Feeds:
Posts

## Five top facts about Jupiter – no. 3

Continuing with my series of the five top facts about Jupiter which BBC Radio 5 posted around the time of my interview on the morning show a few weeks ago, the third fact that I listed was

it [Jupiter] takes only 10 hours to rotate but 12 years to orbit the Sun

The tweet from BBC Radio 5 with the five most interesting facts about Jupiter.

The list of the five facts

As I mentioned in my blog on the 2nd fact last week, when we look at Jupiter we are seeing the tops of the clouds. So, when we talk about its rotation, we are talking about how long it takes for e.g. the red spot to go around once and come back to the same place as we look at Jupiter. The rotation rate of Jupiter is about 10 hours, much quicker than the Earth and the quickest of all of the planets. However, the picture is not quite as simple as this, because it rotates differentially, the gases at the equator go around faster than the gases near the poles. There is not much difference in the rotation periods, only about 5 minutes, but there is a difference.

Also, because of this rapid rotation, Jupiter exhibits what we call an equatorial bulge which is basically the flattening out of a sphere due to rotation, so its diameter at the equator is more than its diameter at the poles, and it is flattened out by its rotation away from being a perfect sphere.

Jupiter is about five times further away from the Sun than the Earth is (compare this to Mars, which is about 1.5 times further away than the Earth). As it is further away the force of gravity from the Sun is weaker, and so it moves more slowly in its orbit than the Earth does. Added to this, it has further to go (its orbit is longer, as it is further from the Sun), so these two things combined lead to Jupiter taking much longer to go around the Sun than the Earth does. It doesn’t take five times longer, but rather about twelve times longer.

As I mentioned in my blog about the position of Jupiter and the Moon in early February, Jupiter appears to wander through the background stars from week to week and month to month. As it takes about 12 years to go once around the Sun, in one year it will move about one twelfth of the way around the sky, and as there are twelve constellations in the zodiac, it moves from one constellation to the next in a 1-year period.

At the moment Jupiter is in Cancer moving into Leo. By this time next year it will be in Leo and moving into the next constellation to the East of Leo, which is Virgo. In 6-years’ time it will have moved halfway around the 12 zodiac constellations, and so will be visible in the summer months and not the winter months as it is presently. The series of diagrams below show Jupiter’s position at 8:24pm on the 5th of March 2015, 2016 and 2017 as seen from London. As you can see, the stars stay in the same place at the same time each year, but Jupiter has moved eastwards, roughly one constellation further East in each twelve month period.

The position of Jupiter as seen from London on the 5th of March 2015 at 8:24pm. As you can see, Jupiter is at the eastern end of Cancer.

The position of Jupiter as seen from London on the 5th of March 2016 at 8:24pm. As you can see, Jupiter is now in Leo.

The position of Jupiter as seen from London on the 5th of March 2017 at 8:24pm. As you can see, Jupiter has now moved into Virgo and has not even risen at 8:24pm.

## Five top facts about Jupiter – no. 1

Last week I said I would blog in more detail about the “five top facts” I gave to BBC Radio 5 when I was on the breakfast show back on the 3rd of February. Here is the tweet that BBC Radio 5 sent out, and below it the “five top facts” that I chose. Today I am going to blog about the first fact.

The tweet from BBC Radio 5 with the five most interesting facts about Jupiter.

The list of the five facts

## Fact 1 – Jupiter is a failed star

Jupiter is, like the Sun, mainly comprised of hydrogen and helium. In the Universe as a whole, about 75% of the Universe is hydrogen, about 24% is helium, and the remaining 1% is everything else (carbon, oxygen, nitrogen etc.). It was George Gamow and his co-workes Ralph Alpher and Robert Hermann who showed in the late 1940s that the hydrogen and helium we find in the Universe was all be created in the first few minutes after the Big Bang. In fact, the observed abundances of hydrogen and helium are one of the main pieces of evidence for the “hot big bang” theory, there is just too much helium to have all been created in stars as e.g. Fred Hoyle argued.

The other elements beyond hydrogen and helium have all been created within stars. A star like the Sun is what we call a “main sequence” star, (see my blog on the HR diagram here) and this means that it is burning hydrogen in its core. It is turning the hydrogen into helium via nuclear fusion, and the tiny mass difference between the hydrogen that goes into this reaction and the helium which comes out provides the energy of the Sun, because of Einstein’s famous equation $E=mc^{2}$. Because hydrogen is converted into helium in stars, the abundance of helium is very slowly getting greater in the Universe, and the abundance of hydrogen is very slowly going down.

We believe that stars form from the collapse of huge clouds of hydrogen, things we refer to as “giant molecular clouds” (like the “pillars of creation” that I blogged about here). When these clouds are massive and cold enough they can collapse under their own gravity, and as they do so they fragment. The individual fragments are what form the stars, but a fragment has to be large enough to actually “ignite” as a star. This requires a high enough pressure and density and temperature at the core of the fragment.

Jupiter has all the ingredients to be a star, but what it lacks is the mass (or size). It is too small to create a high enough temperature and pressure at the centre to force hydrogen to fuse into helium. If it were about ten to one hundred times more massive it would have become a star, albeit a very faint one (what we call a “red dwarf”). We believe stars can range in mass from about one tenth the mass of the Sun (smaller than this and they will be “failed stars” like Jupiter) up to about 50 or maybe 100 times the mass of the Sun. Larger than this and they are beyond something called the Hyashi limit, and will just blow themselves apart before they can settle down onto the main sequence.

It was discovered in the 1960s, from observing Jupiter in the infrared, that it was hotter than it should be given its distance from the Sun. This is because the gravitational potential energy lost when the gas from which it formed collapsed was converted to heat, and this heat has been leaking out over since. It emits about twice the energy that it gets from the Sun.

## Jupiter and the Moon last week

On Tuesday morning of last week (the 3rd) I woke up early, which is a habit I am trying to get back into so that I can recommence my running, which I try to do first thing in the morning before work. In a semi-awake state I was listening to BBC Radio 5’s breakfast show co-presenter Rachel Burden and someone else on the show between 6 and 7 (was it George Riley the sports person or someone else??) discussing how they had seen the Moon and Jupiter very bright in the sky on Monday evening.

The following morning during the same 6-7am period, the discussion was resumed with remarks being made about how much Jupiter had moved away from the Moon. I decided to tweet into the show and to Rachel saying that it was the Moon which has moved, not Jupiter. To cut a long story short, I ended up being on the show on the Thursday morning to explain what was moving and why. Here is a link to a recording of my 3-minute stint, where Rachel and Nicky Campbell (Rachel’s co-presenter) interview me.

The evening before my being on the show, one of the show’s researchers had asked me to send in a list of the five most interesting facts about Jupiter. They then put those out as a tweet during my interview. I chose the following five, would you have chosen the same ones?

The tweet from BBC Radio 5 with the five most interesting facts about Jupiter.

The list of the five facts

I will talk about each of these five facts separately in future blogs over the next several weeks. I will also blog in a little more detail about the NASA mission to Europa and the proposed ESA mission Juice; both of which intend to study this fascinating moon in more detail.

Being on the radio it is, of course, impossible to show diagrams about the motion of the Moon and Jupiter; and there really wasn’t enough time to explain it properly. So, I have decided to put together these slides to explain it in a little more detail.

## Jupiter and the Moon during the first week of February 2015

Getting software to show you what is in the sky is easy, and although for PCs and Macs you may end up paying several tens of pounds, for tablet devices the software is much cheaper, with many reasonable ones being free. I use “Skysafari”, which is not free but is not too expensive either. It is made by Carina Software, who made the wonderful Voyager Software on Macs that I used for many many years in my classes.

Below is a screen capture using this software of the sky as seen from London at 20:00 on Thursday the 5th of February 2015. As you can see, I have done the screen capture with Jupiter just to the left (East) of the middle of the window (in the software you can use your finger to move around and look in different directions such as north or north-east if you wish). The Moon is at about 7 o’clock from Jupiter in direction, if you imagine a clock face.

This is a screen capture from an app I use on my iPad called “Skysafari” which can show what is in the sky at any location and at any time and date. There are lots of other similar apps available, but I like this one the most of the ones I’ve tried. This is the screen capture looking south for 20:00 on Thursday the 5th of February 2015 as seen from London.

Below I show a sequence of screen captures of Jupiter and the Moon (I have zoomed in on just enough to show the two) from Monday evening (the 2nd) to Friday evening (the 6th), all at 20:00 to show the motion of the Moon compared to Jupiter’s position.

Jupiter and the Moon as seen on Monday the 2nd of February at 20:00 from London

Jupiter and the Moon as seen on Tuesday the 3rd of February at 20:00 from London

Jupiter and the Moon as seen on Wednesday the 4th of February at 20:00 from London

Jupiter and the Moon as seen on Thursday the 5th of February at 20:00 from London

Jupiter and the Moon as seen on Friday the 6th of February at 20:00 from London

I think these screen captures make it quite easy to see that Jupiter is staying fixed in the same place relative to the stars during this sequence (which spans 5 nights), and it is the Moon which is moving. Why is this?

## The Motion of the Moon in the sky

The reason it is the Moon which appears to move against the background of Jupiter and the stars is because the Moon is orbiting us; whereas Jupiter is orbiting the Sun and the stars are not orbiting the Sun but are, along with the Sun, in fact orbiting the centre of our Milky Way galaxy.

As the Moon takes roughly 30 days to orbit the Earth (see this blog for the more precise figure, and the difference between how long it takes to orbit the Earth – the “sidereal month” – and how long it is between two New Moons – the “synodic months”), then if we divide $360^{\circ}$ by 30 we get that the Moon moves $12^{\circ}$ in its orbit about the Earth each day/night. This figure is only approximate (but good enough for our purposes) because (a) a sidereal month is not exactly 30 days and (b) the Moon moves in an ellipse and not a circle about the Earth, and so changes its speed at different points in the orbit, so does not move the same amount each 24 hour period.

As the Moon is $0.5^{\circ}$ in diameter, $12^{\circ}$ corresponds to 24 times the diameter of the Moon. This is quite a lot, and so the motion of the Moon from night to night against the background planets and stars is very easily seen, as the sequence of diagrams above show.

In fact, Jupiter is also moving against the background stars, but it does so much more slowly. Jupiter takes about 12 years to orbit the Sun, and so each year it moves roughly 1/12th of a full circle. Along with the Sun, the Moon and the other planets, Jupiter moves through the zodiacal constellations during its travels, and so moves roughly into a new zodiacal constellation each year. At the moment it is in Cancer, but by this time next year it will be in Leo, the next constellation along the zodiac to the East (to the left of Cancer in the diagram above). You may be able to notice that Jupiter has moved relative to the background stars in a month or two, but certainly by next February, if you remember where it is now, you will see a difference.

Finally, although we refer to the stars as “the fixed stars”, they are not fixed. They, along with our Sun, are orbiting the centre of our Milky Way galaxy. Our Sun will take 250 million years to do this, stars closer to the centre of the Milky Way will take less time and stars further out from the centre will take longer. This leads to the positions of the stars relative to each other changing, but the change is very very slow, taking tens of thousands of years to be noticeable.

## Space debris and geostationary satellites

Last week, as part of the “tech hour” on BBC Radio 5’s “Afternoon Edition” on Wednesdays, I heard one of their reporters do an item on visiting a facility in Australia which is trying to track space debris and locate their positions precisely. After the item a listener asked about geostationary orbits, and it was clear from the reporter’s answer that he did not really understand them. Then again, if he is not a trained physicist or engineer, why should he?

In the light of the wrong answer given to the listener by the reporter, I’ve decided to blog about satellite orbits to clear up any confusion, and also to show that the mathematics of it is not that hard. This is the kind of problem that a student studying physics would be expected to do in their last year of high-school here in Wales. Mark Thompson, one of the astronomers who has appeared on BBC Stargazing Live, did clear up some of the confusion when he came on the programme later in the same hour, but he did not have time to go into all the details (nor would he want to do so on a radio show!).

## Low-Earth and geostationary orbits

For those of you not familiar with the terminology, a low-Earth orbit is typically just a few hundred kilometres above the Earth’s surface, which as I will show below leads to an orbit about the Earth which takes a couple of hours at most. These are the kinds of orbits which are used by most spy satellites, and also by the International Space Station (ISS) and, when it used to fly, by the Space Shuttle. The famous Hubble Space Telescope is in a low-Earth orbit. Any satellites in low-Earth orbits will experience drag from the Earth’s atmosphere.

Space “officially” begins at 100km above the Earth’s surface, but the atmosphere doesn’t suddenly stop at any particular altitude,it just gets thinner and thiner. In fact, the “scale height” of the atmosphere is about 1 mile, or about 1.5km. For each 1.5km altitude, the amount of atmosphere roughly halves (based on pressure), so at an altitude of 3km the air is $0.5 \times 0.5 = 0.25 (25\%)$ as thick, and if you go up to 10.5km which is roughly the altitude at which commercial aeroplanes fly, then the thickness of the atmosphere is about $0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 = 0.5^{7} = 0.0078$ as thick, or $0.78 \%$ of the thickness at the ground. Even though the thickness of the atmosphere a few hundred kilometres up is very very thin, there is still enough of it to slow satellites down. So, satellites in low-Earth orbits cannot maintain their orbits without some level of thrust, otherwise their orbits would just decay and they would burn up in the atmosphere.

Low-Earth orbit satellites can be in any orbit as long as it is centred on the Earth’s centre. So, for example, many spy satellites orbit the Earth’s poles. If you think about it, if they are taking about 90 minutes to orbit (some take less, they will be even closer to the Earth), then each time they orbit they will see a different part of the Earth, which is often what you want for spy satellites. The ISS orbits the Earth in an orbit which is titled to the Earth’s equator, so it passes over different parts of the Earth on each orbit, but it is not a polar orbit. A satellite cannot, for example, orbit, say, along an orbit which is directly above the tropic of Cancer, and satellites do not orbit the Earth in the opposite direction to the Earth’s rotation, they all orbit the Earth in the same direction as we are rotating.

Geostationary orbits, on the other hand, are a particular orbit where the satellite takes 24 hours to orbit the Earth. So, as seen by a person on the rotating Earth, they appear fixed relative to that person’s horizons. Geostationary orbits are used for communication satellites so that, for example, your TV satellite dish can be fixed to the side of your house and point in the same direction, as the satellite from which it’s getting the TV signals stays fixed in your sky. They are also used by Earth-monitoring satellites such as weather satellites and satellites which measure sea and land temperatures. A geostationary satellite has to be at a particular altitude above the Earth’s surface, and it has to be in orbit about the Earth’s equator (so it sits above the Earth’s equator), and of course orbits in the same direction as the Earth’s rotation.

How do we calculate the altitude of these respective orbits, the low-Earth and the geostationary ones? It is not too difficult, as I will show below.

## The altitude of a satellite which takes 90 minutes to orbit the Earth

To make the mathematics simpler, I will do these calculations assuming a circular orbit. In reality, some satellites have elliptical orbits where they are closer to the Earth’s surface at some times than at others, but that makes the maths a little more complicated, and it’s not necessary to understand the basic ideas.

As I showed in this blog, when an object is moving in a circle it must have a force acting on it, because its velocity (in this case its direction) is constantly changing. This force is called the centripetal force, and as I showed in the blog this is given by

$F_{centripetal} = \frac{ mv^{2} }{ r }$

where $m$ is the mass of the object moving in the circle, $v$ is its speed, and $r$ is the radius of the circle.

For a satellite orbiting the Earth, the centripetal force is provided by gravity, and the force of gravity is given by

$F_{gravity} = \frac{ G M m}{ r^{2} }$

where $M$ is the mass of the Earth (in this case), $m$ is the mass of the satellite, $r$ is the radius of the orbit as measured from the centre of the Earth, and $G$ is a physical constant, known as Big G or the universal gravitational constant.

Because the centripetal force is being provided by gravity, we can set these two equal to each other and so we have

$\frac{ mv^{2} }{ r } = \frac{ G M m }{ r^{2} } \rightarrow v^{2} = \frac{ GM }{ r } \text{ (Equ. 1) }$

The speed of orbit, $v$ is related to how long the satellite takes to orbit via the definition of speed. Speed is defined as distance travelled per unit time, and as we are assuming circular orbits the distance travelled in an orbit is the circumference of a circle. So, we can write

$v = \frac{ 2 \pi r }{ T }$

where $T$ is the time it takes to make the orbit. Substituting this expression for $v$ in Equation (1) we can write

$\left( \frac{ 2 \pi r }{ T } \right)^{2} = \frac{ GM }{ r }$

Remember we are trying to calculate $r$ for a satellite which takes 90 minutes to orbit, so we re-arrange this equation to give

$r^{3} = \frac{ GM T^{2} }{ 4 \pi^{2} } \text{ (Equ. 2)}$

Notice that the only two variables in this equation are $r \text{ and } T$, everything else is a constant (for orbiting the Earth, obviously if the object were orbiting e.g. the Sun we’d replace the mass $M$ by the mass of the Sun). Equation (2) is, in fact, just Kepler’s Third law (which I blogged about here), which is that

$\boxed{ T^{2} \propto r^{3} }$

Now, to calculate $r$ for a satellite taking 90 minutes to orbit, we just need to plug in the values for the mass of the Earth ($5.972 \times 10^{24}$ kg), the value of $G=6.67 \times 10^{-11}$, and convert 90 minutes to seconds which is $T=5400 \text{ s}$. Doing this, we get

$r^{3} = 2.942 \times 10^{20} \rightarrow r = 6.65 \times 10^{6} \text{ m}$

But, remember, this is the radius from the Earth’s centre, so we need to subtract off the Earth’s radius to get the height above the Earth’s surface. The average radius of the Earth is $R = 6.37 \times 10^{6} \text{ m}$ so the height above the Earth’s surface of a satellite which takes 90 minutes to orbit is

$6.65 \times 10^{6} - 6.37 \times 10^{6} = 281 \times 10^{3} \text{ m } \boxed{ = 281 \text{ km} }$

which is roughly 300km.

## The altitude of a geostationary satellite

To calculate the altitude of a geostationary satellite we just need to replace the time $T$ in equation (2) with 24 hours (but put into seconds), instead of using 90 minutes. 24 hours is $24 \times 3600 = 8.64 \times 10^{4} \text{ s}$, and so we now have

$r^{3} = 7.53 \times 10^{22} \rightarrow r = 4.22 \times 10^{7} \text{ m}$

Again, subtracting off the Earth’s radius, we get the altitude (height above the Earth’s surface) to be

$4.22 \times 10^{7} - 6.37 \times 10^{6} = 3.59 \times 10^{7} \text{ m } \boxed{ = 35,900 \text{ km} }$

which is roughly 36,000 km.

## Summary

It surprises most people that geostationary satellites are so far above the Earth’s surface, some 36,000 km. There is no atmosphere at that altitude, so geostationary satellites really pose no threat in terms of man-made space debris. Their orbits will not decay, as there is no atmospheric drag. It is the low-Earth orbit satellites which are the source of our man-made space debris, and if their orbits are not maintained the orbits will decay and they will burn up in the Earth’s atmosphere. Of course, most satellites entering the thicker parts of the atmosphere burn up completely, but larger satellites have been known to have parts of them survive to hit the Earth’s surface. For example this is what happened to the Russian space station MIR when its orbit was decayed by switching off its thrusters.

The time it takes a satellite to orbit the Earth is entirely dependent on how far from the centre of the Earth the satellite is. Low-Earth orbit satellites orbit the Earth in a matter of hours, many less than two hours. Geostationary satellites have to be about 36,000 km from the surface of the Earth in order to take 24 hours to orbit. All satellites orbit the Earth in the same direction as the Earth is rotating, except for satellites in a polar orbit which essentially allow the Earth to rotate below them as they go around the poles. Geostationary satellites have to orbit above the Earth’s equator, but other satellites can be in an orbit which is inclined to the equator, as long as the centre of their orbit is the centre of the Earth.

Finally, coming back to the issue of space debris; one of the reasons that space debris is so dangerous to other satellites and astronauts is that objects in Earth-orbit are moving very quickly. For example, for an object taking 90 minutes to orbit the Earth, such as the orbit the ISS is in, we can work out how quickly (in metres per second) it is moving quite easily.

If $T=90 \text{ minutes } = 5400 \text{ s}$ and the radius of the orbit (as we showed above) is $r = 6.65 \times 10^{6} \text{ m }$ then

$v = \frac{ 2 \pi \cdot 6.65 \times 10^{6} }{ 5400 } \; \; \boxed{= 7737.6 \text{ m/s } }$

so, nearly 7,740 m/s or 7.7 km/s or nearly 28 thousand km/h, which by anyone’s standards is FAST!

The damage a piece of space debris can make is dependent on its kinetic energy (KE), which is given by

$\text{Kinetic energy } = \frac{ 1 }{ 2 } mv^{2}$

An object with a mass of 1kg moving at 7,740 m/s will thus have a KE of

$KE = \frac{ 1 }{ 2 } 1 \cdot (7740)^{2} = 30.0 \times 10^{6} \text{ Joules}$

(or 30 Mega Joules). This is a huge amount of energy, because the object is moving so fast. By comparison, a car, which we will assume to have a mass of 1,000 kg, moving at 100 km/h will have a kinetic energy of $0.77 \times 10^{6} \text{ Joules}$, or nearly 0.8 Mega Joules, so about a factor of 40 (forty!) less than a 1kg object in low-Earth orbit. This is why such small objects can cause such damage!