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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”.

### 11 Responses

1. […] In part 1 of this blog, I showed how Kepler’s third law, could be written as Today, in part 2, I will show how this can also be written as 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. […]

2. […] the mass by the mass of the Sun). Equation (2) is, in fact, just Kepler’s Third law (which I blogged about here), that is […]

3. on 12/04/2015 at 10:55 | Reply Necat Tasdelen

Kepler’s period law and Newton’s period law have the same equation,but not the same meaning.Newton says period is valid for circular orbi with constant peripheral velocity,while Kepler says period law is valid for non circular orbits with accelerated peripheral velocity.

For kepler and Newton
P^2/r^3=K is linear for eternity
Unfortunately it is not linear.It is parabolic.
So that for a given date(era) it seems to be linear
Then Newton and Kepler are wrong for this period law.

Kepler is wrong for area and orbit laws and period law.
In 1311 Kutbettin of Shiraz explained cycloidal,spiraled orbits
It is impossible to prove Kepler laws with Newton Laws.
Newton laws are for refusing Kepler Laws.

• on 08/05/2015 at 02:04 | Reply RhEvans

I’m afraid that I don’t follow your logic. What does “P^2/r^3=K for eternity” mean? Kepler derived this relationship (his 3rd law), Newton showed that it comes about as a natural consequence of an object moving about a larger mass with the force being an inverse square law (this is both true for a circular or an elliptical orbit).

• on 08/05/2015 at 14:18 Necat Taşdelen

Hi,RhEvans,
Newton’s period law is (r1/r2)^3=(P1/P2)^2.This is valid only and only for r=Constant1 and P=Constant2.In other words period law is valid for circular mouvement with constant peripheral velocity.Kepler law is (r1/r2)^3=(P1/P2)^2.The same equation as for Newton.But orbits are not circular,and peripheral velocity is variable.(r1/r2)^3=(P1/P2)^2 means r*Vp^2.According Newton r*Vp^2=Constant for eternity.Because both (r and Vp) are constant for eternity.With Kepler (r,Vp) are variable.Unfortunately,in the univers,the distance of a body to the sun is not constant.So,r*Vp^2=Constant is not linearly constant.İt is parabolic.So Kepler’s third law is wrong.Also Newton’s law application to the univers is wrong.(for mechanical media Newton’s period law is correct).Please see Newton’s PRINCIPIA,page 290,tarduction by A.Motte.

4. on 02/08/2015 at 18:16 | Reply Necat Tasdelen

Hi,RhEvans,
Kepler’s and Nwton’s,period law is (r1/r2)^3=(P1/P2)^2.This gives r1^3/P1^2=K=Constant.When it is said constant it is constant for eternity.
Spiraled theory says r*Vp^2=Ct,derived from Newton^s period law.As Vp=CT,y=r*Vp^2 has the same graph as for (r).And this graph is parabolic.So r*Vp^2=Constant is constant for a time,for an era,not for eternity.By this occasion spiraled theory says Vp=Ct and not r*Vp=Ct as says Kepler with his area law and r=-4*t^2++*t*T-4*T^2/6 where (t= real time.T life time) of the planet.

5. on 15/12/2015 at 09:47 | Reply Necat Taşdelen

Hi,
Kepler’s laws should be wrong.He says area law is r*Vp=Ct.But Newton’s univesal attraction law says:attraction force is radial.A side force componenet does not exist.That is Fp=m*dVp/dt=0 and integrating we have Vp=Ct.Now is r*Vp=Ct or only Vp=Ct

6. on 10/07/2016 at 17:17 | Reply Necat Tasdelen

After all comments ,I am deciding that Newton’s equation about the mechanical law F*dt=m*dV should be wrong.Because universal attraction force is only radial,a side force component,perpendicular to the radial does not exist,So Fp=m*dVp/dt=0 means Vp=CT.But kepler says r*Vp=Ct.As kepler is correct since 400 years,Newton should be wrong.Or you choose which one is correct.

7. on 29/11/2017 at 15:55 | Reply Harish

I didn’t get equation 3…how did you replaced time t with radius r…….like….it should have been like this….F=mv^2/t not…..mv^2/r…….if im not wrong that is…….plz reply.

• on 30/11/2017 at 15:01 | Reply RhEvans

8. on 14/09/2018 at 20:04 | Reply Roy Lotz

Thanks for this! It was quite helpful.