<p>The correct version of Kepler's third law is
<span class="math-container">$$(M + m)P^2 = a^3$$</span>
where <span class="math-container">$Ma' = ma''$</span> and <span class="math-container">$a = a' +a''$</span>, where <span class="math-container">$a''$</span> is the distance of mass <span class="math-container">$m$</span> from the centre of mass.</p>
<p>The result follows from eliminating <span class="math-container">$m$</span> and <span class="math-container">$a''$</span> using the latter two equations.</p>
<p>NB You cannot assume <span class="math-container">$M\gg m$</span> to simplify Kepler's third law since that implies <span class="math-container">$a' = 0$</span>.</p>

CC BY-SA 4.0
2023-07-28T22:36:33.877Z
2023-07-30T09:01:12.707Z
2023-07-30T09:01:12.707Z
ProfRob
<p><a href="https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law" rel="nofollow noreferrer">Kepler's third law</a> is too much like the equation in the exercise to be useful in proving it.
Instead, assume a circular orbit (<em>a'</em> &gt; 0) and let the centripetal force on the planet equal the gravitational force.
<span class="math-container">$$ \frac{m v^2}{a - a'} = \frac{GMm}{a^2} $$</span>
where
<span class="math-container">$$ v = \frac{2 \pi (a - a')}{P} $$</span>
To determine a value for <em>G</em>, note that in exercise units, the Sun-Earth system has <em>M</em>=1, <em>P</em>=1, and <em>a</em>=1.</p>

CC BY-SA 4.0
2023-07-27T15:47:30.520Z
wait so the star remains stationary to assume a circular orbit or should I assume the star and planet both move circularly around a common center of mass.
Moiz khokhar
CC BY-SA 4.0
2023-07-27T15:30:14.773Z
2023-07-30T20:03:00.013Z
2023-07-30T20:03:00.013Z
Mike G
<p>Kepler's third law is in general given by</p>
<p><span class="math-container">$$G (M+m)\cdot P^2 = 4 \pi^2 a^3$$</span></p>
<p>where <span class="math-container">$G$</span> is the gravitational constant, <span class="math-container">$M$</span> and <span class="math-container">$m$</span> the masses of the two bodies, <span class="math-container">$P$</span> their orbital period and <span class="math-container">$a$</span> the semi-major axis of the orbit.</p>
<p>Now for the distance <span class="math-container">$a-a'$</span> from the planet to the center of mass we have according to the definition of the center of mass</p>
<p><span class="math-container">$$a-a'=\frac{M}{M+m} a$$</span></p>
<p>so</p>
<p><span class="math-container">$$a =\frac{M+m}{M} (a-a')$$</span></p>
<p>Substitute just one of the <span class="math-container">$a$</span> in <span class="math-container">$a^3$</span> in Kepler's law with this expression and you get by multiplying through with <span class="math-container">$M/(M+m)$</span></p>
<p><span class="math-container">$$G M \cdot P^2 = 4 \pi^2 a^2\cdot(a-a')$$</span></p>
<p>This holds in full generality and in any physical units, so not only for <span class="math-container">$m&lt;&lt;M$</span> (the <span class="math-container">$m$</span> does not appear explicitly anymore, but it is still contained in <span class="math-container">$a'$</span>).</p>
<p>In the exercise, this equation has merely been reduced such that <span class="math-container">$G/4\pi^2=1$</span>, <span class="math-container">$M$</span> in solar masses, <span class="math-container">$a$</span> in AU and <span class="math-container">$P$</span> in years. It holds only in these particular units. In other units it would be wrong and the physical dimensions in fact be inconsistent.</p>

CC BY-SA 4.0
2023-07-30T16:17:06.303Z
@ProfRob I think this is a vicious homework question. It warrants a proper answer in order to avoid confusion. This is certainly a correct form of Kepler's third law as well (ignoring the omission of the gravitational constant), but it is highly unusual and and has probably just been formulated in order to cause students problems rather than helping them to understand Kepler's laws.
Thomas
CC BY-SA 4.0
2023-07-30T16:23:55.013Z
$G=1$ in the units given in the question.
ProfRob
CC BY-SA 4.0
2023-07-30T20:33:01.920Z
In the equation written in this answer,  $G=1$ @MikeG That is why it is omitted in the OP.
ProfRob
CC BY-SA 4.0
2023-07-30T20:40:35.423Z
@MikeG but you are correct in pointing out that $G$ as used in this answer is not *the* "gravitational constant" normally assigned the symbol $G$.
ProfRob
CC BY-SA 4.0
2023-07-31T18:41:18.803Z
@ProfRob "G=1 in the units given in the question" The value of G depends on the units used obviously, but you should not omit it from the equation even if its numerical value is 1. It is sloppy and dangerous (as some of the comments prove here).
Thomas
CC BY-SA 4.0
2023-07-31T21:45:52.303Z
I note that you have corrected your equations so that G *is* now the gravitational constant and added the last paragraph.
ProfRob
CC BY-SA 4.0
2023-08-01T19:19:18.403Z
Best to phrase it: In the first equation one sees that with M and m in units solar mass, a in AU, and P in years, then $G/4\pi^2$  is approximately 1.
eshaya
CC BY-SA 4.0
2023-07-30T15:31:12.173Z
2023-07-31T21:11:25.600Z
2023-07-31T21:11:25.600Z
Thomas
<p>A planet (mass m) is orbiting a star
(mass M) at a distance a. The distance of the star from
the centre of gravity of the system is a'
. Show that</p>
<p><span class="math-container">$$
MP^2 = a^2(a-a')
$$</span></p>
<p>So first we can assume that <span class="math-container">$M &gt;&gt; m$</span></p>
<p>so now we get keplers third law as <span class="math-container">$MP^2 = \text{distance to the center mass}$</span> of the two objects</p>
<p>so if we then solve for the planet-Center of mass distance it equals <span class="math-container">$(a-a')$</span></p>
<p>wont that make the final equation <span class="math-container">$MP^2 = (a-a')^3$</span></p>
<p>and not <span class="math-container">$MP^2 = a^2(a-a')$</span></p>
<p>so what seems to be the problem in my attempt and why cant I get the correct answer.</p>
<p>the question is taken from fundamentals of astronomy 5th edition springer publication exercise 9.2</p>
<p><a href="https://i.sstatic.net/hDPWG.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/hDPWG.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-07-28T18:43:20.430Z
Your version of Kepler's third law is incorrect. And you can't assume $M\gg m$ or $a'=0$.
ProfRob
CC BY-SA 4.0
2023-07-30T16:19:14.337Z
@ProfRob This version if Kepler;s third law is correct (see my own answer below) although highly unusual and likely confusing.
Thomas
CC BY-SA 4.0
2023-07-30T16:23:26.837Z
@Thomas THEIR version ("$MP^2 = $ distance to the center mass [sic]") is incorrect. Your answer had already been presented (by me) but without giving away the full solution to a simple homework problem.
ProfRob
CC BY-SA 4.0
2023-07-30T17:03:27.737Z
@ProfRob The error the OP made was to assume that $m<<M$ because he did not see the mass $m$ in the equation even though it is mentioned in the problem. With this assumption (which you certainly can make if the masses are sufficiently different) his equation $MP^2 = (a-a')^3$ is actually correct (as $a'=0$), as are $MP^2 =a (a-a')^2$ and $MP^2 =a^2 (a-a')$. Had you pointed out to the OP how $a'$ depends on the masses he would have realized himself where he was going wrong.
Thomas
<binary-star><keplers-laws>
Keplers laws and this question
Moiz khokhar
# Keplers laws and this question

(Asked by: username_0 on 2023-07-27)

A planet (mass m) is orbiting a star
(mass M) at a distance a. The distance of the star from
the centre of gravity of the system is a'
. Show that

$$
MP^2 = a^2(a-a')
$$

So first we can assume that $M >> m$

so now we get keplers third law as $MP^2 = \text{distance to the center mass}$ of the two objects

so if we then solve for the planet-Center of mass distance it equals $(a-a')$

wont that make the final equation $MP^2 = (a-a')^3$

and not $MP^2 = a^2(a-a')$

so what seems to be the problem in my attempt and why cant I get the correct answer.

the question is taken from fundamentals of astronomy 5th edition springer publication exercise 9.2

[![enter image description here](https://i.sstatic.net/hDPWG.png)](https://i.sstatic.net/hDPWG.png)

**Tags:** binary-star, keplers-laws
> **Comments:**
> **username_1**: Your version of Kepler's third law is incorrect. And you can't assume $M\gg m$ or $a'=0$.  
> **username_2**: @ProfRob This version if Kepler;s third law is correct (see my own answer below) although highly unusual and likely confusing.  
> **username_1**: @Thomas THEIR version ("$MP^2 = $ distance to the center mass [sic]") is incorrect. Your answer had already been presented (by me) but without giving away the full solution to a simple homework problem.  
> **username_2**: @ProfRob The error the OP made was to assume that $m<<M$ because he did not see the mass $m$ in the equation even though it is mentioned in the problem. With this assumption (which you certainly can make if the masses are sufficiently different) his equation $MP^2 = (a-a')^3$ is actually correct (as $a'=0$), as are $MP^2 =a (a-a')^2$ and $MP^2 =a^2 (a-a')$. Had you pointed out to the OP how $a'$ depends on the masses he would have realized himself where he was going wrong.  

---

## 3 answers

### Answer by username_1 on 2023-07-28. Score: 3

The correct version of Kepler's third law is
$$(M + m)P^2 = a^3$$
where $Ma' = ma''$ and $a = a' +a''$, where $a''$ is the distance of mass $m$ from the centre of mass.

The result follows from eliminating $m$ and $a''$ using the latter two equations.

NB You cannot assume $M\gg m$ to simplify Kepler's third law since that implies $a' = 0$.


---

### Answer by username_3 on 2023-07-27. Score: 2

[Kepler's third law](https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law) is too much like the equation in the exercise to be useful in proving it.
Instead, assume a circular orbit (*a'* > 0) and let the centripetal force on the planet equal the gravitational force.
$$ \frac{m v^2}{a - a'} = \frac{GMm}{a^2} $$
where
$$ v = \frac{2 \pi (a - a')}{P} $$
To determine a value for *G*, note that in exercise units, the Sun-Earth system has *M*=1, *P*=1, and *a*=1.

> **Comments:**
> **username_0**: wait so the star remains stationary to assume a circular orbit or should I assume the star and planet both move circularly around a common center of mass.  

---

### Answer by username_2 on 2023-07-30. Score: 1

Kepler's third law is in general given by

$$G (M+m)\cdot P^2 = 4 \pi^2 a^3$$

where $G$ is the gravitational constant, $M$ and $m$ the masses of the two bodies, $P$ their orbital period and $a$ the semi-major axis of the orbit.

Now for the distance $a-a'$ from the planet to the center of mass we have according to the definition of the center of mass

$$a-a'=\frac{M}{M+m} a$$

so

$$a =\frac{M+m}{M} (a-a')$$

Substitute just one of the $a$ in $a^3$ in Kepler's law with this expression and you get by multiplying through with $M/(M+m)$

$$G M \cdot P^2 = 4 \pi^2 a^2\cdot(a-a')$$

This holds in full generality and in any physical units, so not only for $m< (the $m$ does not appear explicitly anymore, but it is still contained in $a'$).

In the exercise, this equation has merely been reduced such that $G/4\pi^2=1$, $M$ in solar masses, $a$ in AU and $P$ in years. It holds only in these particular units. In other units it would be wrong and the physical dimensions in fact be inconsistent.

> **Comments:**
> **username_2**: @ProfRob I think this is a vicious homework question. It warrants a proper answer in order to avoid confusion. This is certainly a correct form of Kepler's third law as well (ignoring the omission of the gravitational constant), but it is highly unusual and and has probably just been formulated in order to cause students problems rather than helping them to understand Kepler's laws.  
> **username_1**: $G=1$ in the units given in the question.  
> **username_1**: In the equation written in this answer,  $G=1$ @MikeG That is why it is omitted in the OP.  
> **username_1**: @MikeG but you are correct in pointing out that $G$ as used in this answer is not *the* "gravitational constant" normally assigned the symbol $G$.  
> **username_2**: @ProfRob "G=1 in the units given in the question" The value of G depends on the units used obviously, but you should not omit it from the equation even if its numerical value is 1. It is sloppy and dangerous (as some of the comments prove here).  
> **username_1**: I note that you have corrected your equations so that G *is* now the gravitational constant and added the last paragraph.  
> **username_4**: Best to phrase it: In the first equation one sees that with M and m in units solar mass, a in AU, and P in years, then $G/4\pi^2$  is approximately 1.  

---

