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