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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

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    $\begingroup$ Your version of Kepler's third law is incorrect. And you can't assume $M\gg m$ or $a'=0$. $\endgroup$
    – ProfRob
    Commented Jul 28, 2023 at 18:43
  • $\begingroup$ @ProfRob This version if Kepler;s third law is correct (see my own answer below) although highly unusual and likely confusing. $\endgroup$
    – Thomas
    Commented Jul 30, 2023 at 16:19
  • $\begingroup$ @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. $\endgroup$
    – ProfRob
    Commented Jul 30, 2023 at 16:23
  • $\begingroup$ @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. $\endgroup$
    – Thomas
    Commented Jul 30, 2023 at 17:03

3 Answers 3

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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 $a' = 0$.

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Kepler's 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.

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  • $\begingroup$ 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. $\endgroup$ Commented Jul 27, 2023 at 15:47
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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 $m<<M$ (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.

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  • $\begingroup$ @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. $\endgroup$
    – Thomas
    Commented Jul 30, 2023 at 16:17
  • $\begingroup$ $G=1$ in the units given in the question. $\endgroup$
    – ProfRob
    Commented Jul 30, 2023 at 16:23
  • $\begingroup$ In the equation written in this answer, $G=1$ @MikeG That is why it is omitted in the OP. $\endgroup$
    – ProfRob
    Commented Jul 30, 2023 at 20:33
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    $\begingroup$ @MikeG but you are correct in pointing out that $G$ as used in this answer is not the "gravitational constant" normally assigned the symbol $G$. $\endgroup$
    – ProfRob
    Commented Jul 30, 2023 at 20:40
  • $\begingroup$ @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). $\endgroup$
    – Thomas
    Commented Jul 31, 2023 at 18:41

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