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## Newton’s equations of motion – revisited

Last week, I showed how one could derive 3 of Newton’s equations of motion. As a colleague of mine pointed out to me on FaceBook, the 3rd equation I showed can, in fact, be derived using algebra from the 1st and 2nd. Also, if you go to the Wikipedia page on the equations of motion, you will find 5 listed, not just the 3 I showed. It turns out that the only two “fundamental” ones are the 1st and 2nd that I showed, the other three (including my 3rd, shown as the 4th in the image below) can be derived from the 1st and 2nd ones.

The 5 equations of motion as shown on the Wikipedia page

This is often the case in mathematics, there is more than one way to do something. So, although the method I showed to derive the 3rd equation is perfectly correct, and it also shows how we can rewrite $dv/dt \text{ as } (dv/ds \cdot ds/dt)$, which is a useful technique, I will today show how it can also be derived by algebraically combining equations (1) and (2).

Remember, equations (1) and (2) were

$v = u + at \text{ (Equ. 1)}$

$s = ut + \frac{1}{2} at^{2} \text{ (Equ. 2)}$

The first step is to square equation (1), which gives us

$v^{2} = (u + at)^{2} = u^{2} + 2uat + a^{2}t^{2} \text{ (Equ. 3a)}$

Next, we multiply equation (2) by 2 to get rid of the fraction

$2s = 2ut + at^{2} \text{ (Equ. 3b)}$

We next multiply each term in equation (3b) by $a$ to give

$2as = 2uat + a^{2}t^{2} \text{ (Equ. 3c)}$

Comparing equations (3a) and (3c) we can see that we can substitute the 2nd and 3rd terms of equation (3a) by $2as$ and so we have

$v^{2} = u^{2} + 2as \text{ (Equ. 3)}$

which is our equation (3) from the previous blog. As they say in mathematics, Quod erat domonstradum (QED) ðŸ™‚

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## Derivation of Newton’s form of Kepler’s 3rd law – part 2

In part 1 of this blog, I showed how Kepler’s third law, $T^{2} \propto a^{3}$ could be written as

$T^{2} = \frac{ 4\pi^{2} }{ GM } a^{3}$

Today, in part 2, I will show how this can also be written as

$(M + m)T^{2} = a^{3}$

The key to doing this is to use something called the reduced mass, but to understand the reduced mass we first need to know about the centre of mass of a two-body system.

## The centre of mass

If you have two objects of masses $m_{1} \text{ and } m_{2}$, the centre of mass is the point between them where, if the two masses were on a balance beam, the beam would be balanced. When the two masses are equal, this is just the mid-point between them.

The centre of mass for two objects with the same mass is the mid-point between them.

If the masses are not equal common sense tells us the centre of mass will be closer to the more massive object.

If we have a baby and a cow, the centre of mass will be much closer to the cow, at is much more massive.

The exact position of the centre of mass can be found from the fact that

$m_{1}r_{1} = m_{2}r_{2}$

where $r_{1} \text{ and } r_{2}$ are the distances of objects 1 and 2 respectively from the centre of mass. Re-arranging this gives that

$\frac{ r_{1} }{ r_{2} } = \frac{ m_{2} }{ m_{1} }$

Clearly, when the masses are equal $\frac{ r_{1} }{ r_{2} } =1$ and the centre of mass is midway between the two masses. In the case of the baby and the cow, let us suppose the baby has mass of 5kg and the cow a mass of 500kg, then

$\frac{ r_{1} }{ r_{2} } = \frac{ 500 }{ 5 } = 100$

Suppose the baby and the cow were separated by 3 metres, this is the vlaue of $r = r_{1} + r_{2}$. We can write

$\frac{ r_{1} }{ r_{2} } = 100 \text{ and } (r_{1} + r_{2}) = 3$

and so

$r_{1} = 100 r_{2} \rightarrow (100+1)r_{2} = 3. \text{ So } r_{2} = (3/101) = 0.03 \text{m, and } r_{1} = (300/101) = 2.97 \text{m}$

The centre of mass is only 3cm from the centre of mass of the cow, and 297cm from the centre of mass of the baby!

## The reduced mass

When two objects are acted upon by the same central force, such as in the case of gravity, it is useful to use a concept called the reduced mass.

If we have two objects acted upon by a force towards the centre of mass, we can use the concept of the reduced mass.

To determine the expression for the reduced mass, we note that the two forces, which I will call $\vec{F_{1}} \text{ and } \vec{F_{2}}$ which act on objects 1 and 2 are equal and opposite (from Newton’s 3rd law). We can therefore write that $\vec{F_{1}} = -\vec{F_{2}}$.

But, we can also write

$\vec{F_{1}} = m_{1} \vec{a_{1}} \text{ and } \vec{F_{2}} = m_{2} \vec{a_{2}}$

Although the two forces are equal, the accelerations in general are not, they will only be equal if the two masses are equal. We can write that the relative acceleration between the two objects is just the difference in their accelerations, that is $\vec{a} = \vec{a_{1}} - \vec{a_{2}}$. But, we can also write this as

$\vec{a} = \frac{ \vec{F_{1}} }{ m_{1} } - \frac{ \vec{F_{2}} }{ m_{2} } = \vec{F_{1}} \left( \frac{ 1 }{ m_{1} } + \frac{ 1 }{ m_{2} } \right) = \vec{F} \left( \frac{ m_{1} + m_{2} }{ m_{1}m_{2} } \right)$

(as $\vec{F_{2}} = -\vec{F_{1}}$). This then leads to

$\vec{F_{1}} = \vec{F} = \vec{a} \left( \frac{ m_{1}m_{2} }{ m_{1} + m_{2} } \right) = \mu \vec{a}$
where $mu$ is the so-called reduced mass. The reduced mass is the effective inertial mass appearing in two-body problems, such as when a planet orbits a star (if we ignore the gravitational influence of any other planets).

## Newton’s form of Kepler’s third law

We are now ready to derive what is known as Newton’s form of Kepler’s third law. Remember, we are talking about a planet orbiting the Sun. Our starting point is to realise that, strictly speaking, the Earth does not orbit the Sun. More correctly, they each orbit their common centre of mass. Because the mass of the Sun is about 1 million times the mass of the Earth, the centre of mass of the Earth-Sun system is pretty close to the centre of the Sun, in fact it lies within the body of the Sun.

But, the centre of the Sun does orbit this centre of mass point. Or, it would if the Earth were the only planet in the Solar System. In fact, the Sun orbits a point which is dominated by the centre of mass in the Jupiter-Sun system, but you get the idea.

The figure below shows a diagram of our two objects, 1 and 2 (the Sun and one of the planets), orbiting their common centre of mass. They orbit with speeds $v_{1} \text{ and } v_{2}$.

Any two objects in orbit actually orbit their common centre of mass. To make things easier we will assume the orbits are circular about this centre of mass, but the same equations can be derived for elliptical orbits too.

As can be seen from the diagram, the distance between the two objects, which we will call $r$, is given by $r = r_{1} + r_{2}$. We can write an equation separately for each object equating the gravitational force felt by each object to the centripetal force.

$\frac{ G m_{1} m_{2} }{ (r_{1}+r_{2})^{2} } = \frac{ m_{1} v_{1}^{2} }{ r_{1} } = \frac{ m_{2} v_{2}^{2} }{ r_{2} }$

But, we can also write this as

$\frac{ G m_{1} m_{2} }{ r^{2} } = \frac{ \mu v^{2} }{ r }$

where $\mu$ is the reduced mass. Re-writing $\mu$ and remembering that $v = (2 \pi r)/T$ we have

$\frac{ G m_{1} m_{2} }{ r^{2} } = \left( \frac{ m_{1}m_{2} }{ m_{1} + m_{2} } \right) \left( \frac{ 4 \pi^{2} r }{ T^{2} } \right)$

$T^{2} (m_{1} + m_{2}) = \left( \frac{ 4 \pi^{2} }{ G } \right) a^{3}$

where we have replaced $r \text{ with } a$. If we express $T$ in years and $a$ in Astronomical Units, this reduces to

$\boxed{ T^{2}(m_{1} + m_{2}) = a^{3} }$

which is Newton’s form of Kepler’s third law.

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## Derivation of Newton’s form of Kepler’s 3rd law – part 1

In 1619, Johannes Kepler published a relationship between how long a planet takes to orbit the Sun and the size of that orbit, something we now call his 3rd law of planetary motion, or just “Kepler’s 3rd law”. It states that

$T^{2} \propto a^{3}$

where $T$ is the period of the orbit and $a$ is the size of the orbit. Kepler also found that the planets orbit the Sun in elliptical orbits (his 1st law), and so the size of the orbit $a$ that we refer to is actually something called the “semi-major axis”, half the length of the long axis of an ellipse.

Any proportionality can be written as an equality if we introduce a constant, so we can write

$T^{2} = k a^{3} \text{ (equation 1)}$

where $k$ is our constant of proportionality.

Kepler found that the planets orbit the Sun in ellipses, with the Sun at one of the foci. The long axis of an ellipse is called its major axis. The $a$ in Kepler’s 3rd law refers to the length of the semi-major axis of a planet’s ellipse.

Newton was able to show in his Principia, published in 1687, that this law comes about as a natural consequence of his laws of motion and his law of gravity. How can this be shown?

## Why is Kepler’s law true?

To show how Kepler’s law comes from Newton’s laws of motion and his law of gravitation, we will first of all make two simplifying assumptions, to make the mathematics easier. First we will assume that the orbits are circular, rather than elliptical. Secondly, we will assume that the Sun is at the centre of a planet’s circular orbit. Neither of these assumptions is strictly true, but they will make the derivation much simpler.

Newton’s law of gravity states that the gravitational force between two bodies of masses $M \text{ and } m$ is given by

$F = \frac { G M m }{ r^{2} } \text{ (equation 2)}$

where $r$ is the distance between the two bodies and $G$ is a constant, known as Newton’s universal gravitational constant, usually called “big G”. In the case we are considering here, $r$ is of course the radius of a planet’s circular orbit about the Sun.

When an object moves in a circle, even at a constant speed, it experiences an acceleration. This is because the velocity is always changing, as the direction of the velocity vector is always changing, even if its size is constant. From Newton’s 2nd law, $F=ma$, which means if there is an acceleration there must be a force causing it, and for circular motion this force is known as the centripetal force. It is given by

$F = \frac{ m v^{2} }{ r } \text{ (equation 3)}$

where $m$ is the mass of the moving body, $v$ is its speed, and $r$ is the radius of the circular orbit. This centripetal force in this case is provided by gravity, so we can say that

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

With a little bit of cancelling out we get

$\frac{ G M }{ r } = v^{2} \text{ (equation 4)}$

But the speed $v$ is given by the distance the body moves divided by the time it takes. For one full circle this is just

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

where $2 \pi r$ is the circumference of a circle and $T$ is the time it takes to complete one full orbit, its period. Substituting this into equation (4) gives

$\frac{ G M }{ r } = \frac{ 4 \pi^{2} r^{2} }{ T^{2} }$

Doing some re-arranging this gives

$\boxed{ T^{2} = \frac{4 \pi^{2} }{ GM } a^{3} } \text{ (equation 5)}$

where we have substituted $a$ for $r$. This, as you can see, is just Kepler’s 3rd law, with the constant of proportionality $k$ found to be $(4 \pi^{2})/(GM)$. So, Kepler’s 3rd law can be derived from Newton’s laws of motion and his law of gravity. The value of $k$ above is true if we express $a$ in metres and $T$ in seconds. But, if we express $a$ in Astronomical Units and $T$ in Earth years, then $k$ actually comes out to be 1!

## Newton’s form of Kepler’s 3rd law

A web search for Newton’s form of Kepler’s 3rd law will turn up the following equation

$(M + m) T^{2} = a^{3} \text{ (equation 6)}$

How can we derive this? I will show how it is done in part 2 of this blog, as we will need to learn about something called “reduced mass”, and also the “centre of mass”.

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## The 10 best physicists – no. 5 – James Clerk Maxwell

At number 5 in The Guardian’sten best physicists” is the Scottish theoretical physicist James Clerk Maxwell.

As the caption above says, James Clerk Maxwell is probably one of the least known of the physicists in this list, and yet one of the most important. It was through his theoretical work that a full understanding of electromagnetism became possible. His four equations, known as Maxwell’s equations, are as important to understanding electromagnetism as Newton’s laws of motion are to understanding mechanics. In some ways, Maxwell was to Faraday what Newton was to Galileo. Neither Galileo nor Faraday had the mathematical ability to write in mathematics the experimental results they obtained in mechanics and electricity and magnetism respectively. Newton wrote the mathematical equations to explain Galileo’s experiments in mechanics; and similarly Maxwell wrote the mathematical equations to explain the experiments Faraday had been doing on electricity and magnetism.

## Maxwell’s brief biography

James Clerk was born in 1831 in Edinburgh, Scotland. He was born into a wealthy family, and later added Maxwell to his name when he inherited an estate as part of his wider family’s wealth. His mother died when Maxwell was eight years old, and at ten years of age his father sent him to the prestigious Edinburgh Academy.

At the age of 16, in 1847, Maxwell left Edinburgh Academy and started attending classes at Edinburgh University. His ability in mathematics quickly became apparent. At the age of only 18 he contributed two papers to the Transactions of the Royal Society of Edinburgh, and in 1850, at the age of 19, he transferred from Edinburgh University to Cambridge. Initially he attended Peterhouse College, but transferred to the richer and better known Trinity College. He graduated from Trinity College Cambridge in 1854 with a degree in mathematics. He graduated second in his class.

In 1855 he was made a fellow of Trinity College, allowing him to get on with his research. But in November 1856 he left Cambridge, being appointed Professor of Natural Philosophy at the University of Aberdeen in Scotland. He was 25 years of age, and head of the Natural Philosophy department, some 15 years younger than any of his colleagues there.

Maxwell left Aberdeen in 1860 to become Professor of Natural Philosophy at Kings College, University of London. He resigned in 1865 to return to his estate in Scotland, and for the next 6 years he got on with his research, and due to being independently wealthy he didn’t need a formal paying position at a university. But in 1871 he was tempted back into university life, he was appointed the first Cavendish Professor of Physics at Cambridge University, and given the role of establishing the Cavendish Laboratory, which opened in 1875.

Maxwell died in 1879 of abdominal cancer, and is buried in Galloway, Scotland, near where he grew up.

## Maxwell’s scientific legacy

Maxwell is best known amongst physicists for two main areas of work, his work on electromagnetism and his work on the kinetic theory of gases. As I mentioned above, his work on electromagnetism can be likened to a certain extent to Newton’s work on mechanics. Newton was able to take the experimental results Galileo had found on the behaviour of moving bodies, and find the underlying laws of mechanics to explain them, and to express these in mathematical form. This is essentially what Maxwell did for the work of Faraday, he was able to find the laws which explained the observations Faraday had been making, and was a sufficiently skilled mathematician to write these laws in equation form.

Maxwell first presented his thoughts on electromagnetism in a paper in 1855, when he was 24 years old. In this he reduced all the known knowledge of electromagnetism as it existed in 1855 into a set of differential equations with 20 equations and 20 variables. In 1862 he showed that the speed of propagation of an electromagnetic field was approximately the same as the speed of light, something Maxwell thought was unlikely to be a coincidence. This was the first indication that light was a form of electromagnetism, but a form to which are eyes are senstive.

In 1873 Maxwell reduced the 20 differential equations of his 1855 paper into the four differential equations which are familiar to every undergraduate physics student. These equations, expressed in their differential form are
$\boxed{ \begin{array}{lcll} \nabla \cdot \vec{D} & = & \rho & (1) \\ & & & \\ \nabla \cdot \vec{B} & = & 0 & (2) \\ & & & \\ \nabla \times \vec{E} & = & - \frac{\partial \vec{B}}{\partial t} & (3) \\ & & & \\ \nabla \times \vec{H} & = & - \frac{\partial \vec{D}}{\partial t} + \vec{J} & (4) \end{array} }$

The symbol $\nabla$ is known as the vector differential operator, or del, and I have explained del in this blog.

The four equations can also be written in integral form, which many people find easier to understand. In integral form, the equations become

$\boxed{ \begin{array}{lcll} \iint_{\partial \Omega} \vec{D} \cdot d\vec{S}& = & Q_{f}(V) & (5) \\ & & & \\ \iint_{\partial \Omega} \vec{B} \cdot d\vec{S} & = & 0 & (6) \\ & & & \\ \oint_{\partial \Sigma} \vec{E} \cdot d\vec{\l} & = - & \iint_{\Sigma} \frac{\partial \vec{B} }{\partial t} \cdot d\vec{S} & (7) \\ & & & \\ \oint_{\partial \Sigma} \vec{H} \cdot d\vec{l} & = & I_{f} + \iint_{\Sigma} \frac{\partial \vec{D} }{\partial t} \cdot d\vec{S} & (8) \end{array} }$

## Kinetic theory of gases

Maxwell also did important work on one of the other main areas of physics research in the 19th Century – the kinetic theory of gases, a branch of thermodynamics. Maxwell and the Austrian physicist Ludwig Boltzmann independently came up with an expression which describes the distribution of speeds of the atoms or molecules in a gas. The speed depends on the temperature, the higher the temperature the faster the atoms or molecules move. But Maxwell and Boltzmann showed that not all the atoms will be moving at this speed; some will be moving faster and some will be moving slower. The distribution of speeds is known as the Maxwell-Boltzmann distribution.

## The theory of colour vision

Less well known to most physicists is the work Maxwell did on optics, and on colour vision. He was the first person to show how a colour image could be created from combining three images taken through red, green and blue filters. This lies at the foundation of practical colour photography, and is also how our colour television sets work.

I think it is true to say that most physicists would place Maxwell in their top four physicists. His contributions to physics are immense, creating the formal framework for our understanding of electromagnetism, without which the modern world would not be possible. That he is so unknown outside of the world of physics is strange, but whereas Newton had his $F=ma$ and Einstein his $E=mc^{2}$, Maxwell’s equations are far too complicated to have permeated into the consciousness of the public. But they are every bit as important as Newton and Einstein’s.

You can read more about James Clerk MaxwellÂ and the other physicists in this “10 best” list in our bookÂ 10 Physicists Who Transformed Our Understanding of the Universe.Â Click here for more detailsÂ and to read some reviews.

Ten Physicists Who Transformed Our Understanding of Reality is available now. Follow this link to order

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## The 10 best Physicists?

In an interesting exercise, The Guardian newspaper recently drew up a list of the “10 best physicists”. I don’t think the list they compiled is in any particular order, but here it is.

1. Isaac Newton (1643-1727)
2. Niels Bohr (1885-1962)
3. Galileo Galilei (1564-1642)
4. Albert Einstein (1879-1955)
5. James Clerk Maxwell (1831-1879)
6. Michael Faraday (1791-1867)
7. Marie Curie (1867-1934)
8. Richard Feynman (1918-1988)
9. Ernest Rutherford (1871-1937)
10. Paul Dirac (1902-1984)

How many of these names do you recognise? Whilst some are “household names”, others are maybe only known to physicists.

Over the next several months I will post a blog about each of these entries, giving more details of what their contribution(s) to physics were. Any such list is, of course, bound to promote discussion and disagreement, and I can also see that “The Guardian” have also allowed readers to nominate their own names.

You can read more about the physicistsÂ in this “10 best” list in our bookÂ 10 Physicists Who Transformed Our Understanding of the Universe.Â Click here for more detailsÂ and to read some reviews.

Ten Physicists Who Transformed Our Understanding of Reality is available now. Follow this link to order

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## Vector basics

I wanted to get back to explaining Maxwell’s equations, which I mentioned in this blog of the statue to James Clerk Maxwell that is in Edinburgh. Before I do that I thought I would cover some very basic mathematics, namely the basic idea of vectors.

In physics and mechanics, a vector is something which has both size (magnitude) and direction. If the quantity has only a size, we call it a scalar. So, for example, in physics we differentiate between speed and velocity, even though in everyday language they are used interchangeably.

If we say a car is moving at 50 km/h, that is a speed, and hence is a scalar. But, if we were to say it is moving at 50 km/h due North, that is a velocity, and hence a vector as we have given it both a size and direction.

Vectors are very important in physics, and one of the most important and widely used vectors is force. We measure force in Newtons (named after Sir Isaac Newton), but a force has both a size and a direction. We usually denote a vector by either putting a line or arrow above the symbol, or by using a bold font. I will use an arrow above the symbol, so for example $v$ would be a scalar, but $\vec{v}$ would be a vector.

Any vector in 3-dimensional space can be split into 3-components. This is often useful, as unlike scalars which add simply, when we add vectors we need to take into account their directions. As an example, suppose we have two forces $\vec{F_{1}}$ which has a size of 7N in the x-direction, and $\vec{F_{2}}$ which has a size of 5N at $25^{\circ}$ to the x-axis. The combined force is not 12N, as the two forces are not acting in exactly the same direction.

A 7N and a 5N force at 25 degrees to each other, with the 7N force acting along the x-direction.

To find the combined (resultant) force, we need to split the two forces up into their x and y-components. To do this we simply use trigonometry, and note the unit vectors $\hat{x}$ and $\hat{y}$, which are vectors with a size of unity (1) in the x and y-directions respectively. We can then split each of the forces $\vec{F_{1}}$ and $\vec{F_{2}}$ into their respective x and y-components.

$\vec{F_{1}} = 7\cos(0) \hat{x} + 7\sin(0) \hat{y} = 7\hat{x} + 0\hat{y}$

$\vec{F_{2}} = 5\cos(25) \hat{x} + 5\sin(25) \hat{y} = (5 \times 0.9063)\hat{x} + (5 \times 0.4226)\hat{y} = 4.53 \hat{x} + 2.11 \hat{y}$

From this we can write that the total force in the x-direction is $7 + 4.53 = 11.53 \hat{x}$ and the total force in the y-direction is $0 + 2.11 = 2.11 \hat{y}$. To then find the resultant force, we need to combine these two components as follows

The size of the resultant force R can be calculated using Pythagoras’ theorem, its direction using trigonometry.

To calculate the size of the resultant vector we just use $R^{2} = x^{2} + y^{2}$ so here $R^{2} = (11.53)^{2} + 2.11^{2} = 157.0 + 4.45 = 137.39$ so $R=\sqrt{137.39}=11.72 N$. To calculate the angle $\theta$ we note that $\theta = \arctan \left( \frac {2.11} {11.72} \right) = 10.2^{\circ}$.

So, as we can see, the resultant of these two forces is a force of size 11.72N which is at an angle of $10.2^{\circ}$ to the x-axis.

In a series of future blogs I will go on from these basic ideas of vectors to talk about vector fields, which we need to understand in order to understand Maxwell’s equations.

A vector field. The length of each arrow represents the strength of the field at that location, the direction is given by the orientation of the arrow.

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## The Analemma (part 2)

A few weeks ago I showed a photograph an an Analemma. As the Analemma in the photograph was vertical, I explained that it must have been taken at midday. Here is the photograph again, just to remind you.

A Solar analemma. Because it is vertical, this analemma was taken at midday.

In part 1 of my series on the Analemma, I also explained how the North-South motion of the Sun in the photograph was due to the changing elevation of the Sun at midday. This is, of course, due to the tilt of the Earth’s axis, as explained in this video below.

But, what about the East-West (left-right) motion? What is this due to?

It turns out that the East-West motion is due to two effects. One is the same inclination of the Earth’s axis in its orbit around the Sun which produces the North-South variation in the Sun’s elevation at different times of the year. I will explain how this affects the East-West position of the Sun at midday in part 3 of this blog.

But, the second effect is unrelated to this, it has to do with the details of the Earth’s path around the Sun.

## The Heliocentric Universe

When Copernicus suggesed in 1547 that the Earth and the other planets went around the Sun, he argued that they would do so in perfect circles. He was, in fact, not the first to suggest that the Sun and not the Earth was at the centre of things. Aristarchus had suggested the same thing in the 3rd Century B.C., but his work had been largely ignored in preference to the teachings of Plato and Aristotle, who firmly held that the Earth was the natural centre of all things.

Building on Aristotle’s Geocentric Universe model, the Greek-Roman astronomer Ptolemy developed a sophisticated model of the Sun, Moon, planets and stars orbiting the Earth. This model was incredibly successful, and able to predict the positions of the celestial objects to a good degree of accuracy for some 1500 years.

In Copernicus’ 16th Century model, the planets orbited the Sun in circles. In the latter part of the 16th Century, the greatest observational astronomer was a Danish man, Tycho Brahe. Brahe had his own Observatory and research institute, Uraniborg, on the Danish (now Swedish) island of Hven, with a Royal patronage to fund his observing programme.

Tycho Brahe (1546-1601)

Brahe produced the most accurate observations of the planetary positions, and he found he got better agreement with Ptolemy’s geocentric model than he did with Copernicus’ heliocentric one. Towards the end of his career, a young mathematician by the name of Johannes Kepler came to work with him.

## A very particular kind of curve

After Brahe’s death, Kepler set about seeing whether he could get a heliocentric model to agree with the observations. After over a decade of trial and error, he eventually found that, if he allowed the planets to move in ellipses rather than perfect circles, that very good agreement could be obtained. As Richard Feynman once said, “an ellipse is a very particular kind of curve”. To be more precise, it is the curve obtained when one passes a string about two drawing pins (“thumb tacks” as Americans call them) and draws the ensuing locus of points.

How to draw an ellipse

Kepler’s 2nd law states that a planet will “sweep out equal areas in equal times” in its orbit. What this means is that it will speed up when near the Sun (perihelion) and slow down when further from the Sun (aphelion).

Part of the East-West motion of the Sun in the Analemma is due to Kepler’s 2nd law, the fact that the Earth changes its speed of orbit as it goes around the Sun. The Earth moves quicker when it is closer to the Sun (perihelion), and slower when it is further from the Sun (aphelion). Kepler did not know why this happens, but it is a natural consequence of Newton’s law of gravity.

## A mean Solar day

How do we measure the length of the day? It seems like a simple question. Surely, the answer is that it is the time it takes for the Earth to turn once on its axis. This is, in fact, the wrong answer. The time it takes for the Earth turn once on its axis is the sidereal day, and this is not how we measure our day. Why? The diagram below explains it.

The Earth has to turn a little bit extra for the Sun to cross the local meridian. We call this the Solar day. The sidereal day is the time for the Earth to rotate 360 degrees.

Because the Earth moves about the Sun in its orbit, the Earth has to rotate a little bit extra for the Sun to cross the local meridian on two successive occasions. This is how we define 24 hours, the solar day. But, there is an additional complication; because the Earth’s speed of orbit changes, the extra angle the Earth needs to turn to bring the Sun back over the meridian also changes. As the Earth approaches perihelion (closer to the Sun), it speeds up and so moves through a larger angle each 24 hours than when it is further from the Sun.

Of course, we cannot keep changing the length of the day, we fix it at 24 hours, which is what we call the mean solar day. This is the midday our watches will show. But, near perihelion, the Earth has to turn that little bit extra as I’ve explained, so when our watches say it is midday the Sun will still be to the East of the local meridian. So, if we were taking a photograph of where the Sun was at midday as shown by our watches, the Sun would have shifted eastwards of the mid-point.

The opposite effect happens near aphelion, when the Earth is moving more slowly in its orbit. This time, the Earth moves slightly less in 24 hours than it does in other parts of its orbit, and so the Earth does not have to turn through such a large angle to bring the Sun back over the local meridian. When our watch says midday, the Sun will have gone past the local meridian, and be to the West of it.

Hopefully, this now explains why there should be some East-West motion in the Solar Analemma. However, it turns out that there is a second effect which also causes an East-West motion, the tilt of the Earth’s axis in its orbit. I will explain this component and how it affects things in part 3 of this series.

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