Travelling on the surface of a star

As a follow-up to my last post calculating the distance to the Moon I was asked to calculate the following:

"Here is a question for you Mr Science. How long would it take a plane flyng at approximately 900km per hour across the surface of the biggest known star in our galaxy to travel full circle?"

Good question.  Let's assume that the star is perfectly spherical, which seems reasonable enough.  Now, to find the largest known star in our galaxy.  According to Wikipedia, the largest star in our galaxy is VY Canis Majoris found in the constellation Canis Major (meaning 'large dog').  It's a particularly bright star, which appears faint in the night sky because it's so far away.

VY Canis Majoris has a radius of 1800~2100 solar radii (it's 1800-2100 times wider than the Sun) - the figure varies as the star is surrounded by a nebula, which also makes it difficult to get an exact figure.  Taking 1800 solar radii as a minimum figure, to give us an approximate idea, this means that the star has a radius of:

1 solar radius = 695,500 km
1800 solar radii = 1.251 billion kilometres

Now, the circumference (i.e. the distance around the edge of the star) is 2 π r which gives a circumference of 7.866 billion kilometres.

And that's just the distance around the star's equator...  it's enormous.

Travelling at 900 km per hour, this would take 7.866 billion / 900 = 8.74 million hours.
8.74 million hours = 364,163 Earth days = 997 Earth years (assuming 365.25 days per year).

The exact figure depends on the value of solar radius, the rest is maths, but a round figure would be 1000 years.  Having said that, 900 km/h is not that fast - the speed of sound (Mach 1) is 1193 km/h.  The land speed record is held by Thrust SSC which achieved 1240 km/h in 1997, while Concorde used to reach 2170 km/h.

Still, doubling the speed from 900 km/h to 2170 km/h is only going to reduce the journey time to 500 years... so perhaps the question of time should be put aside.   The real question should be, if you're going to fly or travel on the surface of a star with a temperature of 3000 K, how are you going to keep the pilot flying, and stop him from frying?

Calculating the Earth-Moon distance

This post follows up my previous post on geostationary satellites.  Long before we were launching satellites (even non-geostationary ones), our natural satellite, the Moon, was orbiting the Earth.  As the moon goes around the earth, its phase (shape) changes, and in fact, the word "month" derives from "moonth", the time taken for the moon to go from new to full to new again.  This time is the time taken for one complete orbit around the Earth - the different phases of the moon are a result of us seeing a different amount of the lit half of the moon (I once based a very neat science lesson on this principle - in fact I used it in my interview lesson  and subsequently got the job).

One of my photographs of the moon, taken through a telescope.
The darkening at the bottom of the image is the edge of the
telescope's field of view

We can use physics, and our knowledge of the mass of the Earth, the value of pi and the time the Moon takes to complete one orbit, to work out how far it is from the Moon to the Earth.

Back to the two key equations that we'll need, which are the force on a body moving in a circular path:

where

And Newton's Law of Gravity

Equating the two, and rearranging to find r, gives us

This is the same equation used for geostationary satellites, and describes the basic relationship between the distance between two bodies (a planet and a moon, for example, or a star and a planet).  This gives it great power as it can be used in many different situations.

Turning to the current situation, then:

Calculation of the Earth-Moon distance:

G is the universal gravitational constant, 6.67300 × 10^-11 m³ kg^-1 s^-2

M is the mass of  the Earth, 5.9742 × 10²⁴ kg

T is the time to complete one orbit, which for the Moon is 27.32166 days, which is 2,360,591 seconds.

Plugging the numbers into the formula above gives the distance as 383,201 km

However, this is not the distance to the Moon from the Earth's surface.  Newton's law of gravity gives the distance between the centres of gravity of the two bodies.  I'm ignoring the radius of the Moon (which is perhaps an oversight on my part, you decide) but we must subtract the radius of the Earth from this value, to give the orbital height.  Radius of Earth = 6378.1 km, so the distance to the Moon is calculated as  = 376,823 km, or, if you prefer, (at 1.61 km to the mile), 234,147 miles.

Previously, I've learned that the Moon is about a quarter of a million miles away, so I'm glad the method I've used shows a figure which is 'about right' without any checking.  Looking at other sources, it looks like my figures are close enough, considering the assumptions I've made.  One key assumption I've made is to suggest that the moon travels in a circular orbit, and it doesn't.  It has an elliptical orbit, which means the distance from Earth to Moon changes during the orbit - so I've calculated an average distance.  Still, my figure is pretty close, and not a million miles away (and next time somebody reliably informs you that their opinion is not a million miles away, you can tell them that not even the Moon is that far away).

Other astronomy related articles you may find interesting:

Calculating the height of geostationary satellites
Calculating the angle of elevation of geostationary satellites
The Earth-Moon distance is increasing (published 1 April 2011)
What are constellations?

Calculating (and explaining) escape velocity
Measuring g using a pendulum

Chemistry 3: Made-up names for chemicals

Okay, so last time I ranted about badly-defined science, and deliberately-misnamed scientific phrases, like liquid calcium in toothpaste, and silver molecules in deodorants.  But don't get me started on made-up chemicals.  Not make-up chemicals, made-up chemicals.  And not in science fiction, either, where they belong (kryptonite, dilithium and so on).  Crazy, thrown-together prefixes and suffices that sound like they're chemicals, but are nothing of the sort.  No, they aren't.  Nutrileum, nutrisse, proxylane... they're all nonsense.  Not to mention the use of correct chemical names in incorrect ways.  I mentioned this last time, but 'active oxygen' is another that gets me.  Why is it so different from 'inactive oxygen', or 'lethargic oxygen'?  No, adverts must contain 'active oxygen' because inactive oxygen would probably make us all go to sleep.

Let's take a look at a classic acronym:  AHAs.  Alpha-hydroxy acids.  Yes, that's acids - and by applying this product to your cheeks, you'll be putting acids on your face.  Doesn't this worry you?  Of course it does, because acids are renowned among the uninformed as being bad for your health (they cause tooth decay, among other things), so they get labelled AHAs instead.  What are they?  Acid molecules... organic acid molecules at that (more on 'organic' either today or some other time, but definitely soon).

Pentapeptides - or leftover bits of protein - penta meaning 'having five' and peptides meaning 'made from amino acids'.  So a pentapeptide is a molecule made up of five amino acids (most probably).  A typical protein has thousands of amino acid parts, so having five is a bit small.  Applying this goo to your skin and expecting it to automatically improve your complexion is a bit like trying to construct a piece of literature by randomly adding five-letter chunks to the end of it.  To be fair, skin is a living, well-designed material, and is able to pick out the right letters to keep spelling 'skin' the five-letter chunks, but I wouldn't be surprised if it was proven that eating more healthily does wonders for your skin.  After all, the digestive system is designed to absorb the amino acids to develop and maintain a healthy body, moreso than just throwing sticky creams at your skin, at any rate.

Oxygen is best breathed in through the lungs, carried out in red blood cells in the blood, and used to release the energy found in the food we eat.  Otherwise, oxygen is a gas that's capable of causing damage to cells, and that's why it has to be transported so carefully around the body.  A search for, 'dangers of oxygen' will demonstrate what unrestrained oxygen can do in the human body.  For 'active' in this context, I think it's probably best to read 'fizzy' and producing a tingling sensation.  I can't help wondering what they'll come up with next - perhaps turbo-charged water?  There's much discussion of active oxygen around, and the truth be told, I can't find (and don't know) of the exact definition of active oxygen, but from what I gather, it's all about the electrons in the oxygen molecule... suffice it to say, it's probably not all it's cracked up to be!

As for anything that ends in 'ane', 'ene' or 'eum'... oh dear.  How about methane (natural gas, and very smelly), benzene (I mentioned last time, a cancer-causing irritant) and the ileum (the last part of the small intestine, a dark and unpleasant place)?  And the additions of "oxy" or "xylane" make advertisers look as if they've been building words to obtain good Scrabble scores, instead of communicating science.

Not to mention the lazy addition of 'pro' to the start of a chemical name.  How about 'pro-retinol'?  Sounding like a concoction of pro, retina and alcohol, perhaps it's supposed to help with treating the bleary eyes that follow a late night out?  What's its chemical composition?  Nothing of the sort. Pro-argin (Colgate), Pro-retinol and Pro-gen (L'Oreal), not to mention Pantene Pro-V (where the V stands for...?).

Better still are numbers at the end of chemical names.  Pro-retinol-nine sounds even better than pro-retinol-one, although I'm sure the numbers are carefully chosen to sound scientific.  And if you don't think numbers can sound scientific, consider seven compared to four.  I don't know why pro-chemical-seven should sound better than pro-chemical-four, but Chanel certainly rates No 7, for some unknown reason!

I could go on... and let's be honest, in a future blog post I probably will.  In addition to more made-up chemical names, I'll look at the list of components in shampoo, soap, shower gel and so on, and begin to explain what they actually do, and why (if I can work it out) they're given the strange names that they have.  Until then, I'll keep smiling and laughing at the cosmetics adverts, and browsing the Advertising Standards Authority website for the latest promotional blunders!

Chemistry 2: A rant at pseudo-science adverts

"The only toothpaste with liquid calcium to strengthen your teeth"
"When a car brakes, some of the energy it produces is lost"
"Anti-perspirant with silver molecules"
"Contains pro-oxylane to give your hair extra shine"

Don't get me started on pseudo-science in television adverts.  The voice-over begins, sounding professional, authentic and a leading authority on all things scientific.  The picture zooms in with hexagons flying around all over the place, and stick-and-ball-model molecules start being absorbed into your hair, skin and teeth.

What is it with the hexagons anyway?  Do these chemicals contain honey from a honeycomb?  Do the treatments contain a hexagonal molecule, like benzene (causes cancer, leukemia and is fatal if absorbed in even small doses) or cyclohexane, say (harmful if inhaled or swallowed)?  Perhaps "hexagons" means "scientifically clever", and not "very dangerous".  Still, just watch a TV advert for the latest shampoo or face cream and count the hexagons.  Double your score if the hexagons are gold-coloured or shiny.

Liquid calcium, featured recently in a toothpaste advert (and an advert which thankfully has not been seen recently) is an interesting concept.  Calcium is a metal, and as is generally know, metals have high melting points; in the case of calcium, you have to heat it up to 842 to 848°C in order to melt it.  Now, if toothpaste actually contains liquid calcium, I wouldn't want to put it in my mouth - in fact I wouldn't want to hold the toothpaste tube (and I wouldn't even want to contemplate squeezing the tube, which would have to be made of something other than the typical plastic material).  Still, I'm glad the advert was taken off air.  The truth (a strange concept for advertising, I accept) is that it will contain a calcium compound in an emulsion.  It's a bit like saying that the sea is liquid salt:  interesting, but patently untrue.  Actually, a closer comparison would be to say that the sea is liquid chlorine: dramatic and thankfully untrue.

Next:  silver molecules in anti-perspirant.  Yes, some anti-perspirants contain some molecules that contain a silver ion in a larger molecule.  Here's an example of one such anti-perspirant and for those paying attention, please note the shapes on the front of the can.  The advert for the anti-perspirant features a silver truncated icosahedron - another scientific shape that's not relevant here - that strongly suggests that the product contains a molecule composed entirely and uniquely of silver atoms.

Silver molecule?  Or just a football?

The truth about 'silver molecules' is much more prosaic; the compound in question is probably a variation of the molecule silver sulfadiazine, which, incidentally also contains sulphur - again, not something you'll hear in  advert.  Here's silver sulfadiazine - I figured it was time for a diagram with some genuine scientific basis.  The silver ion is shown by the Ag+ as the chemical symbol for silver is Ag (from the Latin argentum, which also provides the French argent).

Next time, I'll look at made-up scientific names - if only for the fun of it.  Pro-xylane, AHAs, nutrileum, pentapeptides and the rest of it.  In the meantime, should advertisers use proper science?  Probably not.  Will they?  No.  Why not?  Because they're not worth it.

Chemistry 1: The Periodic Table

Okay, I'm temporarily setting aside my first chemical rant to provide something a little more useful and objective: a neat way to remember the first twenty or so elements of the periodic table in order.  This mnemonic was given to me by my uncle (thanks Graham!) and I had to extend it in order to complete the first row of the transition elements, which came in useful from my A-levels onwards.

The mnemonic goes like this:

Ha, Here Lies Benjamin Bones; Cry Not Old Friend Needlessly, Nature Magnifies All Simple People, Sometimes Clowns Are Kings.  Callous Scoundrels Tickle Viciously; Crumbling Magnolias Fear Cold Nights, Cute Zones Gather Geraniums.

As you can see, it's a little uneven, but it works.  One important note:  the words are expansions of the chemical symbols for the elements, and are not always helpful in remembering the element names where the symbol and name aren't connected in English.  For example, "nature" is an expansion of Na, which is sodium; similarly, "kings" is the keyword for K, the symbol for potassium.

I'd suggest consulting a periodic table, like this one, to help remember where to put them all (and where the line breaks go :-)

The only other mnemonic for learning the periodic table in sequence that I've found while skimming the web is this one, which makes some sense but doesn't cover much territory:

Harry He Likes Beer But Can Not Obtain Food

Others seem far more complicated, more interested in electronic configurations (which are far too complicated and easier to just sit down and work out, then learn by sight).

So, long live Benjamin Bones... except he's died already!

Mathematical Problems 4B: Close-packed spheres

In a previous post, I looked at close-packed circles - arranging circles hexagonally and calculating how much of the available area they fill.  That was fairly straightforward, and gave me an idea on how to calculate the volume occupancy of close-packed spheres.  Put it another way - how many Maltesers could I fit in a box if I could fill it to capacity (minimum spare volume left over)?

Let's start by putting them in a square arrangement - so that they're all in columns and rows.  How much space will they fill?  The easiest way to solve this is to think of one sphere inside its cube-shaped box.

In a cube-shaped box

The volume of a sphere S = 4/3 π r³
And the volume of the cube it sits in is 2r x 2r x 2r (since the cube has to be twice the radius of the sphere in height, width and depth)  C = 8 r³

Therefore, the ratio of the two volumes is S/C which is 4 π / 24 (notice that the r cancels - it doesn't matter how big the sphere is) and this is equal to 52.36% - only half of the available volume.

Next, let's look at hexagonal packing - arranging the circles so that they form hexagon patterns, instead of squares.

The hexagonal box

If we consider just one of the spheres, enclosed in a regular hexagonal prism, then we have this arrangement:

The volume of a sphere is...

And the volume of the hexagonal prism it occupies is found by multiplying the area of the base by the height.  The height is 2r (it's twice the radius of the sphere in height) and we can look at the base as being made up of six equilateral triangles...

So the volume of the hexagon, H, is 

And now that we have the volume of the hexagonal prism, H, and the volume of the sphere, S, we can work out how much of the prism is being filled by the sphere.

So, if we want to maximise the number of Maltesers in a box of chocolates, it makes sense to arrange them hexagonally, and not cubically.  80% volume coverage, compared to just 52% for the cubic arrangement, is definitely worth having!

Hexagonally arranged Maltesers (a Christmas present!)

On a more theoretical note, science textbooks regularly quote that close-packed spheres fill 80% of their volume, but I've never noticed any of them prove it.  So, this blog post counts as closure from a figure that's been drifting around since my A-level chemistry days, and which has recurred frequently since then.  None of the books seemed bothered enough to spend time on it - perhaps it's too much like maths and not enough like chemistry!

Next time - something different.  It might be projectiles, or escape velocity (conveniently related to each other) or it might be something about chemistry - in which case it'll be a chemistry rant (consider this advance notice!).

Further reading on circles and spheres

A previous post on the radius of a circle in the corner of a circle
Hexagonal close packing - 2D-filling calculation

Physics Experiment: Determine g with a pendulum

Having done some work on determining pi by mathematical methods, I'm now going to use it in conjunction with some experimental work to determine the value of g, which is acceleration due to gravity. Any reference book will tell you the value of g is approximately 9.81 ms-2, but I'm going to do an experiment to show what it is. It's not a difficult experiment, and it doesn't require any specialised scientific equipment. To give you an idea, I did this using a toddler fireguard for my vertical surface, a piece of sewing cotton for my pendulum, and in the absence of any respectable small mass, used a small pine cone tied to the end of it. I also used a standard stopwatch on a digital watch (it's accurate to 1/100th of a second, although I'm not).

The relationship between a pendulum and g is described in the following limerick:

If a pendulum's swinging quite free
Then it's always a marvel to me
That each tick plus each tock
Of the grandfather clock
Is 2 pi root L over g

In order to improve the accuracy of my results, I counted the time taken for ten complete 'swings' or periods, and quoted this. I also repeated each ten-swing measurement three times, so that I could take an average and identify any anomalies. And somehow, saying that, I feel like I'm writing up a GCSE science coursework piece!

Here are my results...

length (l, in metres)	10 swings (10T, seconds)	Run 2	Run 3
0.162	8.45	8.31	8.35
0.237	10.09	9.90	10.04
0.321	11.63	11.67	11.66
0.344	12.09	11.90	12.17
0.410	12.98	12.95	13.00
0.475	14.23	14.13	14.03

I calculated the average value of 10T, and hence T and then T², which I can use to determine g, with the following rearrangement:

An alternative, if I'd wanted to plot a graph of my data, is to determine g by finding the slope of the appropriate plot.  Using the following rearrangement, it's possible to plot T²  against l and have a slope of 4pi²/g

However, I'm going at it in number-crunching form, using the formula above.  My results for g are as follows:

length (l, in metres)	g in ms-2
0.162	9.136
0.237	9.338
0.321	9.332
0.344	9.3436
0.410	9.612
0.475	9.395

So, not perfect, but given the nature of the experiment - me with a fireguard and a pine cone - it's not too bad at all, and I feel quite pleased at having worked out something so massively significant with such basic equipment, and I feel it proves that science isn't just for big-budget departments!

Next time, determining the distance to the moon using the same principle as for geostationary satellites (except that this one is a bit bigger, a bit further away and not geostationary!).