped Roche lobe, which defines the region of space within which matter "belongs" to the star. As a star in a binary evolves into the giant phase it may overflow its Roche lobe, forming a mass-transfer binary as gas flows from the giant onto its companion. If both stars overflow their Roche lobes, a contact binary results. Stellar evolution in binaries can produce states not achievable in single stars. In a sufficiently wide binary, both stars evolve as though they were isolated.
SELF-TEST: TRUE OR FALSE?
1. "Low-mass" stars are conventionally taken to have masses less than about 8 solar masses. (Hint)
2. All the red dwarf stars that ever formed are still on the main sequence today. (Hint)
3. Once on the main sequence, gravity is no longer important in determining a star's internal structure. (Hint)
4. The Sun will get brighter as it begins to run out of fuel in its core. (Hint)
5. As a star evolves away from the main sequence it gets larger. (Hint)
6. As a star evolves away from the main sequence it gets hotter. (Hint)
7. As a red giant the Sun will have a core that is smaller than it was when the Sun was on the main sequence. (Hint)
8. When helium starts to fuse inside a solar-mass red giant, it does so slowly at first the rate of fusion increases gradually over many years. (Hint)
9. With the onset of helium fusion a red giant gets brighter. (Hint)
10. A planetary nebula is the disk of matter around a star that will eventually form a planetary system. (Hint)
11. For a high-mass star there is no helium flash. (Hint)
12. High-mass stars can fuse carbon and oxygen in their cores. (Hint)
13. A star cluster with an age of 100 million years will still contain many O-type stars. (Hint)
14. In a binary star system it is never possible for the lower-mass star to be more evolved than the higher-mass companion. (Hint)
15. In a mass-transfer binary one of the stars has filled its Roche lobe. (Hint)
SELF-TEST: FILL IN THE BLANK
1. A main-sequence star doesn't collapse because of the outward _____ produced by hot gases in the stellar interior. (Hint)
2. The Sun will leave the main sequence in about _____ years. (Hint)
3. While a star is on the main sequence _____ is slowly depleted and _____ builds up in the core. (Hint)
4. A temperature of at least _____ is needed to fuse helium. (Hint)
5. At the end of its main-sequence lifetime, a star's core starts to _____. (Hint)
6. When helium fuses, it produces _____ and releases _____.
7. Just before helium fusion begins in the Sun, the core's outward pressure will be provided mainly by electron _____. (Hint)
8. As a star ascends the asymptotic giant branch its _____ core is shrinking. (Hint)
9. As a red supergiant the Sun will eventually become about ____ times its present size. (Hint)
10. The various stages of stellar evolution predicted by theory can be tested using observations of stars in _____. (Hint)
11. By the time the envelope of a red supergiant is ejected, the core has shrunk down to a diameter of about _____ . (Hint)
12. A typical white dwarf has the following properties: about half a solar mass, fairly _____ surface temperature, small size, and _____ luminosity. (Hint)
13. As time goes by, the temperature and the luminosity of a white dwarf both _____. (Hint)
14. As a star cluster ages, the luminosity of the main-sequence turnoff _____. (Hint)
15. Whether being a member of a binary star system will affect the evolution of a star depends largely on the _____ of the two stars in the binary. (Hint)
REVIEW AND DISCUSSION
1. Why don't stars live forever? Which types of stars live the longest? (Hint)
2. What is main-sequence equilibrium? (Hint)
3. How long can a star like the Sun keep burning hydrogen in its core? (Hint)
4. Why is the depletion of hydrogen in the core of a star such an important event? (Hint)
5. What makes an ordinary star become a red giant? (Hint)
6. How big (in A.U.) will the Sun become when it enters the red-giant phase? (Hint)
7. How long does it take for a star like the Sun to evolve from the main sequence to the top of the red-giant branch? (Hint)
8. What is the helium flash? (Hint)
9. Describe an important way in which winds from red giant stars are linked to the interstellar medium. (Hint)
10. How do the late evolutionary stages of high-mass stars differ from those of low-mass stars? (Hint)
11. What is a planetary nebula? Why do many planetary nebulae appear as rings? (Hint)
12. What are white dwarfs? What is their ultimate fate? (Hint)
13. Do many black dwarfs exist in the Galaxy? (Hint)
14. What are the Roche lobes of a binary system? (Hint)
15. Why is it odd that the binary system Algol consists of a low-mass red giant orbiting a high-mass main-sequence star? How did Algol come to be in this configuration? (Hint)
1. The Sun will evolve off the main sequence when roughly 10 percent of its hydrogen has been fused into helium. Using the data given in Section 16.5, calculate the total amount of mass destroyed (that is, converted into energy) and the total energy released. (Hint)
2. Use the radiusluminositytemperature relation (L R 2 T 4 see Section 17.3) to calculate the radius of a red supergiant with temperature 3000 K (half the solar value) and luminosity 10,000 solar luminosities. How many planets of our solar system would this star engulf? (b) Repeat your calculation for a 12,000 K (twice the temperature of the Sun), 0.0004 solar luminosity white dwarf.
3. A main sequence star at a distance of 20 pc is barely visible through a certain telescope. The star subsequently ascends the giant branch, during which time its temperature drops by a factor of 3 and its radius increases 100-fold. What is the new maximum distance at which the star will still be visible using the same telescope? (Hint)
4. The Sun will reside on the main sequence for 10 10 years. If the luminosity of a main-sequence star is proportional to the cube of the star's mass, what mass star is just now leaving the main sequence in a cluster that formed 400 million years ago? (Hint)
5. A Sun-like star goes through its most rapid luminosity change between stages 8 and 9, when the luminosity increases by about a factor of 100 in 10 5 years. On average, how rapidly does the star's absolute magnitude change, in magnitudes per year? Do you think this change would be noticeable in a distant star within a human lifetime?
6. Calculate the average density of a red-giant core of mass 0.25 solar mass and radius 15,000 km. Compare this with the average density of the giant's envelope, if its mass is 0.5 solar mass and its radius is 0.5 A.U. Compare each with the average core density of the Sun.
7. How many years are spent by a Sun-like star during its postmain-sequence evolution? (Use Table 20.1 and consider stages 8 through 12.) What is the probability that oberservers will happen to catch a star during one of its rapid periods of evolutionsay, between stages 8 and 9if they examine a large number of randomly chosen postmain-sequence stars? How would the odds change if all stars (including those on the main sequence) were included in the search?
8. How long will it take the Sun's planetary nebula, expanding at a speed of 50 km/s, to reach the orbit of Neptune? How long to reach the nearest star?
9. What are the escape speed and surface gravity of Sirius B? (See Table 20.2.) (Hint)
10. From the discussions presented in More Precisely 2-3 and More Precisely 15-1, it may be shown that the angular momentum of a circular binary, of separation R and component masses m1 and m2, is proportional to m1 m2 (if the total mass is held fixed). A circular binary has component masses one and two times the mass of the Sun, respectively, and an orbital period of 2 years. (a) What is its orbital separation (semi-major axis)? (b) Mass transfer moves 0.2 solar mass of material from the more massive to the less massive star, keeping the total mass of the system fixed and conserving angular momentum. If the binary remains circular, calculate its new separation and orbital period.
1. You can tour the Galaxy without ever leaving Earth, just by looking up. In the winter sky, you'll find the red supergiant Betelgeuse in the constellation Orion. It's easy to see because it's one of the brightest stars visible in our night sky. Betelgeuse is a variable star, with a period of about 6.5 years. Its brightness changes as the star expands and contracts. At maximum size, Betelgeuse fills a volume of space that would extend from the Sun beyond the orbit of Jupiter. Betelgeuse is thought to be about 10 to 15 times more massive than our Sun. It is probably between 4 and 10 million years oldand in the final stages of its evolution.
A similar star can be found shining prominently in midsummer. This is the red supergiant Antares in the constellation Scorpius. Depending on the time of year, can you find one of these stars? Why are these stars red?
2. Find a library that has the Astrophysical Journal. Find an article from the late 1950s and 1960s that gives the photometry of a star cluster like the Pleiades or Hyades. Plot a colormagnitude diagram (V vs. B-V see Section 17.6). Determine the V magnitude of the main-sequence turnoff, and hence estimate the age of the cluster. Compare your age with that given in the article.
Kepler's Third Law
A decade after announcing his First and Second Laws of Planetary Motion in Astronomica Nova, Kepler published Harmonia Mundi ("The Harmony of the World"), in which he put forth his final and favorite rule:
It turns out that this relationship will serve as the basis for our attempts to derive stellar masses from observations of binary stars . but notice how the Third Law itself never mentions mass!
Let's check to see just how well Kepler's Third Law works. I'll do the first example:
Now, you verify that if you continue to use the same units -- period in years, semimajor axis in AU -- these other orbits also satisfy the same equation:
Hmmmm. this Third Law doesn't seem to work all the time, does it? Or is there something we're missing?
Just what is that constant, really?
It turns out that the constant in Kepler's Third Law depends on the total mass of the two bodies involved. Kepler himself, studying the motion of the planets around the Sun, always dealt with the 2-body system of Sun-plus-planet. The Sun is so much more massive than any of the planets in the Solar System that the mass of Sun-plus-planet is almost the same as the mass of the Sun by itself. Thus, the constant in Kepler's application of his Third Law was, for practical purposes, always the same.
But in the case of the Moon's orbit around the Earth, the total mass of the two bodies is much, much smaller than the mass of Sun-plus-planet that means that the value of the constant of proportionality in Kepler's Third Law will also be different. On the other hand, if we compared the period and semimajor axis of the orbit of the Moon around the Earth to the orbit of a communications satellite around the Earth, we would once again have (almost) the same total mass in each case and thus we would end up with the same relationship between period-squared and semimajor-axis-cubed.
To make a long story short -- we'll tell the whole story later, including a derivation of the formula below from Newton's Law of Gravitation -- one can write Kepler's Third Law in the following way:
- period P in days
- semimajor axis a in AU
- mass Mtot in solar masses
The key point here is that the only measured quantity we need to find k is time: the period of the Earth's orbit around the Sun. Now, it's not quite so easy as it sounds, but it can be done without too much trouble. Moreover, because we can average over many, many, many years, we can determine the length of the year very accurately -- to many significant figures. Therefore, we can also determine the value of k to many significant figures. If all we want to do is calculate the orbits of objects around the Sun, then k is all we need and with a very accurate value of k, we can calculate very accurate planetary orbits.
For example, it was this constant k that Adams and Leverrier used in their computations of the as-yet-unknown planet VIII, aka Neptune.
- Yes, the two constants are closely related
- No, they don't stand for EXACTLY the same thing
The Gaussian constant, k, is defined in terms of the Earth's orbit around the Sun. The Newtonian constant, G, is defined in terms of the force between two two masses separated by some fixed distance. In order to measure k, all you need to do is count days in order to measure G, you need to know very precisely the masses, separation, and forces between test objects in a laboratory. The Gaussian constant is obviously much easier to determine. Look at a sample pair of values from recent compilations:
Applying Kepler's Third Law to stars
- period of the orbit is 900 days, to a precision of 1 percent
- apparent angular size of the orbit is 0.10 arcseconds, to a precision of 5 percent
- distance to the star is 20 parsecs, to a precision of 10 percent
What is the mass of the star in this system? What is the uncertainty in the mass? Express your result in absolute terms (solar masses), and in percentage terms.
Copyright © Michael Richmond. This work is licensed under a Creative Commons License.
1 Answer 1
Question 2, 3 and 4 all come down to that we assume that there is no external force acting on the two masses and then according to Newton's laws their center of mass should remain stationary (or move at a constant velocity, but I will initially pick this to be zero).
If $T_A eq T_B$ then eventually both masses would be both an the same side of $F_<1A>$/$F_<2B>$ and some time later this point would be in between the two, which would imply that the center of mass would be moving.
For the other two it might be easier if I would first answer your first question. If we represent the distance between $m_A$ and the center of mass as $r_A$ and similar the distance between $m_B$ and the center of mass as $r_B$. By definition of the center of mass, it should always be in between and inline with $m_A$ and $m_B$, and
The gravitational force from $m_B$ on $m_A$ could also be caused by another fictional mass fixed at the center of mass. The mass of this fictional object, denoted with $hat using equation $(1)$ then $r_B$ can be expressed in $r_A$, $m_A$ and $m_B$, In a similar way you could also do this for $m_B$ by replacing $m_A$ with $hat Using these masses you can now use your initial expression for the orbital period, Equation $(1)$ should also hold for the semi-major axes as well, so for the expression for $T_B$ could also be written as, which is the same as the expression for $T_A$ in equation $(6)$. Calling this expression $T$ and rewriting it to a form similar as stated in your question yields, By again applying equation $(1)$ to the semi-major axes, then the left hand side of equation $(9)$ can be written as, which is indeed the relation that you where after. Since I already showed that relative to $m_A$ you could replace $m_B$ with $hat Now assuming that the resulting orbit of $m_A$ looks like, then by using equation $(1)$ an expression for $r_B$ can be found to be, So the eccentricities of both orbits should also be the same, only the point from where you measure $ heta$ should be rotated 180°.
Watch the video: Atlantis Is Calling Unknown Rockpop Music Hall To be deleted! (September 2021).
using equation $(1)$ then $r_B$ can be expressed in $r_A$, $m_A$ and $m_B$,
In a similar way you could also do this for $m_B$ by replacing $m_A$ with $hat Using these masses you can now use your initial expression for the orbital period, Equation $(1)$ should also hold for the semi-major axes as well, so for the expression for $T_B$ could also be written as, which is the same as the expression for $T_A$ in equation $(6)$. Calling this expression $T$ and rewriting it to a form similar as stated in your question yields, By again applying equation $(1)$ to the semi-major axes, then the left hand side of equation $(9)$ can be written as, which is indeed the relation that you where after. Since I already showed that relative to $m_A$ you could replace $m_B$ with $hat Now assuming that the resulting orbit of $m_A$ looks like, then by using equation $(1)$ an expression for $r_B$ can be found to be, So the eccentricities of both orbits should also be the same, only the point from where you measure $ heta$ should be rotated 180°.
Using these masses you can now use your initial expression for the orbital period,
Equation $(1)$ should also hold for the semi-major axes as well, so for the expression for $T_B$ could also be written as,
which is the same as the expression for $T_A$ in equation $(6)$. Calling this expression $T$ and rewriting it to a form similar as stated in your question yields,
By again applying equation $(1)$ to the semi-major axes, then the left hand side of equation $(9)$ can be written as,
which is indeed the relation that you where after.
Since I already showed that relative to $m_A$ you could replace $m_B$ with $hat Now assuming that the resulting orbit of $m_A$ looks like, then by using equation $(1)$ an expression for $r_B$ can be found to be, So the eccentricities of both orbits should also be the same, only the point from where you measure $ heta$ should be rotated 180°.
Now assuming that the resulting orbit of $m_A$ looks like,
then by using equation $(1)$ an expression for $r_B$ can be found to be,