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Planetary and Stellar Motion
Kepler’s Three Laws
Kepler’s first law: Planets orbit in an ellipse shape. An ellipse can be made by two points called foci. The sun is located at one of the focus.
Kepler’s second law: Since planets orbit in an ellipse shape, they will be closer to the Sun at times than other times. When it is closer, the planet moves faster.
So if we were to mark a planet’s position every n days, the marks would be more spread out when the planet was near the sun, and bunched up when it was further away.
The second law states that the areas between two neighbouring points and the sun would be equal to the area between any other two points.
Kepler’s third law: For planets, the square of the period is proportional to the cube of the semi-major axis.
T² ∝ r³
T is the period, how long it takes to complete 1 full orbit. If we think about v = d/t, we can say that v = 2πr/T, rearrange it to get T = 2πr/v.
What is the semi major axis? It’s the average distance from the planet to the sun. In the picture below, I’ve used Pluto as an example, because Earth is too overused.
Newton’s Law of Gravitation
If you’ve studied centripetal acceleration you will recognise a = v²/r.
If not, then I will tell you right now that a = v²/r.
You also need to know Newton’s law of Gravitation:
If we focus on circular orbits, not elliptical, we can use laws of centripetal acceleration to find an equation for Kepler’s third law.
Most planets orbit the sun fairly circular, so this equation may come in handy.
We can set Newton’s equation to equal F = ma and sub in a = v²/r.
We can divide by m and multiply by r.
Let’s move it around to get T², and then hopefully it should be in T² = kr³ form, where k is a constant.
Here T squared is proportional to r cubed, showing Kepler’s 3rd law does work for circular motion.
Let’s go through some examples
Io orbits Jupiter with eccentricity 0.0041, so it can be assumed it is circular.
Io lies 351’900 km above Jupiter’s surface and takes 152,850 seconds to complete one orbit. Calculate the radius of Jupiter.
(Mass of Jupiter = 1.8982e+27 kg)
Let’s try another question
Neptune is 4’500 million km from the sun. Its eccentricity is an incredibly small 0.0095, so it can be assumed that it is a perfect circle.
Mars lies 220 million km closer to the sun than Neptune, and takes 687 Earth days to complete one orbit. It also has a fairly circular orbit.
How long is a Neptunian year?
Now hold your horses! First of all we need to calculate Mars’ semi-major axis, which is 4’500 million km – 220 million km = 230 million km.
Now we can use the equations we’ve learnt.
We can rearrange them so the constants are on one side. Regardless of which planet we focus on, M will always be the mass of the sun (unless you’re working with exoplanets). Therefore, for Mars and Neptune, M is the same, so the whole right side of the equation is the same.
We can set T²/r³ of Mars to equal T²/r³ of Neptune.
Above, I’ve changed my period for Mars to be in seconds instead of days, and distances to metres instead of kilometres.
Then, I subbed in all my figures for Mars, then multiplied by Neptune’s r³ to get T² on its own.
From then on all you have to do is square root T.
T² = 2.6387 x 10^19
T = 5136857239 seconds
That’s no good, I don’t count time in seconds, I’d much prefer days or years. Firstly, divide it by 3600 to turn it into hours, 24 to turn it into days, then 365 to get a Neptunian year in Earth years.
T = 59454 days.
T = 163 Earth years.
Okay, one last question.
Mercury takes 88 days to orbit the sun. It’s aphelion (furthest distance from the sun) is 69,820,000 km but its perihelion (closest point) is only 46,000,000 km.
Using Kepler’s third law, calculate the mass of the sun.
We’ve looked at situations where we have one big object and a little one, but what happens when both objects are equally massive?
In the case of a binary system (2 nearly equally massive objects), both are massive enough to affect each other, so they don’t orbit like a regular planet and star.
Take Pluto for example.
Pluto doesn’t affect the sun’s motion because it’s too small, but Charon (Pluto’s biggest moon) is big enough to affect Pluto’s motion.
In case you are a bit confused as to how a binary system works, have a look at some real-life data of Pluto and Charon:
Notice how if you isolate them, they both look like they’re going around in a circle, but when you look at them together, you notice they both kind of look like they’re orbiting each other.
Here’s a computer rendering of the system:
Both orbit around a centre of mass, but if Charon’s mass were to get smaller, the centre of mass would slowly move towards Pluto. The bigger the mass difference, the closer the centre of mass is to the larger object. This is because both objects exert equal and opposite forces on each other (otherwise the orbit wouldn’t be stable). They also have the same angular velocity.
From this knowledge, we can find the centre. Let’s call it C!
To find C, we need:
- r1 and r2, both object’s distance to C;
- M1 and M2, the mass of each object.
Because the distance to C depends on the difference in mass, we can say that:
We can also adapt Kepler’s third law for a binary system.
Orbital Speeds & Dark Matter
Remember F = mv²/r and F = GMm/r² ?
These two equations will tell us enough about orbits for the case of dark matter.
How are we going to use these equations?
In the case of galaxies, m will be the mass of a single star, and M will be the mass of the entire galaxy within that star’s orbit.
To find the velocity one of these stars should be orbiting with, we can equate both equations to make a new one:
mv²/r = GMm/r²
v²/r = GM/r²
v² = GM/r
What can we learn from this new equation? Firstly, M is the mass of the galaxy, so a larger M means a larger velocity. r is the star’s distance from the centre of the galaxy, so when we increase r we decrease the velocity.
Below I’ve graphed the solar system (Mercury-Pluto). All I care about for now is v² (the planet’s velocity) and r (distance to the centre of the sun), so all I’ve done is plot v² against r.
If G and M are unchanging, then v² ∝ 1/r, which is what the graph shows.
But r is limited in the case of planets since they can’t orbit the sun within the sun. However, we still want to know what happens when the semi-major axis is near 0. This isn’t possible with the solar system, but let’s assume it can.
If a planet were orbiting inside the Sun, the only mass of the sun (M) we need to worry about is the mass contained within the planet’s orbit. So if r is decreasing, M is decreasing by a larger magnitude, resulting in the velocity decreasing.
If the planet is in the very centre, r = 0 and M = 0, so the velocity is 0 (don’t tell the pure mathematicians that I’ve divided by 0).
Our curve looks more like this:
Although this doesn’t exactly work for planets orbiting stars, it does work for stars in a galaxy.
From our curve, we can find expected values for the velocity if we sub in the mass and distance, but when astronomers actually observed galaxies through telescopes, they noticed something odd about the speed at which outer stars were moving around the galaxy: they were moving incredibly quickly!
Wait, how do we know they’re orbiting too fast?
Remember the curve we found before? If we were to observe some stars in a galaxy, each increasingly further away from the centre, we expect that their velocities will also follow the curve i.e stars really close are moving slowly, shortly followed by much faster stars and then a drop in velocity as we get to the furthermost stars.
This isn’t what we observe.
The stars’ velocities follow a the expected sharp increase as we move away from the centre, but beyond that their velocities still increase.
If we plot the velocities of the stars, this is what both the observed and expected curves look like:
It turns out these outer stars are zooming around the galaxy! How weird!
Just about all galaxies also seem to have incredibly speedy stars, which baffled astronomers for ages.
How can this be?
With such high speeds, these stars should be flying out the galaxy, but they’re not.
Now, the problem may just be that out understanding of gravity is inaccurate. For centuries we believed that the further an object is, the less gravitational force it experiences. This idea could be wrong or maybe we got the value for G wrong, which isn’t a stretch since we still don’t know much about the fundamental forces that govern physics, especially gravity.
Or, maybe our understanding of gravity is correct, and we’re just missing a piece of information.
Let’s cast our minds back. We noted that M will be the mass of the entire galaxy within the star’s orbit, and that the star’s velocity is governed by v² = GM/r.
G is a constant, so we don’t need to worry about it; it can’t change. v is what we’re measuring, so to try and find v we need to play around with M and r.
If we increase r, we’ll just be moving the star further away, which decreases v. That’s no good!
However, what if we increased M as we increased r? An increase in M will definitely increase v, so by deduction what’s probably happening is that the mass of the galaxy is increasing a lot as we get further out.
Isn’t that obvious? There are more stars as we get further away.
Yes, that is true. But look at where the bulk of most galaxies lie. In the centre, and don’t forget that many have black holes as the centre too (black holes are extremely massive!). Take the Milky way and Andromeda for example (pictures below). Most stars seem to be near the centre, so M won’t be increasing as much as we get further away.
So to obtain such high results for v the further out we go, the mass needs to be increasing much more dramatically and rapidly than what we’re observing. But that’s not what we’re seeing.
Perhaps there’s some missing galaxy mass that we can’t see!
An thus, the idea of dark matter was created!
Dark matter is something entirely different; it doesn’t behave like normal fermions because it doesn’t interact with the electromagnetic force. It does, however, interact with the gravitational force which is why velocities of stars are so massive.
We used to think it was made of MACHOs and WIMPs, short for Massive Astrophysical Compact Halo Objects and Weakly Interacting Massive Particles respectively. As you can tell, astronomers are fantastic at naming.
MACHOs are basically normal matter objects (planets, dust etc) that just isn’t lit up by starlight; no EM waves can get to it. Thankfully, MACHOs have been found and solved the case for dark matter … for about 2 of the few thousand cases.
Dark matter definitely isn’t made of MACHOs, but it could be made of WIMPs. These weakly interacting particles may be an exotic particle that simply doesn’t interact with EM waves. At first, scientists considered the neutrino to be a potential WIMP, but it doesn’t make up a large amount of the universe’s mass.
Another idea stems from the Higgs Boson, a very heavy and unstable force-carrier. In 2012, researchers at the Large Hadron Collider had discovered the Higgs boson by how it decayed.
One theory suggests that the Higgs boson may decay into dark matter WIMPs. Since WIMPs are massive, just like the Higgs boson, but evidence suggests there may be more than just the one type of Higgs boson.
The light from objects moving at high speeds will have distorted wavelengths. On a spectrum, this looks like the light has shifted more towards one side.
A picture best explains Doppler shift:
This happens with sounds too. A good example would be a fire engine’s siren suddenly dropping in pitch as it passes by.
But, let’s keep this to astronomy and stick with light. An object with a great redshift indicates that it is moving away very quickly, and this discovery has opened a huge gateway to cosmology.
Hippolyte Fizeau was the first to notice redshifts in stars’ spectrums, but it was Edwin Hubble who found that there was a relationship between the distance to them and their redshift. Hubble discovered that the most distant objects had the biggest redshifts, essentially they were moving away the fastest.
Redshift or blueshift is basically how relatively shifted the spectrum is. The equation to determine it is:
NGC 4151, better known as the Eye of Sauron, is roughly 62 million light years from Earth. Here is a photo of the beauty combining x-ray, optical, and radiowave data.
We’re going to look at the red areas – the radiowave and microwave data. In these parts is a lot of neutral hydrogen, which has a wavelength of 21cm in a lab on Earth.
The data taken shows neutral hydrogen to have an observed wavelength of 21.07 cm, a very slight shift.
From this, calculate the Doppler shift observed in the Eye of Sauron.
Comment on the direction on movement of the Eye of Sauron.
Here’s the worked answer:
As you saw with the equation, there’s another way to find z: radial velocity and the speed of light.
This is useful, because we can calculate Doppler shift using the wavelength bit, and then use that to find how fast the object is moving. From then on we can find the distance to all these objects.
Hoag’s Object is by far one of the weirdest galaxies known to humanity. It has a large nucleus surrounded by a circle of young stars which starts 38’000 light years away from the centre! In this picture, you can even see another ring galaxy through Hoag’s Object!
From measuring the shifts in wavelengths, astronomers discovered that Hoag’s object has a redshift of 0.0425. From this, calculate the radial velocity of the galaxy.