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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.

### 37 Responses

1. on 22/10/2014 at 16:40 | Reply Patrick labelle

The derivation does not work in general because the orbits are not necessarily circular, which is assumed here (for example, v = 2 Pi r / Period does not work in general)

• on 22/10/2014 at 16:44 | Reply RhEvans

Yes, i know they’re not circular 🙂
I do state that this is a simplification. I did the blog partly because my son needed to know how to derive it for his A-level physics class. Generalising it to elliotical orbits is beyond A-level physics.

• on 22/10/2014 at 17:13 | Reply RhEvans

From part 1 of this two-part post

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.

2. on 08/05/2015 at 14:42 | Reply Necat Taşdelen

Kepler’s period law is about the distance (r) of the body to the sun and its revolving period (P).
So he discovered with observation that (P^2/r^3)=Constant.This has been proved mathematically by Newton for mechanical media.Not for the univers.
And your formula T^2*(m1+m2)=a^3=Constant is totally wrong.This means T^2/a^3=(m1+m2)=Ct and you say this proves that Kepler’s third law is correct.
Not at all.Your formula is wrong To write (m1+m2)=Ct is correct.But it should not be related to distances.Distances are another value than the mass.
m1*a1=m2*a2 is correct.But your formula (m1+m2)=Ct is wrong.

3. on 08/05/2015 at 15:13 | Reply Necat Taşdelen

Hi,Evans
The universal motion equaton is r=-4*t^2+4*t*T-4*T^2/6 Where (r) is the distance of the celestial body to the sun,(t) the real time.(T) the life time of the celestial body.This equation do not indicate an elliptic orbit.but spiraled .So the period law is not valid for these orbits.Because (r) is variable.Then r*Vp^2=Ct is not valid for eternity.While for Nerwton and Kepler r*Vp^2 was accepted constant for eternity.So the period of a celestial bodydoes not change in the eternity.Spiraled theory refuse this eternity.Because (r) is variable with time(t).For Kepler orbits the distance (r) is assumed tobe constant,and also the revolving time is said to be constant for eternity.These are wrong assumption.
When to the derivation of the celestial motion equation,you just write the energy conservation equation and you find r’^2+2*a*r+K=2*r*r” with solution
r=-a*t^2+a*t*T+Z and when Z is solved you have a=4,t=real time (at the date 2009 t was= 4600000000 cycles around the sun,for the Earth).(T) is the life time of the Earth=9327258100 cycles around the sun.Kepler does not mention of life time ,or time.So He thinks that all the celestial movement of the date 1600 is valid for eternity.No end of the Earth.

4. on 01/03/2016 at 01:17 | Reply Timmy

Please can someone explain why expressing T in years and a in AU allows us to ignore the constant from the previous line?

• on 01/03/2016 at 01:19 | Reply RhEvans

Because by expressing T in years and a in AU you know that, for the Earth, both are 1, so the constant then equals 1!

5. on 01/03/2016 at 01:36 | Reply Timmy

If both are 1, we know the constant=m1+m2. That hasn’t proved the constant is 1.

• on 01/03/2016 at 01:42 | Reply RhEvans

You are confusing constants. In Kepler’s form of the law, $P^{2} = k a^{3}$, where $k$ is the constant. If we express $P$ and $a$ in years and AUs respectively, then $k=1$. The sum of the masses is not 1! I’ve never suggested it was.

• on 01/03/2016 at 01:48 | Reply RhEvans

It’s the $(4\pi^{2}/ G)$ which becomes 1 if we express $P$ in years and $a$ in AUs!

• on 01/03/2016 at 01:55 Timmy

I meant the 4pi^2/G all along. If T=1 and a=1, then m1+m2=4pi^2/G. So how can 4pi^2/G=1?

• on 01/03/2016 at 02:03 RhEvans

Because m1+m2 is approximately m1 (the mass of the Sun, m2 is negligible in comparison), and so if we express the masses in Solar masses m1=1 too!

6. on 01/03/2016 at 02:09 | Reply Timmy

So the sum of the masses is 1?

• on 01/03/2016 at 02:11 | Reply RhEvans

When expressed in solar masses, yes (and assuming m1 >> m2)

• on 01/03/2016 at 02:15 Timmy

Cheers pal

7. on 15/03/2016 at 09:16 | Reply Necat Taşdelen

The period law of kepler is similar to the mathematical expression of Newton:
(r1/r2)^3=(P1/P2)^2.When P=2*pi*r/Vp is considered,the period expression is written r*Vp^2=y.(Vp=cycling velocity of a body).Then r*Vp^2=y is found.(y) is not constant as it is function of r.And (r) is variable with time,according the celestial motion equation: r=-4*t^2+4*t*T-4*T^2/6 (which is not an ellipse,but a spiraled orbit).In fact Kepler’s area law r*Vp=Ct should be wrong as r*Vp^2=Ct
is confirmed with celestial observations.Then Kepler’s laws should be wrong.

8. on 24/04/2016 at 14:53 | Reply Don

Shouldn’t ‘mv1^2’ be ‘mv2^2’ right after the text, “gravitational force felt by each object to the centripetal force.”

• on 24/04/2016 at 16:32 | Reply RhEvans

You’re right, one of the v1 should be a v2! I have corrected it. Thanks 🙂

9. on 24/04/2016 at 23:18 | Reply Necat Tasdelen

I am explaining that Kepler’s celestial laws are wrong.Have you any suggestion about this argument.?The area law is an estimation.Not a prove.What is proved is Vp=Ct for ever.Use Newton’s attraction force.Say the force is radial,a perpendicular force component to the radial does not exist,Then you have Fp=m*dVp/dt=0 which means Vp=Ct.Use Vp=Ct in the energy conservation equation,you get the celestial motion proved equation r=-4*T^2+4*t*T-4*T^2/6
and this prove that the orbits are not elliptical,the area law does not exist,the period law being a function of (r),r*Vp^2=Ct is valid for short time interval.Kepler and Newton living time interval was too short for understanding r*Vp^2=variable.At their time r*Vp^2=Ct was agreed.

10. […] affects the calculated size of the orbit. Based on Newton’s form of Kepler’s 3rd law, which I blogged about here, the period of orbit of a planet is given […]

11. on 21/01/2017 at 15:23 | Reply Tim Maloney

Thank you so much for the post! Your derivation was very accesible 🙂

12. on 21/01/2017 at 15:30 | Reply Tim Maloney

Why does the equation reduce when you express it in terms of AU and years? And must mass be in Solar Masses?

• on 21/01/2017 at 17:50 | Reply RhEvans

Yes, because then all the quantities are “1” in those units, so the constant of proportionality also becomes 1.

13. on 21/01/2017 at 20:30 | Reply Necat Taşdelen

The terms First law ,second law,third law of Kepler may be misunderstood according tha pays where it is used. In Turkey we prefer to say,elliptical orbit law,,constant area law,Period law in places of numbred laws.Now, what Iam looking for is whether Kepler’s area law r*Vp=constant is correct or Newton’s only Vp=constant is correct.Kepler’s area law is an estimation.Newton’s Vp=Constant is a pouve.If Kepler area law is correct,then an acceleration should be feeledon the orbits.Mountains and sees sould move.In 1618 when Kepler announced his period law,he was not aware that he was deniying the law of constant area and affirming that only Vp=Constant is acceptable.So: say the law of period is r*Vp^2=Ct2 and the area law is r*Vp=Ct1.Then r*Vp^2/(r*Vp)=Ct2/Ct1=Ct.=Vp.This is a prove and not an estimation.

• on 21/01/2017 at 20:35 | Reply RhEvans

Why do you keep posting this nonsense on my blog? Every few months you do it! This is the last time I’m allowing it to be posted, the next time I’m blocking it. You’ve made your point (which, by the way IS WRONG). So stop repeatedly posting the same incorrect drivel.

Understand?

• on 22/01/2017 at 10:02 | Reply RhEvans

“If Kepler area law is correct,then an acceleration should be feeledon the orbits.Mountains and sees sould move.”

Anyone who can make such a statement as this shows such a fundamental lack of understanding of physics, and in particular gravity and the laws of motion (and Galilean relativity) that I don’t even know where to begin.

I suggest that you take some courses in basic dynamics and classical mechanics, and try to understand the most basic ideas of Newton’s theory of gravity. If you cannot see the flaw in your sentence that I’ve quoted then I’d suggest that physics is not the subject for you.

14. on 22/01/2017 at 07:55 | Reply Necat Taşdelen

Yes,understood.But as long as you insist saying r*Vp=Ct is equal to say to Vp=Ct,you may have my disturbing comments.Unless you explain where I am wrong.Please consider Kepler’s period law r*Vp^2=Ct2 and the area law r*Vp=Ct1..These two expressions result to Vp=Ct.Ct for ever.And the celestial motion equation gives r=-4*t^2+4*t*T-4*T^2/6 when you solve the energy differential equation.

• on 22/01/2017 at 08:01 | Reply RhEvans

This is the LAST time I’m allowing you to post this. Send your theory to a journal, my blog is not the place to REPEATEDLY post your theory. REPEATEDLY! I’m fed up of seeing it. I’m glad that you understand that any future attempts to post it will be BLOCKED. Thank you.

15. on 22/01/2017 at 11:24 | Reply Necat Taşdelen

I see ! You feel a danger for the future of this site if you agree that Kepler’s area law r*Vp=Ct is similar to Newton’s Vp=Ct when (r=1).But the orbits are not circular.And r=1 has no meaning.So with the elliptical orbits of Kepler,we have an acceleration of 0.000064 m/sec^2 and some one should feel it.You may feel it but not me.And I am sure you are in big physical danger when you say (r1/r2)^3=(P1/P2)^2. Replace P=2*pi*r/Vp and you have r*Vp^2=variable.Ct for a short period.This bring that,when Vp=Ct is proved,the period law is not r^3/P^2=Ct.But is variable as the shape of a parabola.r=r(t) has a parabolic shape.Kepler has affirmed this last sentences.But it seems you are not aware. Please study the period law of kepler and the area law together to see the affirmation of Kepler.Vp=Ct

• on 22/01/2017 at 11:30 | Reply RhEvans

“we have an acceleration of 0.000064 m/sec^2 and some one should feel it”

What is the acceleration due to Earth’s gravity at the Earth’s surface? Tell me, as I’d like to see whether you know.

16. on 22/01/2017 at 12:58 | Reply Necat Taşdelen

Accoding Kepler the orbit of the Earth is elliptical and an area law is valid.
Mean r=149598261 km and Vp=29,72 km/sec.Vp is the cycling orbital velocity perpendicular to the radial attraction force direction.It is not the tangential Velocity .So r*Vp=4446392758 km^2/sec.The velocities at aphalion is 29,233 km/sec and at perihelion 30,227 km/sec.(Vikipedia).Then a variation of velocity is realized at least every 6 months,That is an orbital acceleration of 0,000063893 meters/sec^2 is created.Difficult to be feeled by small mass.But huge force creation on big mass.(mountains,seas),An acceleration exist where a difference of velocity exist.This was not a gravity due to Earth’s mass.But a variation of cycling velociytry between aphelion and perihelion.Do I know some physics.Thanks.Compare Kepler’s period law and area law.You will discover what Newton has said:Vp=Ct.And not r*Vp=ct, estimation of Kepler.

• on 22/01/2017 at 13:02 | Reply RhEvans

I asked you a very simple question, which has NOTHING to do with Kepler’s laws of planetary motion.

What is the acceleration due to Earth’s gravity at the Earth’s surface? What is its value? In m/s^2 ??

17. on 22/01/2017 at 13:48 | Reply Necat Taşdelen

The topic of our communication was not “what is the value of the acceleration due to the earth’s gravity”.It was whether r*Vp=ct is correct or only Vp=Ct is correct.Anyhow I remember the acceleration at the surface of the Earth was 10 m/sec^2 at the equator.Did I know.Have you compared the period law and the area law of Kepler.Even Kepler affirmed in 1618 that only Vp=Constant.No one of the astronomer know that.Compare it,compare it.And try to comment.
I may send you an abstract about this subject when you agree the difference between r*Vp=Ct and Vp=ct.Subject of our communication.My mail address is .You should have read about some oriental astronomer:such as Persian Kutbettin,Omer Hayyam,Ali Kuşçu,Takiyyuddin. Or the Timur’s carpentar of Ulu Cami of Bursa constructed before the war of Ankara (1402) between Timur Han and our Yıldırım Han.All before Kepler and Newton.See PRINCIPIA of Newton page 290(translation of A.Motte) for spiraled orbits .And say some words when Neptun was discovered in 1846 and Pluton in 1930,when these celestial objects were sculpted ın Bursa Ulu cami.

• on 22/01/2017 at 14:49 | Reply RhEvans

End of discussion. Ok? Understand? The next time you mention your ridiculous pet theory, I am NOT approving it. So either you stop posting it, or your posts get blocked. Understand?

18. on 22/01/2017 at 14:55 | Reply RhEvans

Correct, g=10 m/s^2. So, explain how one feels an acceleration of a=0.000063893 m/s^2 when g=9.81 m/s^2? That’s an effect which is 651 millionths of a percent LESS than the local acceleration due to gravity. So to suggest that it would have an effect on mountains and seas shows such a lack of basic understanding that’s it’s not worth my time trying to explain your errors.

19. on 23/01/2017 at 12:57 | Reply Don Flood

In physics since the time of Galileo, the rule has been, “Shut-up and calculate”. The science is all about building models and then testing those models against experiment, observation and mathematical consistency.