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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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