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## First ever asteroid from another solar system detected

In late October 2017, astronomers announced the first ever discovery of an asteroid (or comet?) coming into our Solar System from another stellar system. The object was first spotted on 19 October by the University of Hawaii’s Pan-STARRS telescope, during its nightly search for near-earth objects. Based on its extreme orbit and its rapid speed, it was soon determined that the object has come into our Solar System from somewhere else, and this makes it the first ever asteroid/comet with an extra-solar origin to have been discovered. Originally given the designation A/2017 U1, the International Astronomical Union (IAU) have now renamed it 1I/2017 U1, with the I standing for “interstellar”.

The object, given the designation A/2017 U1, was deemed to be extra-solar in origin from an analysis of its motion.

In addition to its strange trajectory, observations suggest that the object also has quite an unusual shape. It is very elongated, being ten times longer than it is wide. It is thought to be at least 400 metres long but only about 40 metres wide. This was determined by the rapid and dramatic changes in its brightness, which can only be explained by an elongated object tumbling rapidly.

The object has also been given the name Oumuamua (pronounced oh MOO-uh MOO-uh), although this is not its official name (yet).  This means “a messenger from afar arriving first” in Hawaiian. In other respects, it seems to be very much like asteroids found in our own Solar System, and is the confirmation of what astronomers have long suspected, that small objects which formed around other stars can end up wandering through space, not attached to any particular stellar system.

## The origin of the elements

A couple of weeks ago this fascinating version of the periodic table of the elements was the NASA Astronomy Picture of the Day (APOD). Most people have seen the periodic table of the elements, it is shown on the wall of most high school chemistry classrooms. But, what is totally fascinating to me about this version is it shows the origin of each element.

It has been a long process of several decades to understand the origin of the elements. In fact, we have not totally finished understanding the processes yet. But, we do know the story for most elements. All the hydrogen in the Universe was formed in the big bang. This is true for nearly all the helium too. A small amount of the 25% or so of helium in the Universe has been created within stars through the conversion of hydrogen into helium. But, not much has been created this way because most of that helium is further converted to carbon.

The only other element to be formed in the big bang is lithium. About 20% of the lithium in the Universe was formed in the big bang, the rest has been formed since,

Together, hydrogen and helium comprise 99% of the elements in the Universe by number (not by mass).

Where Your Elements Came From – from the NASA Astronomy Picture Of the Day (APOD) 24 October 2017.

I have decided to use this fascinating table as the basis for a series of blogs over the next few weeks to explain each of the 6 processes in these six boxes

## How to add velocities in special relativity

As I mentioned in this blogpost, in special relativity any observer will measure the speed of light in a vacuum to be $c$, irrespective of whether the observer is moving towards or away from the source of light. We can think of the speed of light as a cosmic speed limit, nothing can travel faster than it.

But, let us suppose that we have two reference frames $S$ and $S^{\prime}$ moving relative to each other with a speed of $v=0.9c$, 90% of the speed of light. Surely, if someone in frame $S^{\prime}$ fires a high-speed bullet at a speed of $u^{\prime}= 0.6c$, an observer in frame $S$ will think that the bullet is moving away from him at a speed of $u = v + u^{\prime} = 0.9c + 0.6c = 1.5c$, which seemingly violates the comic speed limit.

What have we done wrong?

We cannot simply add velocities, as we would do in Newtonian mechanics. In special relativity we have to use the Lorentz transformations to add velocities. How do we do this? Let us remind ourselves that the Lorentz transformations can be written as

The Lorentz transformations to go either from reference frame $S \text{ to } S^{\prime}$, or to go from $S^{\prime} \text{ to } S$.

## Calculating a velocity in two different reference frames

To calculate the velocity $u$ of some object moving with a velocity $u^{\prime}$ in reference frame $S^{\prime}$ we need to use these Lorentz transformations.

We start off by writing

$x = \gamma \left( x^{\prime} + vt^{\prime} \right) \text{ (1) }$

and

$t = \gamma \left( t^{\prime} + \frac{x^{\prime}v}{c^{2} } \right) \text{ (2) }$

We will now take the derivative of each term, so we have

$dx = \gamma \left( dx^{\prime} + vdt^{\prime} \right)$

and

$dt = \gamma \left( dt^{\prime} + \frac{dx^{\prime}v}{c^{2} } \right)$

We can now write $dx/dt = u$ (the velocity of the object as seen in frame $S$) as

$\frac{dx}{dt} = \frac{ \gamma \left( dx^{\prime} + vdt^{\prime} \right) }{ \gamma \left( dt^{\prime} + \frac{dx^{\prime}v}{c^{2} } \right) }$

The $\gamma$ terms cancel, and dividing each term on the right hand side by $dt^{\prime}$ gives

$\frac{dx}{dt} = \frac{ \left( dx^{\prime}/dt^{\prime} + vdt^{\prime}/dt^{\prime} \right) }{ \left( dt^{\prime}/dt^{\prime} + \frac{dx^{\prime}v}{c^{2} dt^{\prime} } \right) } = \frac{ \left( u^{\prime} + v \right) }{ \left( 1 + \frac{ u^{\prime} v}{c^{2} } \right) }$

$\boxed{ u = \frac{ \left( u^{\prime} + v \right) }{ \left( 1 + \frac{ u^{\prime} v}{c^{2} } \right) } }$

where $u^{\prime}$ was the velocity of the object in reference frame $S^{\prime}$.

Going back to our example of $v = 0.9c$ and $u^{\prime} = 0.6c$, we can see that the velocity $u$ as measured by an observer in reference frame $S$ will be

$u = \frac{ 0.6c + 0.9c }{ \left( 1 + \frac{ (0.6c \times 0.9c) }{ c^{2} } \right) } = \frac{ 1.5c }{ 1 + 0.54 } = \frac{ 1.5c }{ 1.54} = \boxed {0.974c}$, not $1.5c$ as we naively calculated.

## The constancy of the speed of light

What happens if a person in reference frame $S^{\prime}$ shines a light in the same direction as $S^{\prime}$ is moving away from $S$? In this case, $u^{\prime}=1.0c$. Putting this into our equation for $u$ we get

$u = \frac{ 0.6c + 1.0c }{ \left( 1 + \frac{ (0.6c \times 1.0c) }{ c^{2} } \right) } = \frac{ 1.6c }{ 1 + 0.6 } = \frac{ 1.6c }{ 1.6} = 1.0c$

So they both agree that the light is moving away from them with the same speed $c$!

## Is there life on Mars? (part 1)

As those of you following my blog will know, I am currently on a cruise around New Zealand, giving astronomy talks. One of my six talks is about our current understanding of whether there is (or was) life on Mars. I try to only talk about objects which are visible during the cruise, and Mars is currently visible in the evening sky, albeit a lot fainter than it was in May when it was at opposition.

One of the talks I am giving on this cruise is our current understanding of whether there is (or was) life on Mars.

The question of whether there is life on Mars, or whether there ever has been in its history, is a fascinating one. I thought I would do a series of blogs to explore the question. But, I have to begin by saying that ANY search for life beyond Earth is predicated by our understanding of life on Earth. The only thing, it would seem, required by all forms of life which we have found on earth is water. Extremophiles show that life can exist without oxygen, without light, at high pressure, in radioactive environments; in fact in all sorts of environments which humans would find impossible. But, none of the life so far found on Earth can exist without water.

As a consequence, all searches for life in our Solar System tend to begin with the search for water. Now, it may be that life beyond Earth could have evolved to exist without the need for water. I am no chemist, but I don’t think there is anything particularly unique about water in its chemistry which makes it impossible for living cells to use some other substance. Water is the only substance on Earth which can exist in all three forms naturally (solid, liquid and gas), so it does occupy an unique place in the environment found on Earth. But, on Titan for example, methane seems to exist in all three forms. Maybe life has evolved on Titan to metabolise using methane in the same way that life on Earth has evolved to metabolise using water. We don’t know.

So, I thought I would start this series of blogs with the big news in the 1890s, that Martians had built canals on the red planet!

## Schiaparelli and Martian ‘canali’

The Schiaparelli space probe which ESA sadly failed to land on Mars recently was named after Italian astronomer Giovanni Schiaparelli. In the late 1880s he reported seeing ‘canali’ on the surface of Mars. Although this means ‘channels’, it got mis-translated to ‘canals’, and led to a flurry of excitement that this was evidence of an intelligent civilisation on Mars.

The idea grew that Martians had built canals to transport water from the “wet” regions near the poles to the arid equatorial regions. The ice caps of Mars are easily visible through a small telescope, so astronomers had known for decades that Mars had ice caps which they assumed were similar to the ice caps on Earth.

Giovanni Schiaparelli’s map of ‘canali’ on Mars, from 1888.

One person who became particularly taken with this idea of canals on Mars was American Percival Lowell. Lowell came from a rich Bostonian family, and had enough personal wealth to build an observatory in Flagstaff, Arizona. He set about proving the existence of life on Mars, writing several books on the subject. He published Mars (1895), Mars and Its Canals (1906), and Mars As the Abode of Life (1908). But, by 1909 the 60-inch telescope at Mount Wilson Observatory had shown that the ‘canali’ were natural features, and Lowell was forced to abandon his ideas that intelligent life existed on Mars.

However, his Flagstaff Observatory was to go on and make important contributions to astronomy. In the 1910s Vesto Slipher was the first person to show that nearly all spiral nebulae (spiral galaxies as we now call them) showed a redshift, the first bit of observational evidence that the Universe is expanding. And, in 1930 Clyde Tombaugh discovered Pluto at Flagstaff Observatory.

In part 2 of this blog, next week, I will talk about the first space probes sent to Mars, and the first images taken of Mars by a space probe which successfully orbited the planet, Mariner 4.

## Women, Know Your Limits! – Harry Enfield (comedy)

Last week, my attention was drawn to an article entitled “Here’s Why There Ought to be a Cap on Women Studying Science and Maths”, which actually appeared on the alt-right website Breibart back in June 2015. Breibart is a right-wing site founded by Andrew Breibart, and the website’s current executive chairman is Stephen Bannon, who has recently been named by president-elect Donald Trump as his chief strategist in Trump’s new White House team.

Women: Know Your Limits! (a comedy sketch by Harry Enfield)

“Here’s Why There Ought to be a Cap on Women Studying Science and Maths” is what I can only describe as a diatribe, written by Milo Yiannopoulos, and Englishman who seems to have become one of the main voices for the kind of misogynistic nonsense which many in the “contrastive right” like to spout. I don’t want to give any more attention to the utter drivel that Yiannopoulos writes in his article, but I thought I would quote just a few lines.

Even women who graduate with good degrees in science subjects often don’t use them: they switch careers in their twenties, abandoning the hard sciences. In some cases, they simply drop out of the workforce altogether. This is a disaster for the men who missed out on places, and it’s a criminal waste of public funds.

That’s why I think there ought to be a cap on the number of women enrolling in the sciences, maths, philosophy, engineering… and perhaps medicine and the law, too. It’s hugely expensive to train a doctor, but women have something like a third of the career of a man in medicine, despite having equal access to Harvard Med. Women make up the majority of medical students.

In addition to twisting statistics and misquoting feminist academics to support his theory that women are not really capable of doing science, Yiannopoulos even suggests that there should be caps on women studying law! Interestingly, Yiannopoulos himself has started and failed to complete degrees at two different universities, which I guess means that he is a bit of an expert on failure and wasting taxpayers’ money.

Anyway, there is really no way to argue logically with an imbecile like Yiannopoulos, his views are best countered by a bit of humour. So, here is a wonderful sketch by Harry Enfield – “Women: Know Your Limits!”. Enjoy!

## Off to New Zealand

Tomorrow (Friday 25 November) I am boarding a plane which will eventually get me to Brisbane (Australia), via Seoul. Yes, I’m aware that Brisbane is not New Zealand, but in Brisbane I am joining a cruise which is going around New Zealand. The cruise will last for 14 nights, and I will give about 6 talks during the two weeks.

The Princess Cruise leaves Brisbane on 27 November and returns on 11 December. I will be giving astronomy talks on the 14-night cruise.

This will be the 5th cruise which I’ve done with Princess, and the 6th in total. The last time I did a cruise in the southern hemisphere was in February, when I cruised from Buenos Aires to Santiago around Cape Horn. Unfortunately, during that 14-night cruise, we had only one clear night! I am hoping for better weather this time, as in addition to my talks I run star parties to show the guests what is visible in the night-time sky.

Many of the guests will probably be from Europe or the United States, and so will be very keen to see the Southern Cross. I will also show them the Magellanic Clouds if weather permits. The New Moon is on the 29 November, so the first week of the cruise will be ideal to see the Magellanic Clouds if the skies are clear. After that, the brightening moon will render them all by invisible. So, fingers crossed we get some clear skies during the first week!

## A cylinder rolling down a slope

The other week I was asked to explain how a cylinder (or ball) rolling down a slope differs from e.g. a ball being dropped vertically. It is an interesting question, because it illustrates some things which are not immediately obvious. We all know that, if you drop two balls, say a tennis ball and a cannon ball, they will hit the ground at the same time. This is despite their having very different masses (weights). Galileo supposedly showed this idea by dropping objects of different weights from the tower of Pisa (although he probably never did this, see our book Ten Physicists Who Transformed Our Understanding of Reality).

With a tennis ball and a cannon ball, they clearly have very different masses (weights), but will fall to the ground at the same rate. This fact, contrary to the teachings of Aristotle, was one of the key breakthroughs which Galileo made in our understanding of motion. But, what about if we roll the two balls down a slope? If we build a track to keep them going straight, will a tennis ball roll down a slope at the same rate as a cannon ball? The answer is no, and I will explain why.

## Rolling rather than dropping

When a ball rolls down a slope, it starts off at the top of the slope with gravitational potential energy. When it starts rolling down the slope, this gravitational potential energy gets converted to kinetic energy. This is the same as when the ball drops vertically. But, in the case of the ball dropping vertically, the kinetic energy is all in the form of linear kinetic energy, given by

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

where $m$ is the mass of the ball and $v$ is its velocity (which is increasing all the time as it falls and speeds up). The gravitational potential energy is converted to linear kinetic energy as the ball drops; by the time the ball hits the bottom of its fall all of the PE has been converted to KE.

If, instead, we roll a ball down a slope, the kinetic energy is in two forms, linear kinetic energy but also rotational kinetic energy, which is given by

$\text{ rotational kinetic energy } = \frac{ 1 }{ 2 } I \omega^{2}$

where $I$ is the ball’s moment of inertia, and $\omega$ is the ball’s angular velocity, usually measured in radians per second. The key point is that the the moment of inertia for the two balls in this example (a tennis ball and a cannon ball) have a different value, because the distribution of the mass in the two balls is different. For the tennis ball it is all concentrated in the layer of the rubber near the ball’s surface, with a hollow interior. For the cannon ball, the mass is distributed throughout the ball.

## Two cylinders rolling down a slope

Let us, instead, consider the case of two cylinders rolling down a slope. One is a solid cylinder, the other is a hollow one with all of its mass concentrated near the surface. We will make the two cylinders have the same mass; this can be done by making the material from which the hollow cylinder is made denser than the material for the solid cylinder. So, even though the material of the hollow cylinder is all concentrated near the surface of the cylinder, and there is a lot less of it, if it is denser it can have equal mass.

A solid cylinder on an inclined plane. We will make the mass of this solid cylinder the same as that of the hollow cylinder, by making it of less dense material. Although it will have the same mass $m$ and the same radius $R$, it will not have the same moment of inertia $I$.

We will start both cylinders from rest near the top of the slope, and let them roll down. We will observe what happens.

A hollow rolling down an inclined plane. We will make the hollow cylinder denser than the solid one, so that they both have the same mass $m$ and the same outer radius $R$. But, they will not have the same moment of inertia $I$.

When things are dropped, the rate at which they fall is independent of the mass, but when they roll the rate at which they roll is not indpendent of the moment of inertia. In particular, it is not independent of the distribution of mass in the rolling object. As this video shows, the solid cylinder rolls down the slope faster than the hollow one!

But, why??

## Why does the solid cylinder roll down quicker?

The reason that the solid cylinder rolls down faster than the hollow cylinder has to do with the way that the potential energy (PE) is converted to kinetic energy. Because the cylinder is rolling, some of the PE is converted to rotational kinetic energy (RKE), not just to linear kinetic energy (LKE). The only way that a cylinder can roll down a slope is if there is friction between the cylinder and the slope, if the slope were perfectly smooth the cylinder would slide and not roll.

The torque (rotational force) $\tau$ is related to the angular acceleration $\alpha$ in a similar way that the linear force $F$ is related to linear accelerate $a$. From Newton’s second law we know that $F = ma$ where $m$ is the mass of the object. The rotational equivalent of this law is

$\tau = I \alpha$

where $I$ is the moment of inertia. The moment of inertia $I$ is different for a hollow cylinder and a solid cylinder. For the solid cylinder it is given by

$I_{sc} = \frac{ 1 }{ 2 } mR^{2} = 0.5mR^{2}$

where $m$ is the mass of the cylinder and $R$ is the radius of the cylinder. For the hollow cylinder, the moment of inertia is given by

$I_{hc} = \frac{ 1 }{ 2 }m(R_{2}^{2} + R_{1}^{2})$

where $R_{2} \text{ and } R_{1}$ are the outer and inner radii of the annulus of the cylinder. We are going to make the hollow cylinder such that the inner 80% is hollow, so that $R_{1} = 0.8R_{2} = 0.8R$. We will make the mass $m$ of the two cylinders the same.

Thus, for the hollow cylinder, we can now write

$I_{hc} = \frac{ 1 }{ 2 }m(R^{2} + (0.8R)^{2}) = \frac{ 1 }{ 2 }mR^{2}(1+0.64) = \frac{ 1 }{ 2 }mR^{2}(1.64) = 0.82 mR^{2}$

The cylinder accelerates down the slope due to the component of its weight which acts down the slope. This component is $mg sin(\theta)$ where $g$ is the acceleration due to gravity and $\theta$ is the angle of the slope from the horizontal. To make the maths easier, we are going to set $\theta = 30^{\circ}$, as $sin(30) =0.5$.

Friction always acts in the opposite direction to the direction of motion, and in this case the friction $F_{f}$ is related to the torque $\tau$ via the equation

$\tau = F_{f}R \text{ (1)}$

so we can write

$F_{f}R = \tau = I \alpha \text{ (2)}$

where $\alpha$ is the rotational acceleration. Re-arranging this to give $F_{f}$, we have

$F_{f} = \frac{I \alpha}{ R }$

The force down the slope, $F (=ma)$ is just the component of the weight down the slope minus the frictional force $F_{f}$ acting up the slope.

$ma = mg\sin(30) - F_{f} = 0.5mg - \frac{ I \alpha }{ R } \text{ (3)}$

The angular acceleration $\alpha$ is given by $\alpha = a/R$ where $a$ is the linear acceleration. So, we can re-write Eq. (3) as

$ma = 0.5mg - \frac{ Ia }{ R^{2} } \text{ (4)}$

Now we will put in the moments of inertia for the solid cylinder and the hollow cylinder. For the solid cylinder, we can write

$ma = 0.5mg - \frac{ 0.5mR^{2}a }{ R^{2} } = 0.5mg - 0.5ma$

The mass $m$ can be cancelled out, and assuming $g=9.8 \text{ m/s/s}$, we have

$a = 0.5g - 0.5a \rightarrow 1.5a = 4.9 \rightarrow \boxed{ a = 3.27 \text{ m/s/s (5)} }$

Notice that Equation (5) does not have the mass $m$ in it, as this cancels out. It also does not have the radius $R$ of the cylinder in it; the acceleration of the cylinder as it rolls down the slope is independent of both the mass and the radius of the cylinder.

For the hollow cylinder, again using Eq. (4), we have

$ma = 0.5mg - \frac{ 0.82maR^{2} }{ R^{2} } = 0.5mg - 0.82ma$

This simplifies to

$a = 4.9 - 0.82a \rightarrow 1.82a = 4.9 \rightarrow \boxed{ a = 2.69 \text{ m/s/s} (6)}$

As with Equation (5), Equation (6) is independent of both mass and radius.

So, as we can see, the linear acceleration $a$ for the hollow cylinder is 2.69 m/s/s, less than the linear acceleration for the solid cylinder, which was 3.27 m/s/s. This is why the solid cylinder rolls down the slope quicker than the hollow cylinder! And, the result is independent of both the mass and the radius of either cylinder. Therefore, a less massive solid cylinder will roll down a slope faster than a more massive hollow one, which may seem contradictory.

## Summary

All objects falling vertically fall at the same rate, but this is not true for objects which roll down a slope. We have shown above that a solid cylinder will roll down a slope quicker than a hollow one. This is because their moments of inertia are different, it requires a greater force to get the hollow cylinder turning than it does the solid cylinder. Remember, the meaning of the word ‘inertia’ is a reluctance to change velocity, so in this case a reluctance to start rolling from being stationary. A larger moment of inertia means a greater reluctance to start rolling.

The solid cylinder will start turning more quickly from being stationary than the hollow cylinder, and this means that it will roll down the slope quicker. This result is independent of the masses (and radii) of the two cylinders; even a less massive solid cylinder will roll down a slope quicker than a more massive hollow one, which may be counter-intuitive.